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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 $\textit{ans}$ to record the answer and another variable $\textit{cnt}$ to count the current number of $1$s.

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

  • 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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