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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.

Example 1:

Input: s = "1001101"

Output: 4

Explanation:

We can perform the following operations:

  • Choose index i = 0. The resulting string is s = "**001**1101".
  • Choose index i = 4. The resulting string is s = "0011**01**1".
  • Choose index i = 3. The resulting string is s = "001**01**11".
  • Choose index i = 2. The resulting string is s = "00**01**111".

Example 2:

Input: s = "00111"

Output: 0

Constraints:

  • 1 <= s.length <= 105
  • s[i] is either '0' or '1'.
class Solution {
    public int maxOperations(String s) {
        int n = s.length(), cnt = 0, ans = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '0') {
                ans += cnt;
                while (i < n && s.charAt(i) != '1') {
                    i++;
                }
            }
            cnt++;
        }
        return ans;
    }
}