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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 \(s\).

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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

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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;
    }
}

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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;
    }
};

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

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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;
}

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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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int minimumMoves(char* s) {
    int n = strlen(s);
    int ans = 0;
    int i = 0;
    while (i < n) {
        if (s[i] == 'X') {
            ans++;
            i += 3;
        } else {
            i++;
        }
    }
    return ans;
}

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