969. Pancake Sorting

Problem

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).

Solution (Java)

class Solution {
    public List<Integer> pancakeSort(int[] arr) {
        List<Integer> 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;
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).