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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0901_1000.s0969_pancake_sorting;

// #Medium #Array #Sorting #Greedy #Two_Pointers
// #2022_03_31_Time_1_ms_(100.00%)_Space_42.2_MB_(88.99%)

import java.util.ArrayList;
import java.util.List;

/**
 * 969 - Pancake Sorting\.
 *
 * Medium
 *
 * Given an array of integers `arr`, sort the array by performing a series of **pancake flips**.
 *
 * In one pancake flip we do the following steps:
 *
 * *   Choose an integer `k` where `1 <= k <= arr.length`.
 * *   Reverse the sub-array `arr[0...k-1]` ( **0-indexed** ).
 *
 * For example, if `arr = [3,2,1,4]` and we performed a pancake flip choosing `k = 3`, we reverse the sub-array `[3,2,1]`, so `arr = [1,2,3,4]` after the pancake flip at `k = 3`.
 *
 * Return _an array of the_ `k`_\-values corresponding to a sequence of pancake flips that sort_ `arr`. Any valid answer that sorts the array within `10 * arr.length` flips will be judged as correct.
 *
 * **Example 1:**
 *
 * **Input:** arr = [3,2,4,1]
 *
 * **Output:** [4,2,4,3]
 *
 * **Explanation:**
 *
 * We perform 4 pancake flips, with k values 4, 2, 4, and 3.
 *
 * Starting state: arr = [3, 2, 4, 1]
 *
 * After 1st flip (k = 4): arr = [1, 4, 2, 3]
 *
 * After 2nd flip (k = 2): arr = [4, 1, 2, 3]
 *
 * After 3rd flip (k = 4): arr = [3, 2, 1, 4]
 *
 * After 4th flip (k = 3): arr = [1, 2, 3, 4], which is sorted.
 *
 * **Example 2:**
 *
 * **Input:** arr = [1,2,3]
 *
 * **Output:** []
 *
 * **Explanation:** The input is already sorted, so there is no need to flip anything.
 *
 * Note that other answers, such as [3, 3], would also be accepted.
 *
 * **Constraints:**
 *
 * *   `1 <= arr.length <= 100`
 * *   `1 <= arr[i] <= arr.length`
 * *   All integers in `arr` are unique (i.e. `arr` is a permutation of the integers from `1` to `arr.length`).
**/
public class Solution {
    public List pancakeSort(int[] arr) {
        List result = new ArrayList<>();
        for (int i = arr.length; i >= 1; i--) {
            int max = Integer.MIN_VALUE;
            int index = 0;
            for (int j = 0; j < i; j++) {
                if (max < arr[j]) {
                    index = j + 1;
                    max = arr[j];
                }
            }
            result.add(index);
            reverse(arr, index - 1);
            result.add(i);
            reverse(arr, i - 1);
        }
        return result;
    }

    private void reverse(int[] arr, int index) {
        for (int i = 0; i <= (index - 1) / 2; i++) {
            int temp = arr[i];
            arr[i] = arr[index - i];
            arr[index - i] = temp;
        }
    }
}