What is the time complexity of Pancake Sorting?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Pancake Sorting?

This problem uses Array, Two Pointers, Greedy, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Pancake Sorting?

Insight 1: The key is to iteratively place the largest unsorted element at its correct position (the end of the unsorted portion) in each step. Insight 2: To place the largest unsorted element 'x' at its correct position, two flips are needed: first, flip to bring 'x' to the front of the array, and then flip to place it at its correct sorted position.

Which companies ask LeetCode 969. Pancake Sorting in interviews?

This problem is frequently asked at Block.

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Pancake Sorting - LeetCode 969 Solution

Q: How to solve Pancake Sorting?

The provided solution uses a greedy approach to iteratively place the largest unsorted element at its correct position in the array. It iterates from `n` down to 1, where `n` is the size of the array. In each iteration, it finds the index of the current largest element 'x'. If 'x' is not already at its correct position, it performs two flips. First, it flips the array up to the index of 'x' to bring 'x' to the front. Then, it flips the array up to the correct position of 'x' to place 'x' at the end of the sorted portion.

Pancake Sorting - Complete Solution Guide

Pancake Sorting is LeetCode problem 969, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 a

Detailed Explanation

The Pancake Sorting problem requires sorting an array of unique integers from 1 to n using only 'pancake flips'. A pancake flip involves selecting an index `k` (1 <= k <= array length) and reversing the subarray from index 0 up to index `k-1`. The goal is to find a sequence of `k` values that, when used to perform pancake flips, sorts the array. The problem specifies that any valid answer that sorts the array within 10 * arr.length flips is acceptable.

Solution Approach

The provided solution uses a greedy approach to iteratively place the largest unsorted element at its correct position in the array. It iterates from `n` down to 1, where `n` is the size of the array. In each iteration, it finds the index of the current largest element 'x'. If 'x' is not already at its correct position, it performs two flips. First, it flips the array up to the index of 'x' to bring 'x' to the front. Then, it flips the array up to the correct position of 'x' to place 'x' at the end of the sorted portion.

Step-by-Step Algorithm

Step 1: Iterate from `x = n` down to 2, where `n` is the length of the array. 'x' represents the largest unsorted element to be placed in its correct position.
Step 2: Find the index `idx` of the element `x` in the current array.
Step 3: If `idx` is already at its correct position (`idx == x - 1`), no flip is needed for this element. Continue to the next iteration.
Step 4: If `idx` is not 0, it means `x` is not at the beginning of the array. Flip the array from 0 to `idx` to bring `x` to the beginning of the array. Add `idx + 1` to the result list.
Step 5: Flip the array from 0 to `x - 1` to place `x` at its correct sorted position (end of the sorted portion). Add `x` to the result list.
Step 6: Repeat steps 2-5 for each value of `x` from `n` down to 2.
Step 7: Return the list of flip values generated.

Key Insights

Insight 1: The key is to iteratively place the largest unsorted element at its correct position (the end of the unsorted portion) in each step.
Insight 2: To place the largest unsorted element 'x' at its correct position, two flips are needed: first, flip to bring 'x' to the front of the array, and then flip to place it at its correct sorted position.
Insight 3: The input array contains unique numbers from 1 to n, where n is the length of the array. This simplifies the search for the index of the number 'x'.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers, Greedy, Sorting.

Companies

Asked at: Block.

Pancake Sorting | LeetCode 969

Medium

Array Two Pointers Greedy Sorting

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<Integer> pancakeSort(int[] arr) {
        int n = arr.length;
        List<Integer> result = new ArrayList<>();

        for (int x = n; x > 1; x--) {
            int idx = -1;
            for (int i = 0; i < x; i++) {
                if (arr[i] == x) {
                    idx = i;
                    break;
                }
            }

            if (idx == x - 1) {
                continue;
            }

            if (idx != 0) {
                result.add(idx + 1);
                // Reverse arr[0...idx]
                reverse(arr, 0, idx);
            }

            result.add(x);
            // Reverse arr[0...x-1]
            reverse(arr, 0, x - 1);
        }

        return result;
    }

    private void reverse(int[] arr, int start, int end) {
        while (start < end) {
            int temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }
}

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Problem Info

Description

Problem 969: Pancake Sorting - Detailed explanation and solution approaches.

Topics

Array Two Pointers Greedy Sorting

Companies

Block

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:969

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Iterate from `x = n` down to 2, where `n` is the length of the array. 'x' represents the largest unsorted element to be placed in its correct position.
2
Step 2: Find the index `idx` of the element `x` in the current array.
3
Step 3: If `idx` is already at its correct position (`idx == x - 1`), no flip is needed for this element. Continue to the next iteration.
4
Step 4: If `idx` is not 0, it means `x` is not at the beginning of the array. Flip the array from 0 to `idx` to bring `x` to the beginning of the array. Add `idx + 1` to the result list.
5
Step 5: Flip the array from 0 to `x - 1` to place `x` at its correct sorted position (end of the sorted portion). Add `x` to the result list.
6
Step 6: Repeat steps 2-5 for each value of `x` from `n` down to 2.
7
Step 7: Return the list of flip values generated.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n^2) because the dominant operations are searching for the element's index and reversing the subarray. The outer loop runs n-1 times, and within each iteration, we have a linear search (O(n)) and two reversals, each also O(n). The space complexity is O(n) due to the storage of the pancake flip sequence in the `result` list. The number of pancake flips needed can grow linearly with the size of the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a helper data structure for indexing

Create an auxiliary array or hashmap that maps each value in `arr` to its index. This can make the index lookup O(1) instead of O(n).

Time: O(n log n) on average, O(n^2) in worst case. (Due to potentially needing to search the structure)Space: O(n) to store the auxiliary data structure.

Trade-offs: Trades off space for potentially better time complexity if searching were the bottleneck of the algorithm, but pancake sorting inherently involves reversals, thus not substantially improving overall performance, but making code more complex.

✨

Key Insights

1Insight 1: The key is to iteratively place the largest unsorted element at its correct position (the end of the unsorted portion) in each step.
2Insight 2: To place the largest unsorted element 'x' at its correct position, two flips are needed: first, flip to bring 'x' to the front of the array, and then flip to place it at its correct sorted position.
3Insight 3: The input array contains unique numbers from 1 to n, where n is the length of the array. This simplifies the search for the index of the number 'x'.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty array. The code correctly handles an empty array because the loops won't execute, and an empty `result` list will be returned.
Edge case 2: Already sorted array. The code correctly handles this case, as the inner loop will quickly determine that the largest element is in its right spot, thus the algorithm continues.
Edge case 3: Array with only one element. The algorithm correctly skips flipping since the outer loop condition (x > 1) prevents execution. An empty list is returned, which is a valid output.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the reverse operation. The reverse function must correctly reverse the elements within the specified range.

✗Common mistake 2: Failing to account for the 0-based indexing in the array. Adding 1 to the index is crucial when reporting the k-values for the flips.

✗Common mistake 3: Not considering the case where the element 'x' is already in its correct position and needlessly performing flips, which increases the number of flips and can lead to exceeding the maximum number of allowed flips.

✗Common mistake 4: Forgetting to update the array after each reverse operation, leading to incorrect results.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing the movement of physical objects on a conveyor belt where reversing sections of the belt is the only way to reorder the objects. Although direct application is rare, the concept of minimizing reversals could be relevant.

🏦Real-world scenario 2: This specific algorithm doesn't have direct real-world applications. However, the core idea of sorting with limited operations can be extended to optimization problems where certain operations (like reversals) are cheaper or more efficient than others.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Pancake Sorting"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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