# 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.
```

Example 2:

```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);

}
}

return ans;
}
}
``````