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
ins
to convert.
Constraints:
3 <= s.length <= 1000
s[i]
is either'X'
or'O'
.
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
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;
}
};
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
}
}
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;
}