carloscn
commented
1 year ago Analysis
pub fn minimum_moves(s: &str) -> i32
{
let mut ret:i32 = 0;
if s.len() < 1 {
return ret;
}
let s_v:Vec<char> = s.chars().collect();
let mut count:usize = 0;
for i in 0..s_v.len() {
if s_v[i] != 'X' {
count = 0;
continue;
} else {
if count == 0 {
ret += 1;
}
count += 1;
}
}
return ret;
}
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'.