g2001_2100.s2027_minimum_moves_to_convert_string.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2001_2100.s2027_minimum_moves_to_convert_string;

// #Easy #String #Greedy #2022_05_25_Time_0_ms_(100.00%)_Space_40_MB_(97.32%)

/**
 * 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'`.
**/
public class Solution {
    public int minimumMoves(String s) {
        int r = 0;
        int i = 0;
        char[] sArray = s.toCharArray();
        while (i < sArray.length) {
            if (sArray[i] == 'X') {
                r++;
                i += 2;
            }
            i++;
        }
        return r;
    }
}