README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2000-2099/2027.Minimum%20Moves%20to%20Convert%20String/README_EN.md	1346	Weekly Contest 261 Q1	Greedy String

2027. Minimum Moves to Convert String

中文文档

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 $s$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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

TypeScript

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;
}

Rust

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
    }
}

C

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;
}