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 = [`

after the pancake flip at __1__,__2__,__3__,4]`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`

).

class Solution {
public:
vector<int> pancakeSort(vector<int>& arr) {
}
};