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2027. Minimum Moves to Convert String

Description

You are given a string s consisting of n characters which are either 'X' or 'O'.

A move is defined as selecting three consecutive characters of s and converting them to 'O'. Note that if a move is applied to the character 'O', it will stay the same.

Return the minimum number of moves required so that all the characters of s are converted to 'O'.

 

Example 1:

Input: s = "XXX"
Output: 1
Explanation: XXX -> OOO
We select all the 3 characters and convert them in one move.

Example 2:

Input: s = "XXOX"
Output: 2
Explanation: XXOX -> OOOX -> OOOO
We select the first 3 characters in the first move, and convert them to 'O'.
Then we select the last 3 characters and convert them so that the final string contains all 'O's.

Example 3:

Input: s = "OOOO"
Output: 0
Explanation: There are no 'X's in s to convert.

 

Constraints:

• 3 <= s.length <= 1000
• s[i] is either 'X' or 'O'.

Solutions

Solution 1: Greedy Algorithm

Traverse the string $s$. Whenever you encounter 'X', move the pointer $i$ three steps forward and add $1$ to the answer; otherwise, move the pointer $i$ one step forward.

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

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

• class Solution {
      public int minimumMoves(String s) {
          int ans = 0;
          for (int i = 0; i < s.length(); ++i) {
              if (s.charAt(i) == 'X') {
                  ++ans;
                  i += 2;
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int minimumMoves(string s) {
          int ans = 0;
          for (int i = 0; i < s.size(); ++i) {
              if (s[i] == 'X') {
                  ++ans;
                  i += 2;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def minimumMoves(self, s: str) -> int:
          ans = i = 0
          while i < len(s):
              if s[i] == "X":
                  ans += 1
                  i += 3
              else:
                  i += 1
          return ans
  
  

• func minimumMoves(s string) (ans int) {
  	for i := 0; i < len(s); i++ {
  		if s[i] == 'X' {
  			ans++
  			i += 2
  		}
  	}
  	return
  }
  

• function minimumMoves(s: string): number {
      const n = s.length;
      let ans = 0;
      let i = 0;
      while (i < n) {
          if (s[i] === 'X') {
              ans++;
              i += 3;
          } else {
              i++;
          }
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn minimum_moves(s: String) -> i32 {
          let s = s.as_bytes();
          let n = s.len();
          let mut ans = 0;
          let mut i = 0;
          while i < n {
              if s[i] == b'X' {
                  ans += 1;
                  i += 3;
              } else {
                  i += 1;
              }
          }
          ans
      }
  }
  
  

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