2025-11-13 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 13

Description

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 = "000110".

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 = "0011101".
  • Choose index i = 4. The resulting string is s = "0011011".
  • Choose index i = 3. The resulting string is s = "0010111".
  • Choose index i = 2. The resulting string is s = "0001111".

Example 2:

Input: s = "00111"

Output: 0

 

Constraints:

  • 1 <= s.length <= 105
  • s[i] is either '0' or '1'.

Solution

class Solution {
public:
  int maxOperations(string s) {
    int ones = 0;
    bool prevZero = false;
    int answer = 0;
    for(auto n : s) {
      if(n == '0') {
        prevZero = true;
      } else {
        if(prevZero) answer += ones;
        prevZero = false;
        ones += 1;
      }
    }
    if(prevZero) answer += ones;
    return answer;
  }
};

// Accepted
// 694/694 cases passed (11 ms)
// Your runtime beats 14.13 % of cpp submissions
// Your memory usage beats 53.99 % of cpp submissions (15.9 MB)
Algorithm