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

Description

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'.

Solutions

Solution 1: Greedy

We use a variable $\mathit{\text{ans}}$ to record the answer and another variable $\mathit{\text{cnt}}$ to count the current number of $1$s.

Then, we iterate through the string $s$. If the current character is $1$, then we increment $\mathit{\text{cnt}}$. Otherwise, if there is a previous character and the previous character is $1$, then the previous $\mathit{\text{cnt}}$ number of $1$s can be moved backward, and we add $\mathit{\text{cnt}}$ to the answer.

Finally, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int maxOperations(String s) {
          int ans = 0, cnt = 0;
          int n = s.length();
          for (int i = 0; i < n; ++i) {
              if (s.charAt(i) == '1') {
                  ++cnt;
              } else if (i > 0 && s.charAt(i - 1) == '1') {
                  ans += cnt;
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int maxOperations(string s) {
          int ans = 0, cnt = 0;
          int n = s.size();
          for (int i = 0; i < n; ++i) {
              if (s[i] == '1') {
                  ++cnt;
              } else if (i && s[i - 1] == '1') {
                  ans += cnt;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def maxOperations(self, s: str) -> int:
          ans = cnt = 0
          for i, c in enumerate(s):
              if c == "1":
                  cnt += 1
              elif i and s[i - 1] == "1":
                  ans += cnt
          return ans
  
  

• func maxOperations(s string) (ans int) {
  	cnt := 0
  	for i, c := range s {
  		if c == '1' {
  			cnt++
  		} else if i > 0 && s[i-1] == '1' {
  			ans += cnt
  		}
  	}
  	return
  }
  

• function maxOperations(s: string): number {
      let [ans, cnt] = [0, 0];
      const n = s.length;
      for (let i = 0; i < n; ++i) {
          if (s[i] === '1') {
              ++cnt;
          } else if (i && s[i - 1] === '1') {
              ans += cnt;
          }
      }
      return ans;
  }
  
  

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