Formatted question description: https://leetcode.ca/all/2027.html

2027. Minimum Moves to Convert String (Easy)

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'.

Solution 1.

• 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
}
}