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How to improve time complexity of remove all multiplicands from array or list?

I am trying to find elements from array(integer array) or list which are unique and those elements must not divisible by any other element from same array or list.
You can answer in any language like python, java, c, c++ etc.
I have tried this code in Python3 and it works perfectly but I am looking for better and optimum solution in terms of time complexity.
assuming array or list A is already sorted and having unique elements
A = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
while i<len(A)-1:
while j<len(A):
if A[j]%A[i]==0:
A.pop(j)
else:
j+=1
i+=1
j=i+1
For the given array A=[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16] answer would be like ans=[2,3,5,7,11,13]
another example,A=[4,5,15,16,17,23,39] then ans would be like, ans=[4,5,17,23,39]
ans is having unique numbers
any element i from array only exists if (i%j)!=0, where i!=j

I think it's more natural to do it in reverse, by building a new list containing the answer instead of removing elements from the original list. If I'm thinking correctly, both approaches do the same number of mod operations, but you avoid the issue of removing an element from a list.
A = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
ans = []
for x in A:
for y in ans:
if x % y == 0:
break
else: ans.append(x)
Edit: Promoting the completion else.

This algorithm will perform much faster:
A = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
if (A[-1]-A[0])/A[0] > len(A)*2:
result = list()
for v in A:
for f in result:
d,m = divmod(v,f)
if m == 0: v=0;break
if d<f: break
if v: result.append(v)
else:
retain = set(A)
minMult = 1
maxVal = A[-1]
for v in A:
if v not in retain : continue
minMult = v*2
if minMult > maxVal: break
if v*len(A)<maxVal:
retain.difference_update([m for m in retain if m >= minMult and m%v==0])
else:
retain.difference_update(range(minMult,maxVal,v))
if maxVal%v == 0:
maxVal = max(retain)
result = list(retain)
print(result) # [2, 3, 5, 7, 11, 13]
In the spirit of the sieve of Eratostenes, each number that is retained, removes its multiples from the remaining eligible numbers. Depending on the magnitude of the highest value, it is sometimes more efficient to exclude multiples than check for divisibility. The divisibility check takes several times longer for an equivalent number of factors to check.
At some point, when the data is widely spread out, assembling the result instead of removing multiples becomes faster (this last addition was inspired by Imperishable Night's post).
TEST RESULTS
A = [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16] (100000 repetitions)
Original: 0.55 sec
New: 0.29 sec
A = list(range(2,5000))+[9697] (100 repetitions)
Original: 3.77 sec
New: 0.12 sec
A = list(range(1001,2000))+list(range(4000,6000))+[9697**2] (10 repetitions)
Original: 3.54 sec
New: 0.02 sec

I know that this is totally insane but i want to know what you think about this:
A = [4,5,15,16,17,23,39]
prova=[[x for x in A if x!=y and y%x==0] for y in A]
print([A[idx] for idx,x in enumerate(prova) if len(prova[idx])==0])
And i think it's still O(n^2)

If you care about speed more than algorithmic efficiency, numpy would be the package to use here in python:
import numpy as np
# Note: doesn't have to be sorted
a = [2, 2, 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 16, 29, 29]
a = np.unique(a)
result = a[np.all((a % a[:, None] + np.diag(a)), axis=0)]
# array([2, 3, 5, 7, 11, 13, 29])
This divides all elements by all other elements and stores the remainder in a matrix, checks which columns contain only non-0 values (other than the diagonal), and selects all elements corresponding to those columns.

This is O(n*M) where M is the max size of an integer in your list. The integers are all assumed to be none negative. This also assumes your input list is sorted (came to that assumption since all lists you provided are sorted).
a = [4, 7, 7, 8]
# a = [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
# a = [4, 5, 15, 16, 17, 23, 39]
M = max(a)
used = set()
final_list = []
for e in a:
if e in used:
continue
else:
used.add(e)
for i in range(e, M + 1):
if not (i % e):
used.add(i)
final_list.append(e)
print(final_list)
Maybe this can be optimized even further...
If the list is not sorted then for the above method to work, one must sort it. The time complexity will then be O(nlogn + Mn) which equals to O(nlogn) when n >> M.

Pythonic way to check how many items in a list or combination of items match a condition

Say I have a list like this:
l = [20,17,8,7,4,9]
I would like to check how many items or combinations of items match a value condition.
To check the single items this would do:
minimum_value = 12
count = 0
for item in l:
if item >= minimum_value:
count += 1
But I also would like to check combinations of that list. Then the count would be 4: 20, 17, 9+4, 8+7
How could I do this?

You can do that by checking the total value at the each iteration. If the total is larger than the minimum value, increase the count and reset the total. Otherwise, add the number from the next iteration to the total and check again.
l = [20,17,8,7,4,9]
minimum_value = 12
count = 0
sum = 0
for item in l:
sum += item
if sum >= minimum_value:
count += 1
sum = 0
print(count)

If I understand your question, you want to check all values plus all combinations of values against a minimum value and return the count of such. The following code will do this.
Note: there is probably a more concise way to do this, but I feel this is the easiest to read.
l = [20, 17, 8, 7, 4, 9]
minimum_value = 12
count = 0
newl = [] # Create a new list to store all the different combinations
for i in range(len(l)): # Get each index in l
newl.append(l[i]) # Append the value at the current index to newl
for n in l[i+1:]: # Loop through the remaining numbers starting at index + 1
newl.append(l[i] + n) # Combine
count = len([x for x in newl if x >= minimum_value]) # Get the length of a list generated by taking all numbers in newl that are >= minimum_value
print(count)
In the above case, newl contains:
[20, 37, 28, 27, 24, 29, 17, 25, 24, 21, 26, 8, 15, 12, 17, 7, 11, 16, 4, 13, 9]
And thus has 16 values that exceed the minimum_value of 12

Print a line if conditions have not been met

Hello fellow stackoverflowers, I am practising my Python with an example question given to me (actually a Google interview practice question) and ran into a problem I did not know how to a) pose properly (hence vague title), b) overcome.
The question is: For an array of numbers (given or random) find unique pairs of numbers within the array which when summed give a given number. E.G: find the pairs of numbers in the array below which add to 6.
[1 2 4 5 11]
So in the above case:
[1,5] and [2,4]
The code I have written is:
from secrets import *
i = 10
x = randbelow(10)
number = randbelow(100) #Generate a random number to be the sum that we are after#
if number == 0:
pass
else:
number = number
array = []
while i>0: #Generate a random array to use#
array.append(x)
x = x + randbelow(10)
i -= 1
print("The following is a randomly generated array:\n" + str(array))
print("Within this array we are looking for a pair of numbers which sum to " + str(number))
for i in range(0,10):
for j in range(0,10):
if i == j or i>j:
pass
else:
elem_sum = array[i] + array[j]
if elem_sum == number:
number_one = array[i]
number_two = array[j]
print("A pair of numbers within the array which satisfy that condition is: " + str(number_one) + " and " + str(number_two))
else:
pass
If no pairs are found, I want the line "No pairs were found". I was thinking a try/except, but wasn't sure if it was correct or how to implement it. Also, I'm unsure on how to stop repeated pairs appearing (unique pairs only), so for example if I wanted 22 as a sum and had the array:
[7, 9, 9, 13, 13, 14, 23, 32, 41, 45]
[9,13] would appear twice
Finally forgive me if there are redundancies/the code isn't written very efficiently, I'm slowly learning so any other tips would be greatly appreciated!
Thanks for reading :)

You can simply add a Boolean holding the answer to "was at least one pair found?".
initialize it as found = false at the beginning of your code.
Then, whenever you find a pair (the condition block that holds your current print command), just add found = true.
after all of your search (the double for loop`), add this:
if not found:
print("No pairs were found")

Instead of actually comparing each pair of numbers, you can just iterate the list once, subtract the current number from the target number, and see if the remainder is in the list. If you convert the list to a set first, that lookup can be done in O(1), reducing the overall complexity from O(n²) to just O(n). Also, the whole thing can be done in a single line with a list comprehension:
>>> nums = [1, 2, 4, 5, 11]
>>> target = 6
>>> nums_set = set(nums)
>>> pairs = [(n, target-n) for n in nums_set if target-n in nums_set and n <= target/2]
>>> print(pairs)
[(1, 5), (2, 4)]
For printing the pairs or some message, you can use the or keyword. x or y is interpreted as x if x else y, so if the result set is empty, the message is printed, otherwise the result set itself.
>>> pairs = []
>>> print(pairs or "No pairs found")
No pairs found
Update: The above can fail, if the number added to itself equals the target, but is only contained once in the set. In this case, you can use a collections.Counter instead of a set and check the multiplicity of that number first.
>>> nums = [1, 2, 4, 5, 11, 3]
>>> nums_set = set(nums)
>>> [(n, target-n) for n in nums_set if target-n in nums_set and n <= target/2]
[(1, 5), (2, 4), (3, 3)]
>>> nums_counts = collections.Counter(nums)
>>> [(n, target-n) for n in nums_counts if target-n in nums_counts and n <= target/2 and n != target-n or nums_counts[n] > 1]
[(1, 5), (2, 4)]

List your constraints first!
numbers added must be unique
only 2 numbers can be added
the length of the array can be arbitrary
the number to be summed to can be arbitrary
& Don't skip preprocessing! Reduce your problem-space.
2 things off the bat:
Starting after your 2 print statements, the I would do array = list(set(array)) to reduce the problem-space to [7, 9, 13, 14, 23, 32, 41, 45].
Assuming that all the numbers in question will be positive, I would discard numbers above number. :
array = [x for x in array if x < number]
giving [7, 9, 9, 13, 13, 14]
Combine the last 2 steps into a list comprehension and then use that as array:
smaller_array = [x for x in list(set(array)) if x < number]
which gives array == [7, 9, 13, 14]
After these two steps, you can do a bunch of stuff. I'm fully aware that I haven't answered your question, but from here you got this. ^this is the kind of stuff I'd assume google wants to see.

How does one find the largest consecutive set of numbers in a list that are not necessarily adjacent?

For instance, if I have a list
[1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11]
This algorithm should return [1,2,3,4,5,6,7,8,9,10,11].
To clarify, the longest list should run forwards. I was wondering what is an algorithmically efficient way to do this (preferably not O(n^2))?
Also, I'm open to a solution not in python since the algorithm is what matters.
Thank you.

Here is a simple one-pass O(n) solution:
s = [1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11,42]
maxrun = -1
rl = {}
for x in s:
run = rl[x] = rl.get(x-1, 0) + 1
print x-run+1, 'to', x
if run > maxrun:
maxend, maxrun = x, run
print range(maxend-maxrun+1, maxend+1)
The logic may be a little more self-evident if you think in terms of ranges instead of individual variables for the endpoint and run length:
rl = {}
best_range = xrange(0)
for x in s:
run = rl[x] = rl.get(x-1, 0) + 1
r = xrange(x-run+1, x+1)
if len(r) > len(best_range):
best_range = r
print list(best_range)

Not that clever, not O(n), could use a bit of optimization. But it works.
def longest(seq):
result = []
for v in seq:
for l in result:
if v == l[-1] + 1:
l.append(v)
else:
result.append([v])
return max(result, key=len)

You can use The Patience Sort implementation of the Largest Ascending Sub-sequence Algorithm
def LargAscSub(seq):
deck = []
for x in seq:
newDeck = [x]
i = bisect.bisect_left(deck, newDeck)
deck[i].insert(0, x) if i != len(deck) else deck.append(newDeck)
return [p[0] for p in deck]
And here is the Test results
>>> LargAscSub([1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11])
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
>>> LargAscSub([1, 2, 3, 11, 12, 13, 14])
[1, 2, 3, 11, 12, 13, 14]
>>> LargAscSub([11,12,13,14])
[11, 12, 13, 14]
The Order of Complexity is O(nlogn)
There was one note in the wiki link where they claimed that you can achieve O(n.loglogn) by relying on Van Emde Boas tree

How about using a modified Radix Sort? As JanneKarila pointed out the solution is not O(n). It uses Radix sort, which wikipedia says Radix sort's efficiency is O(k·n) for n keys which have k or fewer digits.
This will only work if you know the range of numbers that we're dealing with so that will be the first step.
Look at each element in starting list to find lowest, l and highest, h number. In this case l is 1 and h is 11. Note, if you already know the range for some reason, you can skip this step.
Create a result list the size of our range and set each element to null.
Look at each element in list and add them to the result list at the appropriate place if needed. ie, the element is a 4, add a 4 to the result list at position 4. result[element] = starting_list[element]. You can throw out duplicates if you want, they'll just be overwritten.
Go through the result list to find the longest sequence without any null values. Keep a element_counter to know what element in the result list we're looking at. Keep a curr_start_element set to the beginning element of the current sequence and keep a curr_len of how long the current sequence is. Also keep a longest_start_element and a `longest_len' which will start out as zero and be updated as we move through the list.
Return the result list starting at longest_start_element and taking longest_len
EDIT: Code added. Tested and working
#note this doesn't work with negative numbers
#it's certainly possible to write this to work with negatives
# but the code is a bit hairier
import sys
def findLongestSequence(lst):
#step 1
high = -sys.maxint - 1
for num in lst:
if num > high:
high = num
#step 2
result = [None]*(high+1)
#step 3
for num in lst:
result[num] = num
#step 4
curr_start_element = 0
curr_len = 0
longest_start_element = -1
longest_len = -1
for element_counter in range(len(result)):
if result[element_counter] == None:
if curr_len > longest_len:
longest_start_element = curr_start_element
longest_len = curr_len
curr_len = 0
curr_start_element = -1
elif curr_start_element == -1:
curr_start_element = element_counter
curr_len += 1
#just in case the last element makes the longest
if curr_len > longest_len:
longest_start_element = curr_start_element
longest_len = curr_len
#step 5
return result[longest_start_element:longest_start_element + longest_len-1]

If the result really does have to be a sub-sequence of consecutive ascending integers, rather than merely ascending integers, then there's no need to remember each entire consecutive sub-sequence until you determine which is the longest, you need only remember the starting and ending values of each sub-sequence. So you could do something like this:
def longestConsecutiveSequence(sequence):
# map starting values to largest ending value so far
map = collections.OrderedDict()
for i in sequence:
found = False
for k, v in map.iteritems():
if i == v:
map[k] += 1
found = True
if not found and i not in map:
map[i] = i + 1
return xrange(*max(map.iteritems(), key=lambda i: i[1] - i[0]))
If I run this on the original sample date (i.e. [1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11]) I get:
>>> print list(longestConsecutiveSequence([1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11]))
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
If I run it on one of Abhijit's samples [1,2,3,11,12,13,14], I get:
>>> print list(longestConsecutiveSequence([1,2,3,11,12,13,14]))
[11, 12, 13, 14]
Regrettably, this algorithm is O(n*n) in the worst case.

Warning: This is the cheaty way to do it (aka I use python...)
import operator as op
import itertools as it
def longestSequence(data):
longest = []
for k, g in it.groupby(enumerate(set(data)), lambda(i, y):i-y):
thisGroup = map(op.itemgetter(1), g)
if len(thisGroup) > len(longest):
longest = thisGroup
return longest
longestSequence([1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11, 15,15,16,17,25])

You need the Maximum contiguous sum(Optimal Substructure):
def msum2(a):
bounds, s, t, j = (0,0), -float('infinity'), 0, 0
for i in range(len(a)):
t = t + a[i]
if t > s: bounds, s = (j, i+1), t
if t < 0: t, j = 0, i+1
return (s, bounds)
This is an example of dynamic programming and is O(N)

O(n) solution works even if the sequence does not start from the first element.
Warning does not work if len(A) = 0.
A = [1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11]
def pre_process(A):
Last = {}
Arrow = []
Length = []
ArgMax = 0
Max = 0
for i in xrange(len(A)):
Arrow.append(i)
Length.append(0)
if A[i] - 1 in Last:
Aux = Last[A[i] - 1]
Arrow[i] = Aux
Length[i] = Length[Aux] + 1
Last[A[i]] = i
if Length[i] > Max:
ArgMax = i
Max = Length[i]
return (Arrow,ArgMax)
(Arr,Start) = pre_process(A)
Old = Arr[Start]
ToRev = []
while 1:
ToRev.append(A[Start])
if Old == Start:
break
Start = Old
New = Arr[Start]
Old = New
ToRev.reverse()
print ToRev
Pythonizations are welcome!!

Ok, here's yet another attempt in python:
def popper(l):
listHolders = []
pos = 0
while l:
appended = False
item = l.pop()
for holder in listHolders:
if item == holder[-1][0]-1:
appended = True
holder.append((item, pos))
if not appended:
pos += 1
listHolders.append([(item, pos)])
longest = []
for holder in listHolders:
try:
if (holder[0][0] < longest[-1][0]) and (holder[0][1] > longest[-1][1]):
longest.extend(holder)
except:
pass
if len(holder) > len(longest):
longest = holder
longest.reverse()
return [x[0] for x in longest]
Sample inputs and outputs:
>>> demo = list(range(50))
>>> shuffle(demo)
>>> demo
[40, 19, 24, 5, 48, 36, 23, 43, 14, 35, 18, 21, 11, 7, 34, 16, 38, 25, 46, 27, 26, 29, 41, 8, 31, 1, 33, 2, 13, 6, 44, 22, 17,
12, 39, 9, 49, 3, 42, 37, 30, 10, 47, 20, 4, 0, 28, 32, 45, 15]
>>> popper(demo)
[1, 2, 3, 4]
>>> demo = [1,4,2,3,5,4,5,6,7,8,1,3,4,5,9,10,11]
>>> popper(demo)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
>>>

This should do the trick (and is O(n)):
target = 1
result = []
for x in list:
for y in result:
if y[0] == target:
y[0] += 1
result.append(x)
For any starting number, this works:
result = []
for x in mylist:
matched = False
for y in result:
if y[0] == x:
matched = True
y[0] += 1
y.append(x)
if not matched:
result.append([x+1, x])
return max(result, key=len)[1:]

Permutations in python 2.5.2

I have a list of numbers for input, e.g.
671.00
1,636.00
436.00
9,224.00
and I want to generate all possible sums with a way to id it for output, e.g.:
671.00 + 1,636.00 = 2,307.00
671.00 + 436.00 = 1,107.00
671.00 + 9,224.00 = 9,224.00
671.00 + 1,636.00 + 436.00 = 2,743.00
...
and I would like to do it in Python
My current constrains are:
a) I'm just learning python now (that's part of the idea)
b) I will have to use Python 2.5.2 (no intertools)
I think I have found a piece of code that may help:
def all_perms(str):
if len(str) <=1:
yield str
else:
for perm in all_perms(str[1:]):
for i in range(len(perm)+1):
#nb str[0:1] works in both string and list contexts
yield perm[:i] + str[0:1] + perm[i:]
( from these guys )
But I'm not sure how to use it in my propose.
Could someone trow some tips and pieces of code of help?
cheers,
f.

Permutations are about taking an ordered set of things and moving these things around (i.e. changing order). Your question is about combinations of things from your list.
Now, an easy way of enumerating combinations is by mapping entries from your list to bits in a number. For example, lets assume that if bit #0 is set (i.e. 1), then number lst[0] participates in the combination, if bit #1 is set, then lst[1] participates in the combination, etc. This way, numbers in range 0 <= n < 2**(len(lst)) identify all possible combinations of lst members, including an empty one (n = 0) and the whole lst (n = 2**(len(lst)) - 1).
You need only combinations of 2 items or more, i.e. only those combination IDs that have at least two nonzero bits in their binary representation. Here is how to identify these:
def HasAtLeastTwoBitsSet(x) :
return (x & (x-1)) != 0
# Testing:
>>> [x for x in range(33) if HasAtLeastTwoBitsSet(x)]
[3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31]
Next step is to extract a combination of list members identified by a combination id. This is easy, thanks to the power of list comprehensions:
def GetSublistByCombination(lst, combination_id) :
res = [x for (i,x) in enumerate(lst) if combination_id & (1 << i)]
return res
# Testing:
>>> GetSublistByCombination([0,1,2,3], 1)
[0]
>>> GetSublistByCombination([0,1,2,3], 3)
[0, 1]
>>> GetSublistByCombination([0,1,2,3], 12)
[2, 3]
>>> GetSublistByCombination([0,1,2,3], 15)
[0, 1, 2, 3]
Now let's make a generator that produces all sums, together with their string representations:
def IterAllSums(lst) :
combinations = [i for i in range(1 << len(lst)) if HasAtLeastTwoBitsSet(i)]
for comb in combinations :
sublist = GetSublistByCombination(lst, comb)
sum_str = '+'.join(map(str, sublist))
sum_val = sum(sublist)
yield (sum_str, sum_val)
And, finally, let's use it:
>>> for sum_str, sum_val in IterAllSums([1,2,3,4]) : print sum_str, sum_val
1+2 3
1+3 4
2+3 5
1+2+3 6
1+4 5
2+4 6
1+2+4 7
3+4 7
1+3+4 8
2+3+4 9
1+2+3+4 10

The code below generates all "subsets" of a given list (except the empty set), i.e. it returns a list of lists.
def all_sums(l): #assumes that l is non-empty
if len(l)==1:
return ([[l[0]]])
if len(l)==0:
return []
result = []
for i in range(0,len(l)):
result.append([l[i]])
for p in all_sums(l[i+1:]):
result.append([l[i]]+p)
return result
Now you could just write a short function doit for output also:
def doit(l):
mylist = all_sums(l)
print mylist
for i in mylist:
print str(i) + " = " + str(sum(i))
doit([1,2,3,4])

With itertools (Python >=2.6) would be:
from itertools import *
a=[1,2,3,4]
sumVal=[tuple(imap(sum,combinations(a,i))) for i in range(2,len(a)+1)]