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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 $s$.

  • 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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