969. Pancake Sorting (Medium)

https://leetcode.com/problems/pancake-sorting/

Given an array A, we can perform a pancake flip: We choose some positive integer k <= A.length, then reverse the order of the first k elements of A.  We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array A.

Return the k-values corresponding to a sequence of pancake flips that sort A.  Any valid answer that sorts the array within 10 * A.length flips will be judged as correct.

Example 1:

Input: [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: A = [3, 2, 4, 1]
After 1st flip (k=4): A = [1, 4, 2, 3]
After 2nd flip (k=2): A = [4, 1, 2, 3]
After 3rd flip (k=4): A = [3, 2, 1, 4]
After 4th flip (k=3): A = [1, 2, 3, 4], which is sorted. 

Input: [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.

Note:

1. 1 <= A.length <= 100
2. A[i] is a permutation of [1, 2, ..., A.length]

Solutions

class Solution {

    void flip(int arr[], int idx) {
        int lptr = 0;
        int rptr = idx;

        while (lptr < rptr) {
            int temp = arr[lptr];
            arr[lptr] = arr[rptr];
            arr[rptr] = temp;

            lptr++;
            rptr--;
        }
    }

    int findMaxIndex(int arr[], int idx) {
        int maxIdx = 0;
        int maxVal = 0;
        for (int i = 0; i <= idx; i++) {
            if (arr[i] > maxVal) {
                maxIdx = i;
                maxVal = arr[i];
            }
        }

        return maxIdx;
    }

    public List<Integer> pancakeSort(int arr[]) {
        List<Integer> ans = new ArrayList<>();

        int len = arr.length;

        // Shrink the flipping window from right to left. Every time after flipping, swap
        // window max to the rightmost.
        for (int i = len - 1; i >= 1; i--) {

            // Find index of the maximum element in window [0:i]
            int maxIdx = findMaxIndex(arr, i);

            // Move th maximum to rightmost by flipping the array twice.
            if (maxIdx != i) {
                // Flip the max to the head.
                flip(arr, maxIdx);

                // Reverse the array elements so that the maximum becomes the tail.
                flip(arr, i);

                ans.add(maxIdx+1);
                ans.add(i+1);
            }
        }

        return ans;
    }
}

Incorrect Solutions

References