# Solving Leetcode Interviews in Seconds with AI: Minimum Moves to Convert String


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2027" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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'.  

	# Explanation
	Here's a breakdown of the solution and the Python code:

*   **High-Level Approach:**

    *   Iterate through the string `s`.
    *   If we find an 'X', apply a move by converting the current character and the next two characters to 'O', and increment the move count. Advance the index by 3 to reflect the move.
    *   If we find an 'O', simply advance the index to the next character.

*   **Complexity Analysis:**

    *   Runtime Complexity: O(n), where n is the length of the string `s`. We iterate through the string at most once.
    *   Storage Complexity: O(1). We use a constant amount of extra space.

	
	# Code
	```python
	def minimumMoves(s: str) -> int:
    """
    Calculates the minimum number of moves to convert all characters in a string to 'O'.

    Args:
        s: The input string consisting of 'X' and 'O' characters.

    Returns:
        The minimum number of moves required.
    """
    n = len(s)
    moves = 0
    i = 0
    while i < n:
        if s[i] == 'X':
            moves += 1
            i += 3
        else:
            i += 1
    return moves
	```