Find the number of 6 digit numbers such that each digit appears atleast twice
I tried the below code but it doesn't work:
count = 0
for n in range(10**5,(10**6)-1):
n = str(n).split()
for i in range(len(n)):
n[i] = int(n[i])
if n.count(i) >= 2:
count+=1
print(count)
The original question is of Permutation and Combinations but I want to solve this using python...
Here are some of the mistakes in your code -
The range needs to be range(10**5, 10**6) to cover all 6 digit numbers
str(n).split() won't split the n into a list of characters since there is no delimiter between the string representation of the number. You can use [letter for letter in str(n)] instead.
You want to run the second loop for each of the numbers in the first loop, so the second loop needs to be inside the first loop.
Within the second loop, you are increasing the count if one of the digits occurs more than twice in the number, so you might end up increasing the count multiple times for a single number. e.g. for 112233 you would end up increasing the count by 3 instead of just 1.
here's a simplified code which does what you want in a similar format -
count = 0
for n in range(10**5, 10**6):
n = [letter for letter in str(n)]
count += 1
for letter in n:
if n.count(letter) < 2:
count -= 1
break
print(count)
outputs -
11754
You can use all() function along with list comprehension to simplify the code -
count = 0
for n in range(10**5, 10**6):
n = [letter for letter in str(n)]
count += all(n.count(letter) >= 2 for letter in n)
answers = []
count = 0
my_dict = {}
digits = '0123456789'
for n in range(10**5, (10**6)-1):
my_dict = {}
ns = str(n)
for m in ns:
if m in digits:
my_dict[m] = ns.count(m)
if my_dict[min(my_dict, key=my_dict.get)] > 1:
count += 1
answers.append(ns)
print(count)
Here is an alternative way. It gets the correct number.
Related
How do you write a program to count up and down from a users integer input, using only one while loop and no for or if statements?
I have been successful in counting up but can't seem to figure out how to count down. It needs to look like a sideways triangle also, with spaces before the number (equal to the number being printed).
this is my code so far;
n = int(input("Enter an integer: "))
i = " "
lines = 0
while lines <= n-1:
print(i * lines + str(lines))
lines += 1
You can solve this by making good use of negative numbers and the absolute value: abs() which will allow you to "switch directions" as the initial number passes zero:
n = int(input("Enter an integer: "))
s = " "
i = n * -1
while i <= n:
num = n - abs(i)
print(s * num + str(num))
i += 1
This will produce:
Enter an integer: 3
0
1
2
3
2
1
0
Move the print line to after the while loop. Form a list of lists (or an array) in the while loop using the content from your current print statement. Then create a second statement in your single while loop to count down, which creates a second list of lists. Now you can sequentially print your two lists of lists.
I'm stuck trying to figure out this while loop. It seemed simple at first but I keep running into errors. I think I just haven't used python in a while so it's hard to get back into the gist of things.
Write a while loop that will add 100 numbers to a list, starting from the number 100 and incrementing by 2 each time. For example, start from 100, then the next number added will be 102, then 104 and so on.
I have this so far;
count = 0
while count < 99:
count += 1
numbers.append(len(numbers)+2)
for x in numbers:
print(x)
But that is simply giving me 0-100 printed out when I want it to be 2,4,6,8,10,...
numbers = []
index = 0
number = 100
while index < 100:
number += 2
index += 1
numbers.append(number)
for x in numbers:
print(x)
With a few modifications, and using a while loop as per your exercise:
numbers = []
count = 0
while count < 100:
numbers.append(100 + (2*count))
count += 1
for x in numbers:
print(x)
Or with a for loop:
numbers = []
for i in range(100):
numbers.append(100 + (2*i))
for x in numbers:
print(x)
Or as a list comprehension:
numbers.extend([(100 + (2*el)) for el in range(100)])
Or as a recursion:
numbers = []
def rec(num):
if num < 100:
numbers.append(100 + (2*num))
rec(num + 1)
rec(0)
Something like this:
numbers = []
while len(numbers) != 100:
if len(numbers) > 0:
numbers.append(numbers[-1] + 2)
else:
numbers.append(100)
Little explanation:
You loop until your list has 100 elements. If list is empty (which will be at the beginning), add first number (100, in our case). After that, just add last number in list increased by 2.
Try numbers.extend(range(100, 300, 2)). It's a much shorter way to do exactly what you're looking for.
Edit: For the people downvoting, can you at least leave a comment? My initial response was prior to the question being clarified (that a while loop was required). I'm giving the pythonic way of doing it, as I was unaware of the requirement.
for homework in an introductory python class, one of the questions is to count the number of even numbers in n. here's my code so far:
def num_even_digits(n):
i=0
count = 0
while i < n:
i+=1
if n%2==0:
count += 1
return count
print(num_even_digits(123456))
Pythonic answer:
def num_even_digits(x):
return len([ y for y in str(x) if int(y) % 2 == 0])
print(num_even_digits(123456))
Disclaimer: I recognized that, for an introductory Python class, my answer may not be appropriate.
you are comparing the whole number every time. you need to convert it to a string and then loop over every number in that string of numbers, cast it back to an integer and see if its remainder is 0
def num_even_digits(numbers):
count = 0
numbers = str(numbers)
for number in numbers:
try:
number = int(number)
except ValueError:
continue
if number % 2 == 0:
count += 1
return count
print(num_even_digits(123456))
if you want to actually loop through every possible number in the range of 0 to your large number you can do this.
def num_even_digits(numbers):
count = 0
for number in range(0, numbers):
if number % 2 == 0:
count += 1
return count
print(num_even_digits(10))
problems with your current function:
def num_even_digits(n): # n is not descriptive, try to make your variable names understandable
i=0
count = 0
while i < n: # looping over every number from 0 to one hundred twenty three thousand four hundred and fifty six.
i+=1
if n%2==0: # n hasn't changed so this is always going to be true
count += 1
return count
print(num_even_digits(123456))
How do I get my num_digit function to return 1 instead of 0 whenever I put in 0 as the parameter in the function?
How do I get the function to return any integer such as negative numbers?
def num_digits(n):
count = 0
while n:
count = count + 1
n = abs(n) / 10
return count
I was working on question number 2 first. Even if I put the abs(n) in the line of code where while is, I still get an infinite loop which I still do not really understand why. I figured if I can get my n value to always be positive and input in say -24, it would convert it to 24 and still count the number of values.
On question 1, I do not know where to start, ive tried:
def num_digits(n):
count = 0
while n:
if n == 0:
count = count + 1
n = n / 10
return count
I forgot to add I have limited tools to use since I am still learning python. I have gotten up to iterations and am studying the while loops and counters. I have not gotten to break yet although I have an idea of what it does.
When in doubt, brute force is always available:
def num_digits(n):
if n == 0:
return 1
if n < 0:
return num_digits(abs(n))
count = 0
while n:
count = count + 1
n = n / 10
return count
Process the exceptional cases first, and then you only have to deal with the regular ones.
If you want to avoid the conditionals, I suggest taking abs(n) only once, at the beginning, and using an infinite loop + break for the 0 case:
def num_digits(n):
n = abs(n)
count = 0
while True:
count = count + 1
n = n / 10
if n == 0:
break
return count
For a more practical solution, you can either count the number of digits in the string (something like len(str(n)) for positive integers) or taking log base 10, which is a mathematical way of counting digits.
def num_digits(n):
if n == 0:
return 0
return math.floor(math.log10(math.abs(n)))
I want to write a program that can calculate the sum of an integer as well as count its digits . It will keep doing this until it becomes a one digit number.
For example, if I input 453 then its sum will be 12 and digit 3.
Then it will calculate the sum of 12=1+2=3 it will keep doing this until it becomes one digit. I did the first part but i could not able to run it continuously using While . any help will be appreciated.
def main():
Sum = 0
m = 0
n = input("Please enter an interger: ")
numList = list(n)
count = len(numList)
for i in numList:
m = int(i)
Sum = m+Sum
print(Sum)
print(count)
main()
It is not the most efficient way, but it doesn't matter much here; to me, this is a problem to elegantly solve by recursion :)
def sum_digits(n):
n = str(n)
if int(n) < 10:
return n
else:
count = 0
for c in n:
count += int(c)
return sum_digits(count)
print sum_digits(123456789) # --> 9 # a string
A little harder to read:
def sum_digits2(n):
if n < 10:
return n
else:
return sum_digits2(sum(int(c) for c in str(n))) # this one returns an int
There are a couple of tricky things to watch out for. Hopefully this code gets you going in the right direction. You need to have a conditional for while on the number of digits remaining in your sum. The other thing is that you need to covert back and forth between strings and ints. I have fixed the while loop here, but the string <-> int problem remains. Good luck!
def main():
count = 9999
Sum = 0
m = 0
n = input("Please enter an integer: ")
numList = list(n)
while count > 1:
count = len(numList)
for i in numList:
m = int(i)
Sum = m+Sum
print(Sum)
print(count)
# The following needs to be filled in.
numlist = ???
main()
You can do this without repeated string parsing:
import math
x = 105 # or get from int(input(...))
count = 1 + int(math.log10(x))
while x >= 10:
sum = 0
for i in xrange(count):
sum += x % 10
x /= 10
x = sum
At the end, x will be a single-digit number as described, and count is the number of original digits.
I would like to give credit to this stackoverflow question for a succinct way to sum up digits of a number, and the answers above for giving you some insight to the solution.
Here is the code I wrote, with functions and all. Ideally you should be able to reuse functions, and here the function digit_sum(input_number) is being used over and over until the size of the return value (ie: length, if sum_of_digits is read as a string) is 1. Now you can use the while loop to keep checking till the size is what you want, and then abort.
def digit_sum(input_number):
return sum(int(digit) for digit in str(input_number))
input_number = input("Please enter a number: ")
sum_of_digits = digit_sum(input_number)
while(len(str(sum_of_digits)) > 1):
sum_of_digits = digit_sum(input_number)
output = 'Sum of digits of ' + str(input_number) + ' is ' + str(sum_of_digits)
print output
input_number = sum_of_digits
this is using recursive functions
def sumo(n):
sumof = 0
while n > 0:
r = n%10 #last digit
n = n/10 # quotient
sumof += r #add to sum
if sumof/10 == 0: # if no of digits in sum is only 1, then return
return sumof
elif sumof/10 > 0: #else call the function on the sumof
sumo(sumof)
Probably the first temptation would be to write
while x > 9:
x = sum(map(int, str(x)))
that literally means "until there is only one digit replace x by the sum of its digits".
From a performance point of view however one should note that computing the digits of a number is a complex operation because Python (and computers in general) store numbers in binary form and each digit in theory requires a modulo 10 operation to be extracted.
Thus if the input is not a string to begin with you can reduce the number of computations noting that if we're interested in the final sum (and not in the result of intermediate passes) it doesn't really matter the order in which the digits are summed, therefore one could compute the result directly, without converting the number to string first and at each "pass"
while x > 9:
x = x // 10 + x % 10
this costs, from a mathematical point of view, about the same of just converting a number to string.
Moreover instead of working out just one digit however one could also works in bigger chunks, still using maths and not doing the conversion to string, for example with
while x > 99999999:
x = x // 100000000 + x % 100000000
while x > 9999:
x = x // 10000 + x % 10000
while x > 99:
x = x // 100 + x % 100
while x > 9:
x = x // 10 + x % 10
The first loop works 8 digits at a time, the second 4 at a time, the third two and the last works one digit at a time. Also it could make sense to convert the intermediate levels to if instead of while because most often after processing n digits at a time the result will have n or less digits, leaving while loops only for first and last phases.
Note that however the computation at this point is so fast that Python general overhead becomes the most important part and thus not much more can be gained.
You could define a function to find the sum and keep updating the argument to be the most recent sum until you hit one digit.
def splitSum(num):
Sum = 0
for i in str(num):
m = int(i)
Sum = Sum + m
return str(Sum)
n = input("Please enter an integer: ")
count = 0
while count != 1:
Sum = splitSum(n)
count = len(Sum)
print(Sum)
print(count)
n = Sum