The minimum number of operations to convert strings in leetcode simple problem

subject

Give you a string s , from n Characters make up , Each character is not ‘X’ Namely ‘O’ .
once operation Defined as from s Selected from Three consecutive characters And convert each selected character to ‘O’ . Be careful , If the character is already ‘O’ , Just keep unchanged .
Return to s All characters in are converted to ‘O’ executable least Operating frequency .
Example 1：
Input ：s = “XXX”
Output ：1
explain ：XXX -> OOO
One operation , Select All 3 Characters , And turn them into ‘O’ .
Example 2：
Input ：s = “XXOX”
Output ：2
explain ：XXOX -> OOOX -> OOOO
The first operation , Choose the former 3 Characters , And convert these characters to ‘O’ .
then , After selection 3 Characters , And perform the conversion . The resulting string consists entirely of characters ‘O’ form .
Example 3：
Input ：s = “OOOO”
Output ：0
explain ：s There is no need to convert ‘X’ .
Tips ：
3 <= s.length <= 1000
s[i] by ‘X’ or ‘O’
source ： Power button （LeetCode）

Their thinking

   Traversal string , Once found ‘X’ Just remember one operation , Then the pointer moves back three positions , If you encounter ‘O’ Just move once .

class Solution:
    def minimumMoves(self, s: str) -> int:
        count=0
        i=0
        while i<len(s):
            if s[i]=='X':
                count+=1
                i+=3
            else:
                i+=1
        return count