# Pancake Sorting Problem

## Description

LeetCode Problem 969.

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:

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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:

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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).

## Sample C++ Code

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class Solution {
public:
vector<int> pancakeSort(vector<int>& A) {
vector<int> result;
for (int i = 0; i < A.size(); i++) {
auto maxPosition = max_element(A.begin(), A.end() - i);
result.push_back(maxPosition - A.begin() + 1);
result.push_back(A.size() - i);
reverse(A.begin(), maxPosition+1);
reverse(A.begin(), A.end() - i);
}
return result;
}
};