Problem statement

https://binarysearch.com/problems/Shortest-String/

Solution

We can notice that as we still have 0 and 1 in string, we have place, where they are adjacent and we can delete them. We stop only if we out of 1 or out of 0.

Complexity

It is O(n) for time and space.

Code

class Solution:
    def solve(self, s):
        return abs(s.count("1") - s.count("0"))