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Pancake Sorting

Updated on 09 July, 2025
Pancake Sorting header image

Problem Statement

The objective is to sort an integer array named arr through a series of operations named "pancake flips". A pancake flip is defined as reversing a sub-array within arr. Specifically, a sub-array from the starting index 0 up to an index k-1, where k is chosen such that 1 <= k <= arr.length. The choice of k determines which portion of the array gets reversed during each operation. For instance, if the original array is [3,2,1,4] and if k is chosen to be 3 for a flip, the sub-array [3,2,1] gets reversed to become [1,2,3], and thus the array looks like [1,2,3,4] after the flip. The task is to generate a sequence of such k values for the flips which entirely sorts the array. While the total number of flips must be within 10 * arr.length, varied sequences that result in a sorted array are acceptable.

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).

Approach and Intuition

Understanding the Problem through Examples:

Considering the problem through examples is helpful to illustrate how pancake flips operate.

  1. First Example

    Given the array:

    arr = [3,2,4,1]

    Desired output:

    [4,2,4,3]

    Progression:

    • Start: arr = [3,2,4,1]
    • 1st flip at k=4: Reverse entire array: arr = [1,4,2,3]
    • 2nd flip at k=2: Reverse first two elements: arr = [4,1,2,3]
    • 3rd flip at k=4: Reverse entire array again: arr = [3,2,1,4]
    • 4th flip at k=3: Reverse first three elements: arr = [1,2,3,4] (Sorted)

  2. Second Example

    Given the array:

    arr = [1,2,3]

    Since the array is already sorted, no flips are needed:

    []

Step by Step Strategy:

This problem can be solved by systematically placing the largest unsorted element in its correct position using a series of flips.

  1. Identify the position of the largest element in the unsorted section of the array.
  2. Flip the array from the start up to this position, bringing the largest element to the front.
  3. Flip the entire unsorted section to place this element at its final position.
  4. Reduce the size of the considered unsorted section and repeat the above steps until the whole array is sorted.

Considerations Based on Constraints:

  • The array length limit (1 to 100) and its unique integer nature suggest that an algorithm with a somewhat higher time complexity than linear might still be acceptable.
  • The approach should preferably sort the array within 10 * arr.length flips, providing plenty of room to use a pancake flipping strategy efficiently for arrays of this size.
  • Since all values are unique and are a permutation from 1 to arr.length, each flip operation can be precisely calculated without concerns of value duplications or out-of-bound values.

Solutions

  • Java
java 
class Solution {
    public List<Integer> sortPancakes(int[] arr) {
        List<Integer> result = new ArrayList<>();
    
        for (int currentMax = arr.length; currentMax > 0; currentMax--) {
            int position = this.locate(arr, currentMax);
    
            if (position == currentMax - 1)
                continue;
    
            if (position != 0) {
                result.add(position + 1);
                this.pancakeFlip(arr, position + 1);
            }
    
            result.add(currentMax);
            this.pancakeFlip(arr, currentMax);
        }
    
        return result;
    }
    
    protected void pancakeFlip(int[] subset, int k) {
        int i = 0;
        while (i < k / 2) {
            int temp = subset[i];
            subset[i] = subset[k - i - 1];
            subset[k - i - 1] = temp;
            i++;
        }
    }
    
    protected int locate(int[] data, int target) {
        for (int i = 0; i < data.length; i++) {
            if (data[i] == target) {
                return i;
            }
        }
        return -1;
    }
}

This solution implements the pancake sorting algorithm in Java, where the goal is to sort a given array using a sequence of pancake flips. Pancake flips refer to reversing the order of the first k elements in the array. The primary method, sortPancakes, returns a list of integers representing the sequence of flips needed to sort the array.

  • Start by initializing a result list to store the sequence of flips.
  • Iterate over the array from the last element to the first:
    • Locate the position of the maximum unsorted element.
    • If the maximum element is already in the correct position, continue to the next iteration.
    • If the element is not at the first position, add its index to the result list and perform a pancake flip to move it to the start.
    • Add the current size of the unsorted segment to the result list and perform another pancake flip to move the maximum element to its correct position.

The helper methods include:

  • pancakeFlip, which reverses the order of the first k elements in a subset of the array.
  • locate, which searches for a target value in the array and returns its index.

This algorithm effectively uses flips to sort the array in place, minimizing the number of required operations. Each flip operation involves reversing the segment of the array up to a specified index, directly manipulating the array's order with each iteration.

  • Python
python 
class Solution:
    def sortPancakes(self, pancakes: List[int]) -> List[int]:
        def flipStack(stack, num):
            j = 0
            while j < num / 2:
                stack[j], stack[num-j-1] = stack[num-j-1], stack[j]
                j += 1
    
        result = []
        max_position = len(pancakes)
        while max_position > 0:
            target_index = pancakes.index(max_position)
    
            if target_index != max_position - 1:
                if target_index != 0:
                    result.append(target_index + 1)
                    flipStack(pancakes, target_index + 1)
                result.append(max_position)
                flipStack(pancakes, max_position)
    
            max_position -= 1
    
        return result

The solution provided for pancake sorting involves a Python class named Solution with a method sortPancakes. This method defines a key helper function, flipStack, used to reverse the order of elements in the stack up to a specified number of elements. This mimics flipping a stack of pancakes up to a certain depth.

Here's a step-by-step breakdown of how the pancake sorting process is implemented in the solution:

  1. Define the flipStack function that takes in the stack (list of pancakes) and num, the count of top elements to flip. It uses a simple two-pointer technique to reverse the elements up to the num position.

  2. Initialize an empty list result which will store the sequence of flips made during the sorting process.

  3. Set max_position to the length of the list, indicating the yet-to-be-sorted portion of the stack.

  4. Use a while loop to repeatedly find the position of the maximum unsorted pancake (max_position) and make it reach its correct position by potentially performing two flips:

    • If the maximum pancake is not already in place, check if it's not at the first position. If it isn't, append its index (1-based) to result and flip the stack up to this index to bring the pancake to the top.
    • Append max_position to result and flip the top max_position pancakes to place the maximum pancake at the bottom of the sorted section of the stack.

  5. Decrement max_position to sort the next largest pancake in the next iteration.

  6. Return the result, representing the sequence of flip operations needed to sort the pancakes.

This approach focuses on moving the largest unsorted pancake to its appropriate position from top to bottom sequentially, which efficiently sorts the stack using the minimum number of flips required.

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