Find a substring count in an string

Question

This code is written to find substring count in a string in Python 3. The original problem description is here. Please suggest better ways of solving this.

string =input()
substring = input()
count = 0
i=0
while(i<(len(string)-len(substring))):
    if(string.find(substring,i,len(string))>=0):
        count+=1
        i=string.find(substring,i,len(string))+1
    else:
        break;
print(count)

Answer 1

KISS (Keep it simple, stupid)

You could've done this way simpler than that by just writing:

string.count(substring)

Wrapping this into a function for later use would look like:

def count_substrings(my_string, my_substring):
    return my_string.count(my_substring)

if __name__ == '__main__':
    print(count_substrings('dexter is dexter lala dext dexter', 'dexter'))

Which will print the desired result:

3

About your code:

1. You should always have a space around operators (i=0 should be i = 0).
2. Python is not C. You have way too many parentheses and you shouldn't use ;.

There's something called PEP8 which is the recommended style-guide for Python. I'd strongly recommend you read it.

Your code would look like this if we'd respect the above:

string = input()
substring = input()

count, i = 0, 0
while i < len(string) - len(substring):
    if string.find(substring, i, len(string)) >= 0:
        count += 1
        i = string.find(substring, i, len(string)) + 1
    else:
        break
print(count)

More, I'd let the user know what's about to be entered so I'd write:

my_string = input('Please enter the string: ')
my_substring = input('What substring are you looking for? ')

You should also wrap your code within a function for later use, and add the if __name__ == '__main__' construction, which will allow you you to have that code execute only when you want to run the module as a program, and not have it execute when someone just wants to import your module and call your functions themselves

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Anyway, the first solution does not work for overlapping matches (which I saw it was a requirement after reading the challenge), so you might consider doing it like you did, but a bit different / simpler:

def count_substrings(my_string, my_substring):
    count = start = 0
    while True:
        start = my_string.find(my_substring, start) + 1
        if start > 0:
            count += 1
        else:
            return count

The above might be even faster since it does the comparing in C.

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Bear with the fact that I've only commented on your actual code, and not the entire challenge. You might wanna add some constraints which are clearly specified there (e.g: 1 <= len(my_string) <= 200)