Pancake Sorting

Problem Description

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:

1. Choose an integer k where 1 <= k <= arr.length.
2. 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.

Examples

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).

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Solution for Pancake Sorting Problem

To solve this problem, we perform a series of pancake flips to sort the array in ascending order. A pancake flip reverses the sub-array from the start to a chosen index k. The goal is to bring the largest unsorted element to its correct position iteratively.

Approach

1. Identify Largest Element: Find the largest unsorted element in the array.
2. Bring to Front: If this largest element is not already at the front, flip it to bring it to the front.
3. Move to Correct Position: Flip the entire sub-array up to the correct position of the largest element to place it at its sorted position.
4. Repeat: Repeat the above steps for the next largest unsorted elements, excluding the already sorted part of the array.

Code in Different Languages

C++

Written by @ImmidiSivani

class Solution {
public:
    vector<int> pancakeSort(vector<int>& arr) {
        vector<int> res;
        int n = arr.size();
        for (int i = n; i > 1; --i) {
            int maxIdx = max_element(arr.begin(), arr.begin() + i) - arr.begin();
            if (maxIdx == i - 1) continue;
            if (maxIdx != 0) {
                res.push_back(maxIdx + 1);
                reverse(arr.begin(), arr.begin() + maxIdx + 1);
            }
            res.push_back(i);
            reverse(arr.begin(), arr.begin() + i);
        }
        return res;
    }
};

Java

Written by @ImmidiSivani

class Solution {
    public List<Integer> pancakeSort(int[] arr) {
        List<Integer> res = new ArrayList<>();
        int n = arr.length;
        for (int i = n; i > 1; --i) {
            int maxIdx = findMaxIndex(arr, i);
            if (maxIdx == i - 1) continue;
            if (maxIdx != 0) {
                res.add(maxIdx + 1);
                flip(arr, maxIdx + 1);
            }
            res.add(i);
            flip(arr, i);
        }
        return res;
    }

    private int findMaxIndex(int[] arr, int k) {
        int maxIdx = 0;
        for (int i = 1; i < k; i++) {
            if (arr[i] > arr[maxIdx]) {
                maxIdx = i;
            }
        }
        return maxIdx;
    }

    private void flip(int[] arr, int k) {
        for (int i = 0, j = k - 1; i < j; i++, j--) {
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
    }
}

Python

Written by @ImmidiSivani

class Solution:
    def pancakeSort(self, arr: List[int]) -> List[int]:
        res = []
        n = len(arr)
        for i in range(n, 1, -1):
            max_idx = arr.index(max(arr[:i]))
            if max_idx == i - 1:
                continue
            if max_idx != 0:
                res.append(max_idx + 1)
                arr[:max_idx + 1] = arr[:max_idx + 1][::-1]
            res.append(i)
            arr[:i] = arr[:i][::-1]
        return res

Complexity Analysis

• Time Complexity: $O(n^2)$, where n is the length of the array. Finding the maximum and flipping can each take up to O(n), and we do this for each of the n elements.
• Space Complexity: $O(1)$, excluding the output list.

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Authors:

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