3228. Maximum Number of Operations to Move Ones to the End

You are given a binary string s.

You can perform the following operation on the string any number of times:

• Choose any index i from the string where i + 1 < s.length such that s[i] == '1' and s[i + 1] == '0'.
• Move the character s[i] to the right until it reaches the end of the string or another '1'. For example, for s = "010010", if we choose i = 1, the resulting string will be s = "0**001**10".

Return the maximum number of operations that you can perform.

class Solution {
public:
    int maxOperations(string& s) {
        int res = 0, o = 0, fl = false;
        s.push_back('1');
        for(auto& ch : s) {
            if(ch == '1') {
                if(fl) res += o;
                o++;
                fl = false;
            } else fl = true;
        }
        return res;
    }
};