Maximum difference of zeros and ones in binary string - geeksforgeeks

 Given a binary string S consisting of 0s and 1s. The task is to find the maximum difference of the number of 0s and the number of 1s (number of 0s – number of 1s) in the substrings of a string.

Note: In the case of all 1s, the answer will be -1.

Example 1:

Input : S = "11000010001" 
Output : 6 
Explanatio: From index 2 to index 9, 
there are 7 0s and 1 1s, so number 
of 0s - number of 1s is 6. 

Example 2:

Input: S = "111111"
Output: -1
Explanation: S contains 1s only 

Your task:
You do not need to read any input or print anything. The task is to complete the function maxSubstring(), which takes a string as input and returns an integer.

Expected Time Complexity: O(|S|)
Expected Auxiliary Space: O(|S|)

Constraints:
1 ≤ |S| ≤ 105
S contains 0s and 1s only

Solution(Only function):

class Solution {

    int maxSubstring(String S) {

        // code here

        

        int max_until=0;

        int ma=Integer.MIN_VALUE;

        

        for(int i=0;i<S.length();i++){

            

            int x=(S.charAt(i)=='0')?1:-1;

            

            max_until+=x;

            

            if(max_until>ma) ma=max_until;

            

            if(max_until<0) max_until=0;

        }

        return ma;

        

    }

}