We Keep Coding

What is the complexity of this program. Is it O(n) or 0(1)? [duplicate]

I'd prefer as little formal definition as possible and simple mathematics.

Quick note, my answer is almost certainly confusing Big Oh notation (which is an upper bound) with Big Theta notation "Θ" (which is a two-side bound). But in my experience, this is actually typical of discussions in non-academic settings. Apologies for any confusion caused.
BigOh complexity can be visualized with this graph:
The simplest definition I can give for Big Oh notation is this:
Big Oh notation is a relative representation of the complexity of an algorithm.
There are some important and deliberately chosen words in that sentence:
relative: you can only compare apples to apples. You can't compare an algorithm that does arithmetic multiplication to an algorithm that sorts a list of integers. But a comparison of two algorithms to do arithmetic operations (one multiplication, one addition) will tell you something meaningful;
representation: BigOh (in its simplest form) reduces the comparison between algorithms to a single variable. That variable is chosen based on observations or assumptions. For example, sorting algorithms are typically compared based on comparison operations (comparing two nodes to determine their relative ordering). This assumes that comparison is expensive. But what if the comparison is cheap but swapping is expensive? It changes the comparison; and
complexity: if it takes me one second to sort 10,000 elements, how long will it take me to sort one million? Complexity in this instance is a relative measure to something else.
Come back and reread the above when you've read the rest.
The best example of BigOh I can think of is doing arithmetic. Take two numbers (123456 and 789012). The basic arithmetic operations we learned in school were:
addition;
subtraction;
multiplication; and
division.
Each of these is an operation or a problem. A method of solving these is called an algorithm.
The addition is the simplest. You line the numbers up (to the right) and add the digits in a column writing the last number of that addition in the result. The 'tens' part of that number is carried over to the next column.
Let's assume that the addition of these numbers is the most expensive operation in this algorithm. It stands to reason that to add these two numbers together we have to add together 6 digits (and possibly carry a 7th). If we add two 100 digit numbers together we have to do 100 additions. If we add two 10,000 digit numbers we have to do 10,000 additions.
See the pattern? The complexity (being the number of operations) is directly proportional to the number of digits n in the larger number. We call this O(n) or linear complexity.
Subtraction is similar (except you may need to borrow instead of carry).
Multiplication is different. You line the numbers up, take the first digit in the bottom number and multiply it in turn against each digit in the top number and so on through each digit. So to multiply our two 6 digit numbers we must do 36 multiplications. We may need to do as many as 10 or 11 column adds to get the end result too.
If we have two 100-digit numbers we need to do 10,000 multiplications and 200 adds. For two one million digit numbers we need to do one trillion (1012) multiplications and two million adds.
As the algorithm scales with n-squared, this is O(n2) or quadratic complexity. This is a good time to introduce another important concept:
We only care about the most significant portion of complexity.
The astute may have realized that we could express the number of operations as: n2 + 2n. But as you saw from our example with two numbers of a million digits apiece, the second term (2n) becomes insignificant (accounting for 0.0002% of the total operations by that stage).
One can notice that we've assumed the worst case scenario here. While multiplying 6 digit numbers, if one of them has 4 digits and the other one has 6 digits, then we only have 24 multiplications. Still, we calculate the worst case scenario for that 'n', i.e when both are 6 digit numbers. Hence Big Oh notation is about the Worst-case scenario of an algorithm.
The Telephone Book
The next best example I can think of is the telephone book, normally called the White Pages or similar but it varies from country to country. But I'm talking about the one that lists people by surname and then initials or first name, possibly address and then telephone numbers.
Now if you were instructing a computer to look up the phone number for "John Smith" in a telephone book that contains 1,000,000 names, what would you do? Ignoring the fact that you could guess how far in the S's started (let's assume you can't), what would you do?
A typical implementation might be to open up to the middle, take the 500,000th and compare it to "Smith". If it happens to be "Smith, John", we just got really lucky. Far more likely is that "John Smith" will be before or after that name. If it's after we then divide the last half of the phone book in half and repeat. If it's before then we divide the first half of the phone book in half and repeat. And so on.
This is called a binary search and is used every day in programming whether you realize it or not.
So if you want to find a name in a phone book of a million names you can actually find any name by doing this at most 20 times. In comparing search algorithms we decide that this comparison is our 'n'.
For a phone book of 3 names it takes 2 comparisons (at most).
For 7 it takes at most 3.
For 15 it takes 4.
…
For 1,000,000 it takes 20.
That is staggeringly good, isn't it?
In BigOh terms this is O(log n) or logarithmic complexity. Now the logarithm in question could be ln (base e), log10, log2 or some other base. It doesn't matter it's still O(log n) just like O(2n2) and O(100n2) are still both O(n2).
It's worthwhile at this point to explain that BigOh can be used to determine three cases with an algorithm:
Best Case: In the telephone book search, the best case is that we find the name in one comparison. This is O(1) or constant complexity;
Expected Case: As discussed above this is O(log n); and
Worst Case: This is also O(log n).
Normally we don't care about the best case. We're interested in the expected and worst case. Sometimes one or the other of these will be more important.
Back to the telephone book.
What if you have a phone number and want to find a name? The police have a reverse phone book but such look-ups are denied to the general public. Or are they? Technically you can reverse look-up a number in an ordinary phone book. How?
You start at the first name and compare the number. If it's a match, great, if not, you move on to the next. You have to do it this way because the phone book is unordered (by phone number anyway).
So to find a name given the phone number (reverse lookup):
Best Case: O(1);
Expected Case: O(n) (for 500,000); and
Worst Case: O(n) (for 1,000,000).
The Traveling Salesman
This is quite a famous problem in computer science and deserves a mention. In this problem, you have N towns. Each of those towns is linked to 1 or more other towns by a road of a certain distance. The Traveling Salesman problem is to find the shortest tour that visits every town.
Sounds simple? Think again.
If you have 3 towns A, B, and C with roads between all pairs then you could go:
A → B → C
A → C → B
B → C → A
B → A → C
C → A → B
C → B → A
Well, actually there's less than that because some of these are equivalent (A → B → C and C → B → A are equivalent, for example, because they use the same roads, just in reverse).
In actuality, there are 3 possibilities.
Take this to 4 towns and you have (iirc) 12 possibilities.
With 5 it's 60.
6 becomes 360.
This is a function of a mathematical operation called a factorial. Basically:
5! = 5 × 4 × 3 × 2 × 1 = 120
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
…
25! = 25 × 24 × … × 2 × 1 = 15,511,210,043,330,985,984,000,000
…
50! = 50 × 49 × … × 2 × 1 = 3.04140932 × 1064
So the BigOh of the Traveling Salesman problem is O(n!) or factorial or combinatorial complexity.
By the time you get to 200 towns there isn't enough time left in the universe to solve the problem with traditional computers.
Something to think about.
Polynomial Time
Another point I wanted to make a quick mention of is that any algorithm that has a complexity of O(na) is said to have polynomial complexity or is solvable in polynomial time.
O(n), O(n2) etc. are all polynomial time. Some problems cannot be solved in polynomial time. Certain things are used in the world because of this. Public Key Cryptography is a prime example. It is computationally hard to find two prime factors of a very large number. If it wasn't, we couldn't use the public key systems we use.
Anyway, that's it for my (hopefully plain English) explanation of BigOh (revised).

It shows how an algorithm scales based on input size.
O(n2): known as Quadratic complexity
1 item: 1 operations
10 items: 100 operations
100 items: 10,000 operations
Notice that the number of items increases by a factor of 10, but the time increases by a factor of 102. Basically, n=10 and so O(n2) gives us the scaling factor n2 which is 102.
O(n): known as Linear complexity
1 item: 1 operation
10 items: 10 operations
100 items: 100 operations
This time the number of items increases by a factor of 10, and so does the time. n=10 and so O(n)'s scaling factor is 10.
O(1): known as Constant complexity
1 item: 1 operations
10 items: 1 operations
100 items: 1 operations
The number of items is still increasing by a factor of 10, but the scaling factor of O(1) is always 1.
O(log n): known as Logarithmic complexity
1 item: 1 operations
10 items: 2 operations
100 items: 3 operations
1000 items: 4 operations
10,000 items: 5 operations
The number of computations is only increased by a log of the input value. So in this case, assuming each computation takes 1 second, the log of the input n is the time required, hence log n.
That's the gist of it. They reduce the maths down so it might not be exactly n2 or whatever they say it is, but that'll be the dominating factor in the scaling.

Big-O notation (also called "asymptotic growth" notation) is what functions "look like" when you ignore constant factors and stuff near the origin. We use it to talk about how thing scale.
Basics
for "sufficiently" large inputs...
f(x) ∈ O(upperbound) means f "grows no faster than" upperbound
f(x) ∈ Ɵ(justlikethis) mean f "grows exactly like" justlikethis
f(x) ∈ Ω(lowerbound) means f "grows no slower than" lowerbound
big-O notation doesn't care about constant factors: the function 9x² is said to "grow exactly like" 10x². Neither does big-O asymptotic notation care about non-asymptotic stuff ("stuff near the origin" or "what happens when the problem size is small"): the function 10x² is said to "grow exactly like" 10x² - x + 2.
Why would you want to ignore the smaller parts of the equation? Because they become completely dwarfed by the big parts of the equation as you consider larger and larger scales; their contribution becomes dwarfed and irrelevant. (See example section.)
Put another way, it's all about the ratio as you go to infinity. If you divide the actual time it takes by the O(...), you will get a constant factor in the limit of large inputs. Intuitively this makes sense: functions "scale like" one another if you can multiply one to get the other. That is when we say...
actualAlgorithmTime(N) ∈ O(bound(N))
e.g. "time to mergesort N elements
is O(N log(N))"
... this means that for "large enough" problem sizes N (if we ignore stuff near the origin), there exists some constant (e.g. 2.5, completely made up) such that:
actualAlgorithmTime(N) e.g. "mergesort_duration(N) "
────────────────────── < constant ───────────────────── < 2.5
bound(N) N log(N)
There are many choices of constant; often the "best" choice is known as the "constant factor" of the algorithm... but we often ignore it like we ignore non-largest terms (see Constant Factors section for why they don't usually matter). You can also think of the above equation as a bound, saying "In the worst-case scenario, the time it takes will never be worse than roughly N*log(N), within a factor of 2.5 (a constant factor we don't care much about)".
In general, O(...) is the most useful one because we often care about worst-case behavior. If f(x) represents something "bad" like the processor or memory usage, then "f(x) ∈ O(upperbound)" means "upperbound is the worst-case scenario of processor/memory usage".
Applications
As a purely mathematical construct, big-O notation is not limited to talking about processing time and memory. You can use it to discuss the asymptotics of anything where scaling is meaningful, such as:
the number of possible handshakes among N people at a party (Ɵ(N²), specifically N(N-1)/2, but what matters is that it "scales like" N²)
probabilistic expected number of people who have seen some viral marketing as a function of time
how website latency scales with the number of processing units in a CPU or GPU or computer cluster
how heat output scales on CPU dies as a function of transistor count, voltage, etc.
how much time an algorithm needs to run, as a function of input size
how much space an algorithm needs to run, as a function of input size
Example
For the handshake example above, everyone in a room shakes everyone else's hand. In that example, #handshakes ∈ Ɵ(N²). Why?
Back up a bit: the number of handshakes is exactly n-choose-2 or N*(N-1)/2 (each of N people shakes the hands of N-1 other people, but this double-counts handshakes so divide by 2):
However, for very large numbers of people, the linear term N is dwarfed and effectively contributes 0 to the ratio (in the chart: the fraction of empty boxes on the diagonal over total boxes gets smaller as the number of participants becomes larger). Therefore the scaling behavior is order N², or the number of handshakes "grows like N²".
#handshakes(N)
────────────── ≈ 1/2
N²
It's as if the empty boxes on the diagonal of the chart (N*(N-1)/2 checkmarks) weren't even there (N2 checkmarks asymptotically).
(temporary digression from "plain English":) If you wanted to prove this to yourself, you could perform some simple algebra on the ratio to split it up into multiple terms (lim means "considered in the limit of", just ignore it if you haven't seen it, it's just notation for "and N is really really big"):
N²/2 - N/2 (N²)/2 N/2 1/2
lim ────────── = lim ( ────── - ─── ) = lim ─── = 1/2
N→∞ N² N→∞ N² N² N→∞ 1
┕━━━┙
this is 0 in the limit of N→∞:
graph it, or plug in a really large number for N
tl;dr: The number of handshakes 'looks like' x² so much for large values, that if we were to write down the ratio #handshakes/x², the fact that we don't need exactly x² handshakes wouldn't even show up in the decimal for an arbitrarily large while.
e.g. for x=1million, ratio #handshakes/x²: 0.499999...
Building Intuition
This lets us make statements like...
"For large enough inputsize=N, no matter what the constant factor is, if I double the input size...
... I double the time an O(N) ("linear time") algorithm takes."
N → (2N) = 2(N)
... I double-squared (quadruple) the time an O(N²) ("quadratic time") algorithm takes." (e.g. a problem 100x as big takes 100²=10000x as long... possibly unsustainable)
N² → (2N)² = 4(N²)
... I double-cubed (octuple) the time an O(N³) ("cubic time") algorithm takes." (e.g. a problem 100x as big takes 100³=1000000x as long... very unsustainable)
cN³ → c(2N)³ = 8(cN³)
... I add a fixed amount to the time an O(log(N)) ("logarithmic time") algorithm takes." (cheap!)
c log(N) → c log(2N) = (c log(2))+(c log(N)) = (fixed amount)+(c log(N))
... I don't change the time an O(1) ("constant time") algorithm takes." (the cheapest!)
c*1 → c*1
... I "(basically) double" the time an O(N log(N)) algorithm takes." (fairly common)
c 2N log(2N) / c N log(N) (here we divide f(2n)/f(n), but we could have as above massaged the expression and factored out cNlogN as above)
→ 2 log(2N)/log(N)
→ 2 (log(2) + log(N))/log(N)
→ 2*(1+(log2N)-1) (basically 2 for large N; eventually less than 2.000001)
(alternatively, say log(N) will always be below like 17 for your data so it's O(17 N) which is linear; that is not rigorous nor sensical though)
... I ridiculously increase the time a O(2N) ("exponential time") algorithm takes." (you'd double (or triple, etc.) the time just by increasing the problem by a single unit)
2N → 22N = (4N)............put another way...... 2N → 2N+1 = 2N21 = 2 2N
[for the mathematically inclined, you can mouse over the spoilers for minor sidenotes]
(with credit to https://stackoverflow.com/a/487292/711085 )
(technically the constant factor could maybe matter in some more esoteric examples, but I've phrased things above (e.g. in log(N)) such that it doesn't)
These are the bread-and-butter orders of growth that programmers and applied computer scientists use as reference points. They see these all the time. (So while you could technically think "Doubling the input makes an O(√N) algorithm 1.414 times slower," it's better to think of it as "this is worse than logarithmic but better than linear".)
Constant factors
Usually, we don't care what the specific constant factors are, because they don't affect the way the function grows. For example, two algorithms may both take O(N) time to complete, but one may be twice as slow as the other. We usually don't care too much unless the factor is very large since optimizing is tricky business ( When is optimisation premature? ); also the mere act of picking an algorithm with a better big-O will often improve performance by orders of magnitude.
Some asymptotically superior algorithms (e.g. a non-comparison O(N log(log(N))) sort) can have so large a constant factor (e.g. 100000*N log(log(N))), or overhead that is relatively large like O(N log(log(N))) with a hidden + 100*N, that they are rarely worth using even on "big data".
Why O(N) is sometimes the best you can do, i.e. why we need datastructures
O(N) algorithms are in some sense the "best" algorithms if you need to read all your data. The very act of reading a bunch of data is an O(N) operation. Loading it into memory is usually O(N) (or faster if you have hardware support, or no time at all if you've already read the data). However, if you touch or even look at every piece of data (or even every other piece of data), your algorithm will take O(N) time to perform this looking. No matter how long your actual algorithm takes, it will be at least O(N) because it spent that time looking at all the data.
The same can be said for the very act of writing. All algorithms which print out N things will take N time because the output is at least that long (e.g. printing out all permutations (ways to rearrange) a set of N playing cards is factorial: O(N!) (which is why in those cases, good programs will ensure an iteration uses O(1) memory and doesn't print or store every intermediate step)).
This motivates the use of data structures: a data structure requires reading the data only once (usually O(N) time), plus some arbitrary amount of preprocessing (e.g. O(N) or O(N log(N)) or O(N²)) which we try to keep small. Thereafter, modifying the data structure (insertions/deletions/ etc.) and making queries on the data take very little time, such as O(1) or O(log(N)). You then proceed to make a large number of queries! In general, the more work you're willing to do ahead of time, the less work you'll have to do later on.
For example, say you had the latitude and longitude coordinates of millions of road segments and wanted to find all street intersections.
Naive method: If you had the coordinates of a street intersection, and wanted to examine nearby streets, you would have to go through the millions of segments each time, and check each one for adjacency.
If you only needed to do this once, it would not be a problem to have to do the naive method of O(N) work only once, but if you want to do it many times (in this case, N times, once for each segment), we'd have to do O(N²) work, or 1000000²=1000000000000 operations. Not good (a modern computer can perform about a billion operations per second).
If we use a simple structure called a hash table (an instant-speed lookup table, also known as a hashmap or dictionary), we pay a small cost by preprocessing everything in O(N) time. Thereafter, it only takes constant time on average to look up something by its key (in this case, our key is the latitude and longitude coordinates, rounded into a grid; we search the adjacent gridspaces of which there are only 9, which is a constant).
Our task went from an infeasible O(N²) to a manageable O(N), and all we had to do was pay a minor cost to make a hash table.
analogy: The analogy in this particular case is a jigsaw puzzle: We created a data structure that exploits some property of the data. If our road segments are like puzzle pieces, we group them by matching color and pattern. We then exploit this to avoid doing extra work later (comparing puzzle pieces of like color to each other, not to every other single puzzle piece).
The moral of the story: a data structure lets us speed up operations. Even more, advanced data structures can let you combine, delay, or even ignore operations in incredibly clever ways. Different problems would have different analogies, but they'd all involve organizing the data in a way that exploits some structure we care about, or which we've artificially imposed on it for bookkeeping. We do work ahead of time (basically planning and organizing), and now repeated tasks are much much easier!
Practical example: visualizing orders of growth while coding
Asymptotic notation is, at its core, quite separate from programming. Asymptotic notation is a mathematical framework for thinking about how things scale and can be used in many different fields. That said... this is how you apply asymptotic notation to coding.
The basics: Whenever we interact with every element in a collection of size A (such as an array, a set, all keys of a map, etc.), or perform A iterations of a loop, that is a multiplicative factor of size A. Why do I say "a multiplicative factor"?--because loops and functions (almost by definition) have multiplicative running time: the number of iterations, times work done in the loop (or for functions: the number of times you call the function, times work done in the function). (This holds if we don't do anything fancy, like skip loops or exit the loop early, or change control flow in the function based on arguments, which is very common.) Here are some examples of visualization techniques, with accompanying pseudocode.
(here, the xs represent constant-time units of work, processor instructions, interpreter opcodes, whatever)
for(i=0; i<A; i++) // A * ...
some O(1) operation // 1
--> A*1 --> O(A) time
visualization:
|<------ A ------->|
1 2 3 4 5 x x ... x
other languages, multiplying orders of growth:
javascript, O(A) time and space
someListOfSizeA.map((x,i) => [x,i])
python, O(rows*cols) time and space
[[r*c for c in range(cols)] for r in range(rows)]
Example 2:
for every x in listOfSizeA: // A * (...
some O(1) operation // 1
some O(B) operation // B
for every y in listOfSizeC: // C * (...
some O(1) operation // 1))
--> O(A*(1 + B + C))
O(A*(B+C)) (1 is dwarfed)
visualization:
|<------ A ------->|
1 x x x x x x ... x
2 x x x x x x ... x ^
3 x x x x x x ... x |
4 x x x x x x ... x |
5 x x x x x x ... x B <-- A*B
x x x x x x x ... x |
................... |
x x x x x x x ... x v
x x x x x x x ... x ^
x x x x x x x ... x |
x x x x x x x ... x |
x x x x x x x ... x C <-- A*C
x x x x x x x ... x |
................... |
x x x x x x x ... x v
Example 3:
function nSquaredFunction(n) {
total = 0
for i in 1..n: // N *
for j in 1..n: // N *
total += i*k // 1
return total
}
// O(n^2)
function nCubedFunction(a) {
for i in 1..n: // A *
print(nSquaredFunction(a)) // A^2
}
// O(a^3)
If we do something slightly complicated, you might still be able to imagine visually what's going on:
for x in range(A):
for y in range(1..x):
simpleOperation(x*y)
x x x x x x x x x x |
x x x x x x x x x |
x x x x x x x x |
x x x x x x x |
x x x x x x |
x x x x x |
x x x x |
x x x |
x x |
x___________________|
Here, the smallest recognizable outline you can draw is what matters; a triangle is a two dimensional shape (0.5 A^2), just like a square is a two-dimensional shape (A^2); the constant factor of two here remains in the asymptotic ratio between the two, however, we ignore it like all factors... (There are some unfortunate nuances to this technique I don't go into here; it can mislead you.)
Of course this does not mean that loops and functions are bad; on the contrary, they are the building blocks of modern programming languages, and we love them. However, we can see that the way we weave loops and functions and conditionals together with our data (control flow, etc.) mimics the time and space usage of our program! If time and space usage becomes an issue, that is when we resort to cleverness and find an easy algorithm or data structure we hadn't considered, to reduce the order of growth somehow. Nevertheless, these visualization techniques (though they don't always work) can give you a naive guess at a worst-case running time.
Here is another thing we can recognize visually:
<----------------------------- N ----------------------------->
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
x x x x x x x x x x x x x x x x
x x x x x x x x
x x x x
x x
x
We can just rearrange this and see it's O(N):
<----------------------------- N ----------------------------->
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
x x x x x x x x x x x x x x x x|x x x x x x x x|x x x x|x x|x
Or maybe you do log(N) passes of the data, for O(N*log(N)) total time:
<----------------------------- N ----------------------------->
^ x x x x x x x x x x x x x x x x|x x x x x x x x x x x x x x x x
| x x x x x x x x|x x x x x x x x|x x x x x x x x|x x x x x x x x
lgN x x x x|x x x x|x x x x|x x x x|x x x x|x x x x|x x x x|x x x x
| x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x|x x
v x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x|x
Unrelatedly but worth mentioning again: If we perform a hash (e.g. a dictionary/hashtable lookup), that is a factor of O(1). That's pretty fast.
[myDictionary.has(x) for x in listOfSizeA]
\----- O(1) ------/
--> A*1 --> O(A)
If we do something very complicated, such as with a recursive function or divide-and-conquer algorithm, you can use the Master Theorem (usually works), or in ridiculous cases the Akra-Bazzi Theorem (almost always works) you look up the running time of your algorithm on Wikipedia.
But, programmers don't think like this because eventually, algorithm intuition just becomes second nature. You will start to code something inefficient and immediately think "am I doing something grossly inefficient?". If the answer is "yes" AND you foresee it actually mattering, then you can take a step back and think of various tricks to make things run faster (the answer is almost always "use a hashtable", rarely "use a tree", and very rarely something a bit more complicated).
Amortized and average-case complexity
There is also the concept of "amortized" and/or "average case" (note that these are different).
Average Case: This is no more than using big-O notation for the expected value of a function, rather than the function itself. In the usual case where you consider all inputs to be equally likely, the average case is just the average of the running time. For example with quicksort, even though the worst-case is O(N^2) for some really bad inputs, the average case is the usual O(N log(N)) (the really bad inputs are very small in number, so few that we don't notice them in the average case).
Amortized Worst-Case: Some data structures may have a worst-case complexity that is large, but guarantee that if you do many of these operations, the average amount of work you do will be better than worst-case. For example, you may have a data structure that normally takes constant O(1) time. However, occasionally it will 'hiccup' and take O(N) time for one random operation, because maybe it needs to do some bookkeeping or garbage collection or something... but it promises you that if it does hiccup, it won't hiccup again for N more operations. The worst-case cost is still O(N) per operation, but the amortized cost over many runs is O(N)/N = O(1) per operation. Because the big operations are sufficiently rare, the massive amount of occasional work can be considered to blend in with the rest of the work as a constant factor. We say the work is "amortized" over a sufficiently large number of calls that it disappears asymptotically.
The analogy for amortized analysis:
You drive a car. Occasionally, you need to spend 10 minutes going to
the gas station and then spend 1 minute refilling the tank with gas.
If you did this every time you went anywhere with your car (spend 10
minutes driving to the gas station, spend a few seconds filling up a
fraction of a gallon), it would be very inefficient. But if you fill
up the tank once every few days, the 11 minutes spent driving to the
gas station is "amortized" over a sufficiently large number of trips,
that you can ignore it and pretend all your trips were maybe 5% longer.
Comparison between average-case and amortized worst-case:
Average-case: We make some assumptions about our inputs; i.e. if our inputs have different probabilities, then our outputs/runtimes will have different probabilities (which we take the average of). Usually, we assume that our inputs are all equally likely (uniform probability), but if the real-world inputs don't fit our assumptions of "average input", the average output/runtime calculations may be meaningless. If you anticipate uniformly random inputs though, this is useful to think about!
Amortized worst-case: If you use an amortized worst-case data structure, the performance is guaranteed to be within the amortized worst-case... eventually (even if the inputs are chosen by an evil demon who knows everything and is trying to screw you over). Usually, we use this to analyze algorithms that may be very 'choppy' in performance with unexpected large hiccups, but over time perform just as well as other algorithms. (However unless your data structure has upper limits for much outstanding work it is willing to procrastinate on, an evil attacker could perhaps force you to catch up on the maximum amount of procrastinated work all-at-once.
Though, if you're reasonably worried about an attacker, there are many other algorithmic attack vectors to worry about besides amortization and average-case.)
Both average-case and amortization are incredibly useful tools for thinking about and designing with scaling in mind.
(See Difference between average case and amortized analysis if interested in this subtopic.)
Multidimensional big-O
Most of the time, people don't realize that there's more than one variable at work. For example, in a string-search algorithm, your algorithm may take time O([length of text] + [length of query]), i.e. it is linear in two variables like O(N+M). Other more naive algorithms may be O([length of text]*[length of query]) or O(N*M). Ignoring multiple variables is one of the most common oversights I see in algorithm analysis, and can handicap you when designing an algorithm.
The whole story
Keep in mind that big-O is not the whole story. You can drastically speed up some algorithms by using caching, making them cache-oblivious, avoiding bottlenecks by working with RAM instead of disk, using parallelization, or doing work ahead of time -- these techniques are often independent of the order-of-growth "big-O" notation, though you will often see the number of cores in the big-O notation of parallel algorithms.
Also keep in mind that due to hidden constraints of your program, you might not really care about asymptotic behavior. You may be working with a bounded number of values, for example:
If you're sorting something like 5 elements, you don't want to use the speedy O(N log(N)) quicksort; you want to use insertion sort, which happens to perform well on small inputs. These situations often come up in divide-and-conquer algorithms, where you split up the problem into smaller and smaller subproblems, such as recursive sorting, fast Fourier transforms, or matrix multiplication.
If some values are effectively bounded due to some hidden fact (e.g. the average human name is softly bounded at perhaps 40 letters, and human age is softly bounded at around 150). You can also impose bounds on your input to effectively make terms constant.
In practice, even among algorithms which have the same or similar asymptotic performance, their relative merit may actually be driven by other things, such as: other performance factors (quicksort and mergesort are both O(N log(N)), but quicksort takes advantage of CPU caches); non-performance considerations, like ease of implementation; whether a library is available, and how reputable and maintained the library is.
Programs will also run slower on a 500MHz computer vs 2GHz computer. We don't really consider this as part of the resource bounds, because we think of the scaling in terms of machine resources (e.g. per clock cycle), not per real second. However, there are similar things which can 'secretly' affect performance, such as whether you are running under emulation, or whether the compiler optimized code or not. These might make some basic operations take longer (even relative to each other), or even speed up or slow down some operations asymptotically (even relative to each other). The effect may be small or large between different implementation and/or environment. Do you switch languages or machines to eke out that little extra work? That depends on a hundred other reasons (necessity, skills, coworkers, programmer productivity, the monetary value of your time, familiarity, workarounds, why not assembly or GPU, etc...), which may be more important than performance.
The above issues, like the effect of the choice of which programming language is used, are almost never considered as part of the constant factor (nor should they be); yet one should be aware of them because sometimes (though rarely) they may affect things. For example in cpython, the native priority queue implementation is asymptotically non-optimal (O(log(N)) rather than O(1) for your choice of insertion or find-min); do you use another implementation? Probably not, since the C implementation is probably faster, and there are probably other similar issues elsewhere. There are tradeoffs; sometimes they matter and sometimes they don't.
(edit: The "plain English" explanation ends here.)
Math addenda
For completeness, the precise definition of big-O notation is as follows: f(x) ∈ O(g(x)) means that "f is asymptotically upper-bounded by const*g": ignoring everything below some finite value of x, there exists a constant such that |f(x)| ≤ const * |g(x)|. (The other symbols are as follows: just like O means ≤, Ω means ≥. There are lowercase variants: o means <, and ω means >.) f(x) ∈ Ɵ(g(x)) means both f(x) ∈ O(g(x)) and f(x) ∈ Ω(g(x)) (upper- and lower-bounded by g): there exists some constants such that f will always lie in the "band" between const1*g(x) and const2*g(x). It is the strongest asymptotic statement you can make and roughly equivalent to ==. (Sorry, I elected to delay the mention of the absolute-value symbols until now, for clarity's sake; especially because I have never seen negative values come up in a computer science context.)
People will often use = O(...), which is perhaps the more correct 'comp-sci' notation, and entirely legitimate to use; "f = O(...)" is read "f is order ... / f is xxx-bounded by ..." and is thought of as "f is some expression whose asymptotics are ...". I was taught to use the more rigorous ∈ O(...). ∈ means "is an element of" (still read as before). In this particular case, O(N²) contains elements like {2 N², 3 N², 1/2 N², 2 N² + log(N), - N² + N^1.9, ...} and is infinitely large, but it's still a set.
O and Ω are not symmetric (n = O(n²), but n² is not O(n)), but Ɵ is symmetric, and (since these relations are all transitive and reflexive) Ɵ, therefore, is symmetric and transitive and reflexive, and therefore partitions the set of all functions into equivalence classes. An equivalence class is a set of things that we consider to be the same. That is to say, given any function you can think of, you can find a canonical/unique 'asymptotic representative' of the class (by generally taking the limit... I think); just like you can group all integers into odds or evens, you can group all functions with Ɵ into x-ish, log(x)^2-ish, etc... by basically ignoring smaller terms (but sometimes you might be stuck with more complicated functions which are separate classes unto themselves).
The = notation might be the more common one and is even used in papers by world-renowned computer scientists. Additionally, it is often the case that in a casual setting, people will say O(...) when they mean Ɵ(...); this is technically true since the set of things Ɵ(exactlyThis) is a subset of O(noGreaterThanThis)... and it's easier to type. ;-)

EDIT: Quick note, this is almost certainly confusing Big O notation (which is an upper bound) with Theta notation (which is both an upper and lower bound). In my experience this is actually typical of discussions in non-academic settings. Apologies for any confusion caused.
In one sentence: As the size of your job goes up, how much longer does it take to complete it?
Obviously that's only using "size" as the input and "time taken" as the output — the same idea applies if you want to talk about memory usage etc.
Here's an example where we have N T-shirts which we want to dry. We'll assume it's incredibly quick to get them in the drying position (i.e. the human interaction is negligible). That's not the case in real life, of course...
Using a washing line outside: assuming you have an infinitely large back yard, washing dries in O(1) time. However much you have of it, it'll get the same sun and fresh air, so the size doesn't affect the drying time.
Using a tumble dryer: you put 10 shirts in each load, and then they're done an hour later. (Ignore the actual numbers here — they're irrelevant.) So drying 50 shirts takes about 5 times as long as drying 10 shirts.
Putting everything in an airing cupboard: If we put everything in one big pile and just let general warmth do it, it will take a long time for the middle shirts to get dry. I wouldn't like to guess at the detail, but I suspect this is at least O(N^2) — as you increase the wash load, the drying time increases faster.
One important aspect of "big O" notation is that it doesn't say which algorithm will be faster for a given size. Take a hashtable (string key, integer value) vs an array of pairs (string, integer). Is it faster to find a key in the hashtable or an element in the array, based on a string? (i.e. for the array, "find the first element where the string part matches the given key.") Hashtables are generally amortised (~= "on average") O(1) — once they're set up, it should take about the same time to find an entry in a 100 entry table as in a 1,000,000 entry table. Finding an element in an array (based on content rather than index) is linear, i.e. O(N) — on average, you're going to have to look at half the entries.
Does this make a hashtable faster than an array for lookups? Not necessarily. If you've got a very small collection of entries, an array may well be faster — you may be able to check all the strings in the time that it takes to just calculate the hashcode of the one you're looking at. As the data set grows larger, however, the hashtable will eventually beat the array.

Big O describes an upper limit on the growth behaviour of a function, for example the runtime of a program, when inputs become large.
Examples:
O(n): If I double the input size the runtime doubles
O(n2): If the input size doubles the runtime quadruples
O(log n): If the input size doubles the runtime increases by one
O(2n): If the input size increases by one, the runtime doubles
The input size is usually the space in bits needed to represent the input.

Big O notation is most commonly used by programmers as an approximate measure of how long a computation (algorithm) will take to complete expressed as a function of the size of the input set.
Big O is useful to compare how well two algorithms will scale up as the number of inputs is increased.
More precisely Big O notation is used to express the asymptotic behavior of a function. That means how the function behaves as it approaches infinity.
In many cases the "O" of an algorithm will fall into one of the following cases:
O(1) - Time to complete is the same regardless of the size of input set. An example is accessing an array element by index.
O(Log N) - Time to complete increases roughly in line with the log2(n). For example 1024 items takes roughly twice as long as 32 items, because Log2(1024) = 10 and Log2(32) = 5. An example is finding an item in a binary search tree (BST).
O(N) - Time to complete that scales linearly with the size of the input set. In other words if you double the number of items in the input set, the algorithm takes roughly twice as long. An example is counting the number of items in a linked list.
O(N Log N) - Time to complete increases by the number of items times the result of Log2(N). An example of this is heap sort and quick sort.
O(N^2) - Time to complete is roughly equal to the square of the number of items. An example of this is bubble sort.
O(N!) - Time to complete is the factorial of the input set. An example of this is the traveling salesman problem brute-force solution.
Big O ignores factors that do not contribute in a meaningful way to the growth curve of a function as the input size increases towards infinity. This means that constants that are added to or multiplied by the function are simply ignored.

Big O is just a way to "Express" yourself in a common way, "How much time / space does it take to run my code?".
You may often see O(n), O(n2), O(nlogn) and so forth, all these are just ways to show; How does an algorithm change?
O(n) means Big O is n, and now you might think, "What is n!?" Well "n" is the amount of elements. Imaging you want to search for an Item in an Array. You would have to look on Each element and as "Are you the correct element/item?" in the worst case, the item is at the last index, which means that it took as much time as there are items in the list, so to be generic, we say "oh hey, n is a fair given amount of values!".
So then you might understand what "n2" means, but to be even more specific, play with the thought you have a simple, the simpliest of the sorting algorithms; bubblesort. This algorithm needs to look through the whole list, for each item.
My list
1
6
3
The flow here would be:
Compare 1 and 6, which is biggest? Ok 6 is in the right position, moving forward!
Compare 6 and 3, oh, 3 is less! Let's move that, Ok the list changed, we need to start from the begining now!
This is O n2 because, you need to look at all items in the list there are "n" items. For each item, you look at all items once more, for comparing, this is also "n", so for every item, you look "n" times meaning n*n = n2
I hope this is as simple as you want it.
But remember, Big O is just a way to experss yourself in the manner of time and space.

Big O describes the fundamental scaling nature of an algorithm.
There is a lot of information that Big O does not tell you about a given algorithm. It cuts to the bone and gives only information about the scaling nature of an algorithm, specifically how the resource use (think time or memory) of an algorithm scales in response to the "input size".
Consider the difference between a steam engine and a rocket. They are not merely different varieties of the same thing (as, say, a Prius engine vs. a Lamborghini engine) but they are dramatically different kinds of propulsion systems, at their core. A steam engine may be faster than a toy rocket, but no steam piston engine will be able to achieve the speeds of an orbital launch vehicle. This is because these systems have different scaling characteristics with regards to the relation of fuel required ("resource usage") to reach a given speed ("input size").
Why is this so important? Because software deals with problems that may differ in size by factors up to a trillion. Consider that for a moment. The ratio between the speed necessary to travel to the Moon and human walking speed is less than 10,000:1, and that is absolutely tiny compared to the range in input sizes software may face. And because software may face an astronomical range in input sizes there is the potential for the Big O complexity of an algorithm, it's fundamental scaling nature, to trump any implementation details.
Consider the canonical sorting example. Bubble-sort is O(n2) while merge-sort is O(n log n). Let's say you have two sorting applications, application A which uses bubble-sort and application B which uses merge-sort, and let's say that for input sizes of around 30 elements application A is 1,000x faster than application B at sorting. If you never have to sort much more than 30 elements then it's obvious that you should prefer application A, as it is much faster at these input sizes. However, if you find that you may have to sort ten million items then what you'd expect is that application B actually ends up being thousands of times faster than application A in this case, entirely due to the way each algorithm scales.

Here is the plain English bestiary I tend to use when explaining the common varieties of Big-O
In all cases, prefer algorithms higher up on the list to those lower on the list. However, the cost of moving to a more expensive complexity class varies significantly.
O(1):
No growth. Regardless of how big as the problem is, you can solve it in the same amount of time. This is somewhat analogous to broadcasting where it takes the same amount of energy to broadcast over a given distance, regardless of the number of people that lie within the broadcast range.
O(log n):
This complexity is the same as O(1) except that it's just a little bit worse. For all practical purposes, you can consider this as a very large constant scaling. The difference in work between processing 1 thousand and 1 billion items is only a factor six.
O(n):
The cost of solving the problem is proportional to the size of the problem. If your problem doubles in size, then the cost of the solution doubles. Since most problems have to be scanned into the computer in some way, as data entry, disk reads, or network traffic, this is generally an affordable scaling factor.
O(n log n):
This complexity is very similar to O(n). For all practical purposes, the two are equivalent. This level of complexity would generally still be considered scalable. By tweaking assumptions some O(n log n) algorithms can be transformed into O(n) algorithms. For example, bounding the size of keys reduces sorting from O(n log n) to O(n).
O(n2):
Grows as a square, where n is the length of the side of a square. This is the same growth rate as the "network effect", where everyone in a network might know everyone else in the network. Growth is expensive. Most scalable solutions cannot use algorithms with this level of complexity without doing significant gymnastics. This generally applies to all other polynomial complexities - O(nk) - as well.
O(2n):
Does not scale. You have no hope of solving any non-trivially sized problem. Useful for knowing what to avoid, and for experts to find approximate algorithms which are in O(nk).

Big O is a measure of how much time/space an algorithm uses relative to the size of its input.
If an algorithm is O(n) then the time/space will increase at the same rate as its input.
If an algorithm is O(n2) then the time/space increase at the rate of its input squared.
and so on.

It is very difficult to measure the speed of software programs, and when we try, the answers can be very complex and filled with exceptions and special cases. This is a big problem, because all those exceptions and special cases are distracting and unhelpful when we want to compare two different programs with one another to find out which is "fastest".
As a result of all this unhelpful complexity, people try to describe the speed of software programs using the smallest and least complex (mathematical) expressions possible. These expressions are very very crude approximations: Although, with a bit of luck, they will capture the "essence" of whether a piece of software is fast or slow.
Because they are approximations, we use the letter "O" (Big Oh) in the expression, as a convention to signal to the reader that we are making a gross oversimplification. (And to make sure that nobody mistakenly thinks that the expression is in any way accurate).
If you read the "Oh" as meaning "on the order of" or "approximately" you will not go too far wrong. (I think the choice of the Big-Oh might have been an attempt at humour).
The only thing that these "Big-Oh" expressions try to do is to describe how much the software slows down as we increase the amount of data that the software has to process. If we double the amount of data that needs to be processed, does the software need twice as long to finish it's work? Ten times as long? In practice, there are a very limited number of big-Oh expressions that you will encounter and need to worry about:
The good:
O(1) Constant: The program takes the same time to run no matter how big the input is.
O(log n) Logarithmic: The program run-time increases only slowly, even with big increases in the size of the input.
The bad:
O(n) Linear: The program run-time increases proportionally to the size of the input.
O(n^k) Polynomial: - Processing time grows faster and faster - as a polynomial function - as the size of the input increases.
... and the ugly:
O(k^n) Exponential The program run-time increases very quickly with even moderate increases in the size of the problem - it is only practical to process small data sets with exponential algorithms.
O(n!) Factorial The program run-time will be longer than you can afford to wait for anything but the very smallest and most trivial-seeming datasets.

What is a plain English explanation of Big O? With as little formal definition as possible and simple mathematics.
A Plain English Explanation of the Need for Big-O Notation:
When we program, we are trying to solve a problem. What we code is called an algorithm. Big O notation allows us to compare the worse case performance of our algorithms in a standardized way. Hardware specs vary over time and improvements in hardware can reduce the time it takes an algorithms to run. But replacing the hardware does not mean our algorithm is any better or improved over time, as our algorithm is still the same. So in order to allow us to compare different algorithms, to determine if one is better or not, we use Big O notation.
A Plain English Explanation of What Big O Notation is:
Not all algorithms run in the same amount of time, and can vary based on the number of items in the input, which we'll call n. Based on this, we consider the worse case analysis, or an upper-bound of the run-time as n get larger and larger. We must be aware of what n is, because many of the Big O notations reference it.

Ok, my 2cents.
Big-O, is rate of increase of resource consumed by program, w.r.t. problem-instance-size
Resource : Could be total-CPU time, could be maximum RAM space. By default refers to CPU time.
Say the problem is "Find the sum",
int Sum(int*arr,int size){
int sum=0;
while(size-->0)
sum+=arr[size];
return sum;
}
problem-instance= {5,10,15} ==> problem-instance-size = 3, iterations-in-loop= 3
problem-instance= {5,10,15,20,25} ==> problem-instance-size = 5 iterations-in-loop = 5
For input of size "n" the program is growing at speed of "n" iterations in array. Hence Big-O is N expressed as O(n)
Say the problem is "Find the Combination",
void Combination(int*arr,int size)
{ int outer=size,inner=size;
while(outer -->0) {
inner=size;
while(inner -->0)
cout<<arr[outer]<<"-"<<arr[inner]<<endl;
}
}
problem-instance= {5,10,15} ==> problem-instance-size = 3, total-iterations = 3*3 = 9
problem-instance= {5,10,15,20,25} ==> problem-instance-size = 5, total-iterations= 5*5 =25
For input of size "n" the program is growing at speed of "n*n" iterations in array. Hence Big-O is N2 expressed as O(n2)

A simple straightforward answer can be:
Big O represents the worst possible time/space for that algorithm. The algorithm will never take more space/time above that limit. Big O represents time/space complexity in the extreme case.

Big O notation is a way of describing the upper bound of an algorithm in terms of space or running time. The n is the number of elements in the the problem (i.e size of an array, number of nodes in a tree, etc.) We are interested in describing the running time as n gets big.
When we say some algorithm is O(f(n)) we are saying that the running time (or space required) by that algorithm is always lower than some constant times f(n).
To say that binary search has a running time of O(logn) is to say that there exists some constant c which you can multiply log(n) by that will always be larger than the running time of binary search. In this case you will always have some constant factor of log(n) comparisons.
In other words where g(n) is the running time of your algorithm, we say that g(n) = O(f(n)) when g(n) <= c*f(n) when n > k, where c and k are some constants.

"What is a plain English explanation of Big O? With as little formal
definition as possible and simple mathematics."
Such a beautifully simple and short question seems at least to deserve an equally short answer, like a student might receive during tutoring.
Big O notation simply tells how much time* an algorithm can run within,
in terms of only the amount of input data**.
( *in a wonderful, unit-free sense of time!)
(**which is what matters, because people will always want more, whether they live today or tomorrow)
Well, what's so wonderful about Big O notation if that's what it does?
Practically speaking, Big O analysis is so useful and important because Big O puts the focus squarely on the algorithm's own complexity and completely ignores anything that is merely a proportionality constant—like a JavaScript engine, the speed of a CPU, your Internet connection, and all those things which become quickly become as laughably outdated as a Model T. Big O focuses on performance only in the way that matters equally as much to people living in the present or in the future.
Big O notation also shines a spotlight directly on the most important principle of computer programming/engineering, the fact which inspires all good programmers to keep thinking and dreaming: the only way to achieve results beyond the slow forward march of technology is to invent a better algorithm.

Algorithm example (Java):
public boolean search(/* for */Integer K,/* in */List</* of */Integer> L)
{
for(/* each */Integer i:/* in */L)
{
if(i == K)
{
return true;
}
}
return false;
}
Algorithm description:
This algorithm searches a list, item by item, looking for a key,
Iterating on each item in the list, if it's the key then return True,
If the loop has finished without finding the key, return False.
Big-O notation represents the upper-bound on the Complexity (Time, Space, ..)
To find The Big-O on Time Complexity:
Calculate how much time (regarding input size) the worst case takes:
Worst-Case: the key doesn't exist in the list.
Time(Worst-Case) = 4n+1
Time: O(4n+1) = O(n) | in Big-O, constants are neglected
O(n) ~ Linear
There's also Big-Omega, which represent the complexity of the Best-Case:
Best-Case: the key is the first item.
Time(Best-Case) = 4
Time: Ω(4) = O(1) ~ Instant\Constant

Big O notation is a way of describing how quickly an algorithm will run given an arbitrary number of input parameters, which we'll call "n". It is useful in computer science because different machines operate at different speeds, and simply saying that an algorithm takes 5 seconds doesn't tell you much because while you may be running a system with a 4.5 Ghz octo-core processor, I may be running a 15 year old, 800 Mhz system, which could take longer regardless of the algorithm. So instead of specifying how fast an algorithm runs in terms of time, we say how fast it runs in terms of number of input parameters, or "n". By describing algorithms in this way, we are able to compare the speeds of algorithms without having to take into account the speed of the computer itself.

Big O
f(x) = O(g(x)) when x goes to a (for example, a = +∞) means that there is a function k such that:
f(x) = k(x)g(x)
k is bounded in some neighborhood of a (if a = +∞, this means that there are numbers N and M such that for every x > N, |k(x)| < M).
In other words, in plain English: f(x) = O(g(x)), x → a, means that in a neighborhood of a, f decomposes into the product of g and some bounded function.
Small o
By the way, here is for comparison the definition of small o.
f(x) = o(g(x)) when x goes to a means that there is a function k such that:
f(x) = k(x)g(x)
k(x) goes to 0 when x goes to a.
Examples
sin x = O(x) when x → 0.
sin x = O(1) when x → +∞,
x2 + x = O(x) when x → 0,
x2 + x = O(x2) when x → +∞,
ln(x) = o(x) = O(x) when x → +∞.
Attention! The notation with the equal sign "=" uses a "fake equality": it is true that o(g(x)) = O(g(x)), but false that O(g(x)) = o(g(x)). Similarly, it is ok to write "ln(x) = o(x) when x → +∞", but the formula "o(x) = ln(x)" would make no sense.
More examples
O(1) = O(n) = O(n2) when n → +∞ (but not the other way around, the equality is "fake"),
O(n) + O(n2) = O(n2) when n → +∞
O(O(n2)) = O(n2) when n → +∞
O(n2)O(n3) = O(n5) when n → +∞
Here is the Wikipedia article: https://en.wikipedia.org/wiki/Big_O_notation

You want to know all there is to know of big O? So do I.
So to talk of big O, I will use words that have just one beat in them. One sound per word. Small words are quick. You know these words, and so do I. We will use words with one sound. They are small. I am sure you will know all of the words we will use!
Now, let’s you and me talk of work. Most of the time, I do not like work. Do you like work? It may be the case that you do, but I am sure I do not.
I do not like to go to work. I do not like to spend time at work. If I had my way, I would like just to play, and do fun things. Do you feel the same as I do?
Now at times, I do have to go to work. It is sad, but true. So, when I am at work, I have a rule: I try to do less work. As near to no work as I can. Then I go play!
So here is the big news: the big O can help me not to do work! I can play more of the time, if I know big O. Less work, more play! That is what big O helps me do.
Now I have some work. I have this list: one, two, three, four, five, six. I must add all things in this list.
Wow, I hate work. But oh well, I have to do this. So here I go.
One plus two is three… plus three is six... and four is... I don’t know. I got lost. It is too hard for me to do in my head. I don’t much care for this kind of work.
So let's not do the work. Let's you and me just think how hard it is. How much work would I have to do, to add six numbers?
Well, let’s see. I must add one and two, and then add that to three, and then add that to four… All in all, I count six adds. I have to do six adds to solve this.
Here comes big O, to tell us just how hard this math is.
Big O says: we must do six adds to solve this. One add, for each thing from one to six. Six small bits of work... each bit of work is one add.
Well, I will not do the work to add them now. But I know how hard it would be. It would be six adds.
Oh no, now I have more work. Sheesh. Who makes this kind of stuff?!
Now they ask me to add from one to ten! Why would I do that? I did not want to add one to six. To add from one to ten… well… that would be even more hard!
How much more hard would it be? How much more work would I have to do? Do I need more or less steps?
Well, I guess I would have to do ten adds… one for each thing from one to ten. Ten is more than six. I would have to work that much more to add from one to ten, than one to six!
I do not want to add right now. I just want to think on how hard it might be to add that much. And, I hope, to play as soon as I can.
To add from one to six, that is some work. But do you see, to add from one to ten, that is more work?
Big O is your friend and mine. Big O helps us think on how much work we have to do, so we can plan. And, if we are friends with big O, he can help us choose work that is not so hard!
Now we must do new work. Oh, no. I don’t like this work thing at all.
The new work is: add all things from one to n.
Wait! What is n? Did I miss that? How can I add from one to n if you don’t tell me what n is?
Well, I don’t know what n is. I was not told. Were you? No? Oh well. So we can’t do the work. Whew.
But though we will not do the work now, we can guess how hard it would be, if we knew n. We would have to add up n things, right? Of course!
Now here comes big O, and he will tell us how hard this work is. He says: to add all things from one to N, one by one, is O(n). To add all these things, [I know I must add n times.][1] That is big O! He tells us how hard it is to do some type of work.
To me, I think of big O like a big, slow, boss man. He thinks on work, but he does not do it. He might say, "That work is quick." Or, he might say, "That work is so slow and hard!" But he does not do the work. He just looks at the work, and then he tells us how much time it might take.
I care lots for big O. Why? I do not like to work! No one likes to work. That is why we all love big O! He tells us how fast we can work. He helps us think of how hard work is.
Uh oh, more work. Now, let’s not do the work. But, let’s make a plan to do it, step by step.
They gave us a deck of ten cards. They are all mixed up: seven, four, two, six… not straight at all. And now... our job is to sort them.
Ergh. That sounds like a lot of work!
How can we sort this deck? I have a plan.
I will look at each pair of cards, pair by pair, through the deck, from first to last. If the first card in one pair is big and the next card in that pair is small, I swap them. Else, I go to the next pair, and so on and so on... and soon, the deck is done.
When the deck is done, I ask: did I swap cards in that pass? If so, I must do it all once more, from the top.
At some point, at some time, there will be no swaps, and our sort of the deck would be done. So much work!
Well, how much work would that be, to sort the cards with those rules?
I have ten cards. And, most of the time -- that is, if I don’t have lots of luck -- I must go through the whole deck up to ten times, with up to ten card swaps each time through the deck.
Big O, help me!
Big O comes in and says: for a deck of n cards, to sort it this way will be done in O(N squared) time.
Why does he say n squared?
Well, you know n squared is n times n. Now, I get it: n cards checked, up to what might be n times through the deck. That is two loops, each with n steps. That is n squared much work to be done. A lot of work, for sure!
Now when big O says it will take O(n squared) work, he does not mean n squared adds, on the nose. It might be some small bit less, for some case. But in the worst case, it will be near n squared steps of work to sort the deck.
Now here is where big O is our friend.
Big O points out this: as n gets big, when we sort cards, the job gets MUCH MUCH MORE HARD than the old just-add-these-things job. How do we know this?
Well, if n gets real big, we do not care what we might add to n or n squared.
For big n, n squared is more large than n.
Big O tells us that to sort things is more hard than to add things. O(n squared) is more than O(n) for big n. That means: if n gets real big, to sort a mixed deck of n things MUST take more time, than to just add n mixed things.
Big O does not solve the work for us. Big O tells us how hard the work is.
I have a deck of cards. I did sort them. You helped. Thanks.
Is there a more fast way to sort the cards? Can big O help us?
Yes, there is a more fast way! It takes some time to learn, but it works... and it works quite fast. You can try it too, but take your time with each step and do not lose your place.
In this new way to sort a deck, we do not check pairs of cards the way we did a while ago. Here are your new rules to sort this deck:
One: I choose one card in the part of the deck we work on now. You can choose one for me if you like. (The first time we do this, “the part of the deck we work on now” is the whole deck, of course.)
Two: I splay the deck on that card you chose. What is this splay; how do I splay? Well, I go from the start card down, one by one, and I look for a card that is more high than the splay card.
Three: I go from the end card up, and I look for a card that is more low than the splay card.
Once I have found these two cards, I swap them, and go on to look for more cards to swap. That is, I go back to step Two, and splay on the card you chose some more.
At some point, this loop (from Two to Three) will end. It ends when both halves of this search meet at the splay card. Then, we have just splayed the deck with the card you chose in step One. Now, all the cards near the start are more low than the splay card; and the cards near the end are more high than the splay card. Cool trick!
Four (and this is the fun part): I have two small decks now, one more low than the splay card, and one more high. Now I go to step one, on each small deck! That is to say, I start from step One on the first small deck, and when that work is done, I start from step One on the next small deck.
I break up the deck in parts, and sort each part, more small and more small, and at some time I have no more work to do. Now this may seem slow, with all the rules. But trust me, it is not slow at all. It is much less work than the first way to sort things!
What is this sort called? It is called Quick Sort! That sort was made by a man called C. A. R. Hoare and he called it Quick Sort. Now, Quick Sort gets used all the time!
Quick Sort breaks up big decks in small ones. That is to say, it breaks up big tasks in small ones.
Hmmm. There may be a rule in there, I think. To make big tasks small, break them up.
This sort is quite quick. How quick? Big O tells us: this sort needs O(n log n) work to be done, in the mean case.
Is it more or less fast than the first sort? Big O, please help!
The first sort was O(n squared). But Quick Sort is O(n log n). You know that n log n is less than n squared, for big n, right? Well, that is how we know that Quick Sort is fast!
If you have to sort a deck, what is the best way? Well, you can do what you want, but I would choose Quick Sort.
Why do I choose Quick Sort? I do not like to work, of course! I want work done as soon as I can get it done.
How do I know Quick Sort is less work? I know that O(n log n) is less than O(n squared). The O's are more small, so Quick Sort is less work!
Now you know my friend, Big O. He helps us do less work. And if you know big O, you can do less work too!
You learned all that with me! You are so smart! Thank you so much!
Now that work is done, let’s go play!
[1]: There is a way to cheat and add all the things from one to n, all at one time. Some kid named Gauss found this out when he was eight. I am not that smart though, so don't ask me how he did it.

Not sure I'm further contributing to the subject but still thought I'd share: I once found this blog post to have some quite helpful (though very basic) explanations & examples on Big O:
Via examples, this helped get the bare basics into my tortoiseshell-like skull, so I think it's a pretty descent 10-minute read to get you headed in the right direction.

I've more simpler way to understand the time complexity
he most common metric for calculating time complexity is Big O notation. This removes all constant factors so that the running time can be estimated in relation to N as N approaches infinity. In general you can think of it like this:
statement;
Is constant. The running time of the statement will not change in relation to N
for ( i = 0; i < N; i++ )
statement;
Is linear. The running time of the loop is directly proportional to N. When N doubles, so does the running time.
for ( i = 0; i < N; i++ )
{
for ( j = 0; j < N; j++ )
statement;
}
Is quadratic. The running time of the two loops is proportional to the square of N. When N doubles, the running time increases by N * N.
while ( low <= high )
{
mid = ( low + high ) / 2;
if ( target < list[mid] )
high = mid - 1;
else if ( target > list[mid] )
low = mid + 1;
else break;
}
Is logarithmic. The running time of the algorithm is proportional to the number of times N can be divided by 2. This is because the algorithm divides the working area in half with each iteration.
void quicksort ( int list[], int left, int right )
{
int pivot = partition ( list, left, right );
quicksort ( list, left, pivot - 1 );
quicksort ( list, pivot + 1, right );
}
Is N * log ( N ). The running time consists of N loops (iterative or recursive) that are logarithmic, thus the algorithm is a combination of linear and logarithmic.
In general, doing something with every item in one dimension is linear, doing something with every item in two dimensions is quadratic, and dividing the working area in half is logarithmic. There are other Big O measures such as cubic, exponential, and square root, but they're not nearly as common. Big O notation is described as O ( ) where is the measure. The quicksort algorithm would be described as O ( N * log ( N ) ).
Note: None of this has taken into account best, average, and worst case measures. Each would have its own Big O notation. Also note that this is a VERY simplistic explanation. Big O is the most common, but it's also more complex that I've shown. There are also other notations such as big omega, little o, and big theta. You probably won't encounter them outside of an algorithm analysis course.
See more at: Here

What is a plain English explanation of “Big O” notation?
Very Quick Note:
The O in "Big O" refers to as "Order"(or precisely "order of")
so you could get its idea literally that it's used to order something to compare them.
"Big O" does two things:
Estimates how many steps of the method your computer applies to accomplish a task.
Facilitate the process to compare with others in order to determine whether it's good or not?
"Big O' achieves the above two with standardized Notations.
There are seven most used notations
O(1), means your computer gets a task done with 1 step, it's excellent, Ordered No.1
O(logN), means your computer complete a task with logN steps, its good, Ordered No.2
O(N), finish a task with N steps, its fair, Order No.3
O(NlogN), ends a task with O(NlogN) steps, it's not good, Order No.4
O(N^2), get a task done with N^2 steps, it's bad, Order No.5
O(2^N), get a task done with 2^N steps, it's horrible, Order No.6
O(N!), get a task done with N! steps, it's terrible, Order No.7
Suppose you get notation O(N^2), not only you are clear the method takes N*N steps to accomplish a task, also you see that it's not good as O(NlogN) from its ranking.
Please note the order at line end, just for your better understanding.There's more than 7 notations if all possibilities considered.
In CS, the set of steps to accomplish a task is called algorithms.
In Terminology, Big O notation is used to describe the performance or complexity of an algorithm.
In addition, Big O establishes the worst-case or measure the Upper-Bound steps.
You could refer to Big-Ω (Big-Omega) for best case.
Big-Ω (Big-Omega) notation (article) | Khan Academy
Summary
"Big O" describes the algorithm's performance and evaluates it.
or address it formally, "Big O" classifies the algorithms and standardize the comparison process.

Assume we're talking about an algorithm A, which should do something with a dataset of size n.
Then O( <some expression X involving n> ) means, in simple English:
If you're unlucky when executing A, it might take as much as X(n) operations to
complete.
As it happens, there are certain functions (think of them as implementations of X(n)) that tend to occur quite often. These are well known and easily compared (Examples: 1, Log N, N, N^2, N!, etc..)
By comparing these when talking about A and other algorithms, it is easy to rank the algorithms according to the number of operations they may (worst-case) require to complete.
In general, our goal will be to find or structure an algorithm A in such a way that it will have a function X(n) that returns as low a number as possible.

If you have a suitable notion of infinity in your head, then there is a very brief description:
Big O notation tells you the cost of solving an infinitely large problem.
And furthermore
Constant factors are negligible
If you upgrade to a computer that can run your algorithm twice as fast, big O notation won't notice that. Constant factor improvements are too small to even be noticed in the scale that big O notation works with. Note that this is an intentional part of the design of big O notation.
Although anything "larger" than a constant factor can be detected, however.
When interested in doing computations whose size is "large" enough to be considered as approximately infinity, then big O notation is approximately the cost of solving your problem.
If the above doesn't make sense, then you don't have a compatible intuitive notion of infinity in your head, and you should probably disregard all of the above; the only way I know to make these ideas rigorous, or to explain them if they aren't already intuitively useful, is to first teach you big O notation or something similar. (although, once you well understand big O notation in the future, it may be worthwhile to revisit these ideas)

Say you order Harry Potter: Complete 8-Film Collection [Blu-ray] from Amazon and download the same film collection online at the same time. You want to test which method is faster. The delivery takes almost a day to arrive and the download completed about 30 minutes earlier. Great! So it’s a tight race.
What if I order several Blu-ray movies like The Lord of the Rings, Twilight, The Dark Knight Trilogy, etc. and download all the movies online at the same time? This time, the delivery still take a day to complete, but the online download takes 3 days to finish.
For online shopping, the number of purchased item (input) doesn’t affect the delivery time. The output is constant. We call this O(1).
For online downloading, the download time is directly proportional to the movie file sizes (input). We call this O(n).
From the experiments, we know that online shopping scales better than online downloading. It is very important to understand big O notation because it helps you to analyze the scalability and efficiency of algorithms.
Note: Big O notation represents the worst-case scenario of an algorithm. Let’s assume that O(1) and O(n) are the worst-case scenarios of the example above.
Reference : http://carlcheo.com/compsci

Definition :- Big O notation is a notation which says how a algorithm performance will perform if the data input increases.
When we talk about algorithms there are 3 important pillars Input , Output and Processing of algorithm. Big O is symbolic notation which says if the data input is increased in what rate will the performance vary of the algorithm processing.
I would encourage you to see this youtube video which explains Big O Notation in depth with code examples.
So for example assume that a algorithm takes 5 records and the time required for processing the same is 27 seconds. Now if we increase the records to 10 the algorithm takes 105 seconds.
In simple words the time taken is square of the number of records. We can denote this by O(n ^ 2). This symbolic representation is termed as Big O notation.
Now please note the units can be anything in inputs it can be bytes , bits number of records , the performance can be measured in any unit like second , minutes , days and so on. So its not the exact unit but rather the relationship.
For example look at the below function "Function1" which takes a collection and does processing on the first record. Now for this function the performance will be same irrespective you put 1000 , 10000 or 100000 records. So we can denote it by O(1).
void Function1(List<string> data)
{
string str = data[0];
}
Now see the below function "Function2()". In this case the processing time will increase with number of records. We can denote this algorithm performance using O(n).
void Function2(List<string> data)
{
foreach(string str in data)
{
if (str == "shiv")
{
return;
}
}
}
When we see a Big O notation for any algorithm we can classify them in to three categories of performance :-
Log and constant category :- Any developer would love to see their algorithm performance in this category.
Linear :- Developer will not want to see algorithms in this category , until its the last option or the only option left.
Exponential :- This is where we do not want to see our algorithms and a rework is needed.
So by looking at Big O notation we categorize good and bad zones for algorithms.
I would recommend you to watch this 10 minutes video which discusses Big O with sample code
https://www.youtube.com/watch?v=k6kxtzICG_g

Simplest way to look at it (in plain English)
We are trying to see how the number of input parameters, affects the running time of an algorithm. If the running time of your application is proportional to the number of input parameters, then it is said to be in Big O of n.
The above statement is a good start but not completely true.
A more accurate explanation (mathematical)
Suppose
n=number of input parameters
T(n)= The actual function that expresses the running time of the algorithm as a function of n
c= a constant
f(n)= An approximate function that expresses the running time of the algorithm as a function of n
Then as far as Big O is concerned, the approximation f(n) is considered good enough as long as the below condition is true.
lim T(n) ≤ c×f(n)
n→∞
The equation is read as
As n approaches infinity, T of n, is less than or equal to c times f of n.
In big O notation this is written as
T(n)∈O(n)
This is read as T of n is in big O of n.
Back to English
Based on the mathematical definition above, if you say your algorithm is a Big O of n, it means it is a function of n (number of input parameters) or faster. If your algorithm is Big O of n, then it is also automatically the Big O of n square.
Big O of n means my algorithm runs at least as fast as this. You cannot look at Big O notation of your algorithm and say its slow. You can only say its fast.
Check this out for a video tutorial on Big O from UC Berkley. It is actually a simple concept. If you hear professor Shewchuck (aka God level teacher) explaining it, you will say "Oh that's all it is!".

Preface
algorithm: procedure/formula for solving a problem
How do analyze algorithms and how can we compare algorithms against each other?
example: you and a friend are asked to create a function to sum the numbers from 0 to N. You come up with f(x) and your friend comes up with g(x). Both functions have the same result, but a different algorithm. In order to objectively compare the efficiency of the algorithms we use Big-O notation.
Big-O notation: describes how quickly runtime will grow relative to the input as the input get arbitrarily large.
3 key takeaways:
Compare how quickly runtime grows NOT compare exact runtimes (depends on hardware)
Only concerned with runtime grow relative to the input (n)
As n gets arbitrarily large, focus on the terms that will grow the fastest as n gets large (think infinity) AKA asymptotic analysis
Space complexity: aside from time complexity, we also care about space complexity (how much memory/space an algorithm uses). Instead of checking the time of operations, we check the size of the allocation of memory.

I found a really great explanation about big O notation especially for a someone who's not much into mathematics.
https://rob-bell.net/2009/06/a-beginners-guide-to-big-o-notation/
Big O notation is used in Computer Science to describe the performance
or complexity of an algorithm. Big O specifically describes the
worst-case scenario, and can be used to describe the execution time
required or the space used (e.g. in memory or on disk) by an
algorithm.
Anyone who's read Programming Pearls or any other Computer Science
books and doesn’t have a grounding in Mathematics will have hit a wall
when they reached chapters that mention O(N log N) or other seemingly
crazy syntax. Hopefully this article will help you gain an
understanding of the basics of Big O and Logarithms.
As a programmer first and a mathematician second (or maybe third or
fourth) I found the best way to understand Big O thoroughly was to
produce some examples in code. So, below are some common orders of
growth along with descriptions and examples where possible.
O(1)
O(1) describes an algorithm that will always execute in the same time
(or space) regardless of the size of the input data set.
bool IsFirstElementNull(IList<string> elements) {
return elements[0] == null;
}
O(N)
O(N) describes an algorithm whose performance will grow linearly and
in direct proportion to the size of the input data set. The example
below also demonstrates how Big O favours the worst-case performance
scenario; a matching string could be found during any iteration of the
for loop and the function would return early, but Big O notation will
always assume the upper limit where the algorithm will perform the
maximum number of iterations.
bool ContainsValue(IList<string> elements, string value) {
foreach (var element in elements)
{
if (element == value) return true;
}
return false;
}
O(N2)
O(N2) represents an algorithm whose performance is directly
proportional to the square of the size of the input data set. This is
common with algorithms that involve nested iterations over the data
set. Deeper nested iterations will result in O(N3), O(N4) etc.
bool ContainsDuplicates(IList<string> elements) {
for (var outer = 0; outer < elements.Count; outer++)
{
for (var inner = 0; inner < elements.Count; inner++)
{
// Don't compare with self
if (outer == inner) continue;
if (elements[outer] == elements[inner]) return true;
}
}
return false;
}
O(2N)
O(2N) denotes an algorithm whose growth doubles with each additon to
the input data set. The growth curve of an O(2N) function is
exponential - starting off very shallow, then rising meteorically. An
example of an O(2N) function is the recursive calculation of Fibonacci
numbers:
int Fibonacci(int number) {
if (number <= 1) return number;
return Fibonacci(number - 2) + Fibonacci(number - 1);
}
Logarithms
Logarithms are slightly trickier to explain so I'll use a common
example:
Binary search is a technique used to search sorted data sets. It works
by selecting the middle element of the data set, essentially the
median, and compares it against a target value. If the values match it
will return success. If the target value is higher than the value of
the probe element it will take the upper half of the data set and
perform the same operation against it. Likewise, if the target value
is lower than the value of the probe element it will perform the
operation against the lower half. It will continue to halve the data
set with each iteration until the value has been found or until it can
no longer split the data set.
This type of algorithm is described as O(log N). The iterative halving
of data sets described in the binary search example produces a growth
curve that peaks at the beginning and slowly flattens out as the size
of the data sets increase e.g. an input data set containing 10 items
takes one second to complete, a data set containing 100 items takes
two seconds, and a data set containing 1000 items will take three
seconds. Doubling the size of the input data set has little effect on
its growth as after a single iteration of the algorithm the data set
will be halved and therefore on a par with an input data set half the
size. This makes algorithms like binary search extremely efficient
when dealing with large data sets.

Efficient Way to Repeatedly Split Large NumPy Array and Record Middle

I have a large NumPy array nodes = np.arange(100_000_000) and I need to rearrange this array by:
Recording and then removing the middle value in the array
Split the array into the left half and right half
Repeat Steps 1-2 for each half
Stop when all values are exhausted
So, for a smaller input example nodes = np.arange(10), the output would be:
[5 2 8 1 4 7 9 0 3 6]
This was accomplished by naively doing:
import numpy as np
def split(node, out):
mid = len(node) // 2
out.append(node[mid])
return node[:mid], node[mid+1:]
def reorder(a):
nodes = [a.tolist()]
out = []
while nodes:
tmp = []
for node in nodes:
for n in split(node, out):
if n:
tmp.append(n)
nodes = tmp
return np.array(out)
if __name__ == "__main__":
nodes = np.arange(10)
print(reorder(nodes))
However, this is way too slow for nodes = np.arange(100_000_000) and so I am looking for a much faster solution.

You can vectorize your function with Numpy by working on groups of slices.
Here is an implementation:
# Similar to [e for tmp in zip(a, b) for e in tmp] ,
# but on Numpy arrays and much faster
def interleave(a, b):
assert len(a) == len(b)
return np.column_stack((a, b)).reshape(len(a) * 2)
# n is the length of the input range (len(a) in your example)
def fast_reorder(n):
if n == 0:
return np.empty(0, dtype=np.int32)
startSlices = np.array([0], dtype=np.int32)
endSlices = np.array([n], dtype=np.int32)
allMidSlices = np.empty(n, dtype=np.int32) # Similar to "out" in your implementation
midInsertCount = 0 # Actual size of allMidSlices
# Generate a bunch of middle values as long as there is valid slices to split
while midInsertCount < n:
# Generate the new mid/left/right slices
midSlices = (endSlices + startSlices) // 2
# Computing the next slices is not needed for the last step
if midInsertCount + len(midSlices) < n:
# Generate the nexts slices (possibly with invalid ones)
newStartSlices = interleave(startSlices, midSlices+1)
newEndSlices = interleave(midSlices, endSlices)
# Discard invalid slices
isValidSlices = newStartSlices < newEndSlices
startSlices = newStartSlices[isValidSlices]
endSlices = newEndSlices[isValidSlices]
# Fast appending
allMidSlices[midInsertCount:midInsertCount+len(midSlices)] = midSlices
midInsertCount += len(midSlices)
return allMidSlices[0:midInsertCount]
On my machine, this is 89 times faster than your scalar implementation with the input np.arange(100_000_000) dropping from 2min35 to 1.75s. It also consume far less memory (rougthly 3~4 times less). Note that if you want a faster code, then you probably need to use a native language like C or C++.

Edit:
The question has been updated to have a much smaller input array so I leave the below for historical reasons. Basically it was likely a typo but we often get accustomed to computers working with insanely large numbers and when memory is involved they can be a real problem.
There is already a numpy based solution submitted by someone else that I think fits the bill.
Your code requires an insane amount of RAM just to hold 100 billion 64 bit integers. Do you have 800GB of RAM? Then you convert the numpy array to a list which will be substantially larger than the array (each packed 64 bit int in the numpy array will become a much less memory efficient python int object and the list will have a pointer to that object). Then you make a lot of slices of the list which will not duplicate the data but will duplicate the pointers to the data and use even more RAM. You also append all the result values to a list a single value at a time. Lists are very fast for adding items generally but with such an extreme size this will not only be slow but the way the list is allocated is likely to be extremely wasteful RAM wise and contribute to major problems (I believe they double in size when they get to a certain level of fullness so you will end up allocating more RAM than you need and doing many allocations and likely copies). What kind of machine are you running this on? There are ways to improve your code but unless you're running it on a super computer I don't know that you're going to ever finish that calculation. I only..only? have 32GB of RAM and I'm not going to even try to create a 100B int_64 numpy array as I don't want to use up ssd write life for a mass of virtual memory.
As for improving your code stick to numpy arrays don't change to a python list it will greatly increase the RAM you need. Preallocate a numpy array to put the answer in. Then you need a new algorithm. Anything recursive or recursive like (ie a loop splitting the input,) will require tracking a lot of state, your nodes list is going to be extraordinarily gigantic and again use a lot of RAM. You could use len(a) to indicate values that are removed from your list and scan through the entire array each time to figure out what to do next but that will save RAM in favour of a tremendous amount of searching a gigantic array. I feel like there is an algorithm to cut numbers from each end and place them in the output and just track the beginning and end but I haven't figured it out at least not yet.
I also think there is a simpler algorithm where you just track the number of splits you've done instead of making a giant list of slices and keeping it all in memory. Take the middle of the left half and then the middle of the right then count up one and when you take the middle of the left half's left half you know you have to jump to the right half then the count is one so you jump over to the original right half's left half and on and on... Based on the depth into the halves and the length of the input you should be able to jump around without scanning or tracking all of those slices though I haven't been able to dedicate much time to thinking this through in my head.
With a problem of this nature if you really need to push the limits you should consider using C/C++ so you can be as efficient as possible with RAM usage and because you're doing an insane number of tiny things which doesn't map well to python performance.

Code finding the first triangular number with more than 500 divisors will not finish running

Okay, so I'm working on Euler Problem 12 (find the first triangular number with a number of factors over 500) and my code (in Python 3) is as follows:
factors = 0
y=1
def factornum(n):
x = 1
f = []
while x <= n:
if n%x == 0:
f.append(x)
x+=1
return len(f)
def triangle(n):
t = sum(list(range(1,n)))
return t
while factors<=500:
factors = factornum(triangle(y))
y+=1
print(y-1)
Basically, a function goes through all the numbers below the input number n, checks if they divide into n evenly, and if so add them to a list, then return the length in that list. Another generates a triangular number by summing all the numbers in a list from 1 to the input number and returning the sum. Then a while loop continues to generate a triangular number using an iterating variable y as the input for the triangle function, and then runs the factornum function on that and puts the result in the factors variable. The loop continues to run and the y variable continues to increment until the number of factors is over 500. The result is then printed.
However, when I run it, nothing happens - no errors, no output, it just keeps running and running. Now, I know my code isn't the most efficient, but I left it running for quite a bit and it still didn't produce a result, so it seems more likely to me that there's an error somewhere. I've been over it and over it and cannot seem to find an error.
I'd merely request that a full solution or a drastically improved one isn't given outright but pointers towards my error(s) or spots for improvement, as the reason I'm doing the Euler problems is to improve my coding. Thanks!

You have very inefficient algorithm.
If you ask for pointers rather than full solution, main pointers are:
There is a more efficient way to calculate next triangular number. There is an explicit formula in the wiki. Also if you generate sequence of all numbers it is just more efficient to add next n to the previous number. (Sidenote list in sum(list(range(1,n))) makes no sense to me at all. If you want to use this approach anyway, sum(xrange(1,n) will probably be much more efficient as it doesn't require materialization of the range)
There are much more efficient ways to factorize numbers
There is a more efficient way to calculate number of factors. And it is actually called after Euler: see Euler's totient function
Generally Euler project problems (as in many other programming competitions) are not supposed to be solvable by sheer brute force. You should come up with some formula and/or more efficient algorithm first.

As far as I can tell your code will work, but it will take a very long time to calculate the number of factors. For 150 factors, it takes on the order of 20 seconds to run, and that time will grow dramatically as you look for higher and higher number of factors.
One way to reduce the processing time is to reduce the number of calculations that you're performing. If you analyze your code, you're calculating n%1 every single time, which is an unnecessary calculation because you know every single integer will be divisible by itself and one. Are there any other ways you can reduce the number of calculations? Perhaps by remembering that if a number is divisible by 20, it is also divisible by 2, 4, 5, and 10?
I can be more specific, but you wanted a pointer in the right direction.

From the looks of it the code works fine, it`s just not the best approach. A simple way of optimizing is doing until the half the number, for example. Also, try thinking about how you could do this using prime factors, it might be another solution. Best of luck!

First you have to def a factor function:
from functools import reduce
def factors(n):
step = 2 if n % 2 else 1
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(pow(n,0.5) + 1)) if n % i
== 0)))
This will create a set and put all of factors of number n into it.
Second, use while loop until you get 500 factors:
a = 1
x = 1
while len(factors(a)) < 501:
x += 1
a += x
This loop will stop at len(factors(a)) = 500.
Simple print(a) and you will get your answer.

Asymptotic complexity python

I have a task to find asymptotic complexity of this python code.
for i in range(x):
if i == 0:
for j in range(x):
for k in range(500):
print("A ")
By what I know it should be 500*x. Because the first cycle only goes once (i==0), second one goes x times and third one 500 times, so therefore it should be 500*x, shouldn´t it? However this isn´t the right answer. Could you help me out?

Asymptotic complexity describes the growth of execution time as the variable factors grow arbitrarily large. In short, constants added or multiplied don't count at all, since they don't change with the variable.
Yes, there are 500*x lines printed. You also have x-1 non-functional loop iterations. Your total time would be computed as something like
(x-1)[loop overhead] + x[loop overhead] + 500*x[loop overhead + print
time]
However, the loop overhead, being a constant, is insignificant, and dropped out of the complexity expression. Likewise, the 500 is merely a scaling factor, and is also dropped from the expression.
The complexity is O(x).

It's 501*x since you also have to check x times the line if i == 0.
As the other answer says, usually we don't include the factor. But sometimes we do.

Sum of primes below 2,000,000 in python

I am attempting problem 10 of Project Euler, which is the summation of all primes below 2,000,000. I have tried implementing the Sieve of Erasthotenes using Python, and the code I wrote works perfectly for numbers below 10,000.
However, when I attempt to find the summation of primes for bigger numbers, the code takes too long to run (finding the sum of primes up to 100,000 took 315 seconds). The algorithm clearly needs optimization.
Yes, I have looked at other posts on this website, like Fastest way to list all primes below N, but the solutions there had very little explanation as to how the code worked (I am still a beginner programmer) so I was not able to actually learn from them.
Can someone please help me optimize my code, and clearly explain how it works along the way?
Here is my code:
primes_below_number = 2000000 # number to find summation of all primes below number
numbers = (range(1, primes_below_number + 1, 2)) # creates a list excluding even numbers
pos = 0 # index position
sum_of_primes = 0 # total sum
number = numbers[pos]
while number < primes_below_number and pos < len(numbers) - 1:
pos += 1
number = numbers[pos] # moves to next prime in list numbers
sum_of_primes += number # adds prime to total sum
num = number
while num < primes_below_number:
num += number
if num in numbers[:]:
numbers.remove(num) # removes multiples of prime found
print sum_of_primes + 2
As I said before, I am new to programming, therefore a thorough explanation of any complicated concepts would be deeply appreciated. Thank you.

As you've seen, there are various ways to implement the Sieve of Erasthotenes in Python that are more efficient than your code. I don't want to confuse you with fancy code, but I can show how to speed up your code a fair bit.
Firstly, searching a list isn't fast, and removing elements from a list is even slower. However, Python provides a set type which is quite efficient at performing both of those operations (although it does chew up a bit more RAM than a simple list). Happily, it's easy to modify your code to use a set instead of a list.
Another optimization is that we don't have to check for prime factors all the way up to primes_below_number, which I've renamed to hi in the code below. It's sufficient to just go to the square root of hi, since if a number is composite it must have a factor less than or equal to its square root.
We don't need to keep a running total of the sum of the primes. It's better to do that at the end using Python's built-in sum() function, which operates at C speed, so it's much faster than doing the additions one by one at Python speed.
# number to find summation of all primes below number
hi = 2000000
# create a set excluding even numbers
numbers = set(xrange(3, hi + 1, 2))
for number in xrange(3, int(hi ** 0.5) + 1):
if number not in numbers:
#number must have been removed because it has a prime factor
continue
num = number
while num < hi:
num += number
if num in numbers:
# Remove multiples of prime found
numbers.remove(num)
print 2 + sum(numbers)
You should find that this code runs in a a few seconds; it takes around 5 seconds on my 2GHz single-core machine.
You'll notice that I've moved the comments so that they're above the line they're commenting on. That's the preferred style in Python since we prefer short lines, and also inline comments tend to make the code look cluttered.
There's another small optimization that can be made to the inner while loop, but I let you figure that out for yourself. :)

First, removing numbers from the list will be very slow. Instead of this, make a list
primes = primes_below_number * True
primes[0] = False
primes[1] = False
Now in your loop, when you find a prime p, change primes[k*p] to False for all suitable k. (You wouldn't actually do multiply, you'd continually add p, of course.)
At the end,
primes = [n for n i range(primes_below_number) if primes[n]]
This should be a great deal faster.
Second, you can stop looking once your find a prime greater than the square root of primes_below_number, since a composite number must have a prime factor that doesn't exceed its square root.

Try using numpy, should make it faster. Replace range by xrange, it may help you.

Here's an optimization for your code:
import itertools
primes_below_number = 2000000
numbers = list(range(3, primes_below_number, 2))
pos = 0
while pos < len(numbers) - 1:
number = numbers[pos]
numbers = list(
itertools.chain(
itertools.islice(numbers, 0, pos + 1),
itertools.ifilter(
lambda n: n % number != 0,
itertools.islice(numbers, pos + 1, len(numbers))
)
)
)
pos += 1
sum_of_primes = sum(numbers) + 2
print sum_of_primes
The optimization here is because:
Removed the sum to outside the loop.
Instead of removing elements from a list we can just create another one, memory is not an issue here (I hope).
When creating the new list we create it by chaining two parts, the first part is everything before the current number (we already checked those), and the second part is everything after the current number but only if they are not divisible by the current number.
Using itertools can make things faster since we'd be using iterators instead of looping through the whole list more than once.
Another solution would be to not remove parts of the list but disable them like #saulspatz said.
And here's the fastest way I was able to find: http://www.wolframalpha.com/input/?i=sum+of+all+primes+below+2+million 😁
Update
Here is the boolean method:
import itertools
primes_below_number = 2000000
numbers = [v % 2 != 0 for v in xrange(primes_below_number)]
numbers[0] = False
numbers[1] = False
numbers[2] = True
number = 3
while number < primes_below_number:
n = number * 3 # We already excluded even numbers
while n < primes_below_number:
numbers[n] = False
n += number
number += 1
while number < primes_below_number and not numbers[number]:
number += 1
sum_of_numbers = sum(itertools.imap(lambda index_n: index_n[1] and index_n[0] or 0, enumerate(numbers)))
print(sum_of_numbers)
This executes in seconds (took 3 seconds on my 2.4GHz machine).

Instead of storing a list of numbers, you can instead store an array of boolean values. This use of a bitmap can be thought of as a way to implement a set, which works well for dense sets (there aren't big gaps between the values of members).
An answer on a recent python sieve question uses this implementation python-style. It turns out a lot of people have implemented a sieve, or something they thought was a sieve, and then come on SO to ask why it was slow. :P Look at the related-questions sidebar from some of them if you want more reading material.
Finding the element that holds the boolean that says whether a number is in the set or not is easy and extremely fast. array[i] is a boolean value that's true if i is in the set, false if not. The memory address can be computed directly from i with a single addition.
(I'm glossing over the fact that an array of boolean might be stored with a whole byte for each element, rather than the more efficient implementation of using every single bit for a different element. Any decent sieve will use a bitmap.)
Removing a number from the set is as simple as setting array[i] = false, regardless of the previous value. No searching, not comparison, no tracking of what happened, just one memory operation. (Well, two for a bitmap: load the old byte, clear the correct bit, store it. Memory is byte-addressable, but not bit-addressable.)
An easy optimization of the bitmap-based sieve is to not even store the even-numbered bytes, because there is only one even prime, and we can special-case it to double our memory density. Then the membership-status of i is held in array[i/2]. (Dividing by powers of two is easy for computers. Other values are much slower.)
An SO question:
Why is Sieve of Eratosthenes more efficient than the simple "dumb" algorithm? has many links to good stuff about the sieve. This one in particular has some good discussion about it, in words rather than just code. (Nevermind the fact that it's talking about a common Haskell implementation that looks like a sieve, but actually isn't. They call this the "unfaithful" sieve in their graphs, and so on.)
discussion on that question brought up the point that trial division may be fast than big sieves, for some uses, because clearing the bits for all multiples of every prime touches a lot of memory in a cache-unfriendly pattern. CPUs are much faster than memory these days.