Pancake Sorting

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:

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 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.
Notice that we return an array of the chosen k values of the pancake flips.

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:


Solution:

class Solution {
    public List<Integer> pancakeSort(int[] arr) {
        List<Integer> result = new ArrayList();
        for (int largest = arr.length; largest > 0; largest --) {
            int index = find(arr, largest);
            reverse(arr, index);
            reverse(arr, largest - 1);
            result.add(index + 1);
            result.add(largest);
        }
        return result;
    }
    
    private int find(int[] arr, int target) {
        for (int i = 0; i < arr.length; i ++) {
            if (arr[i] == target) {
                return i;
            }
        }
        return -1;
    }
    
    private void reverse(int[] arr, int right) {
        int left = 0;
        while (left < right) {
            int temp = arr[left];
            arr[left ++] = arr[right];
            arr[right --] = temp;
        }
    }
}