What is the time complexity of Permutations?

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

What data structures are used to solve Permutations?

This problem uses Array, Backtracking. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Permutations?

Insight 1: Backtracking is a suitable approach for generating permutations because it explores different possibilities by making choices and undoing them if they don't lead to a valid solution. Insight 2: Swapping elements in the array is an efficient way to explore different orderings during backtracking. It avoids the need for creating copies of the array repeatedly.

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Permutations - LeetCode 46 Solution

Q: How to solve Permutations?

The provided solutions use a backtracking algorithm to generate all permutations. The core idea is to recursively explore possible arrangements by swapping elements in the input array. The `backtrack` function iterates through the array, swapping the element at the current index with all subsequent elements. After each swap, it recursively calls itself to explore the remaining part of the array. After the recursive call returns, the swap is undone to explore other possibilities. When the recursion reaches the end of the array, it means a complete permutation has been formed, and this permutation is added to the result.

Q: What is the space complexity of Permutations?

The space complexity is O(n*n!).

Q: Which companies ask LeetCode 46. Permutations in interviews?

This problem is frequently asked at Adobe, Amazon, American Express, Apple, Barclays and 19 more companies.

Permutations - Complete Solution Guide

Permutations is LeetCode problem 46, 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 nums of distinct integers, return all the possible permutations . You can return the answer in any order . Example 1: Input: nums = [1,2,3] Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] Example 2: Input: nums = [0,1] Output: [[0,1],[1,0]] Example 3: Input: nums = [1] Output: [[1]] Constraints: 1 <= nums.length <= 6 -10 <= nums[i] <= 10 All the integers of nums are unique .

Detailed Explanation

The problem requires generating all possible orderings (permutations) of a given array of distinct integers. The input is an array `nums` of integers, and the output should be a list of lists, where each inner list represents a unique permutation of the input array. The order of the permutations in the output doesn't matter. The key constraint is that all integers in the input array are distinct, simplifying the permutation generation process.

Solution Approach

The provided solutions use a backtracking algorithm to generate all permutations. The core idea is to recursively explore possible arrangements by swapping elements in the input array. The `backtrack` function iterates through the array, swapping the element at the current index with all subsequent elements. After each swap, it recursively calls itself to explore the remaining part of the array. After the recursive call returns, the swap is undone to explore other possibilities. When the recursion reaches the end of the array, it means a complete permutation has been formed, and this permutation is added to the result.

Step-by-Step Algorithm

Step 1: Initialize an empty list `result` to store the permutations.
Step 2: Define a recursive function `backtrack(start_index)` to generate permutations starting from the given `start_index`.
Step 3: Base Case: If `start_index` is equal to the length of the input array `n`, it means a complete permutation is formed. Add a copy of the current `nums` array to the `result` list and return.
Step 4: Iterate through the elements from `start_index` to `n-1` using a `for` loop.
Step 5: Swap the element at `start_index` with the element at the current index `i` in the loop.
Step 6: Recursively call `backtrack(start_index + 1)` to explore permutations for the rest of the array.
Step 7: After the recursive call returns, swap the elements back to their original positions (undo the swap) to explore other possibilities. This is the 'backtracking' step.
Step 8: After the `for` loop completes, all permutations starting from `start_index` have been explored. The `backtrack` function returns.
Step 9: Call `backtrack(0)` to start the permutation generation process from the beginning of the array.
Step 10: Return the `result` list containing all the generated permutations.

Key Insights

Insight 1: Backtracking is a suitable approach for generating permutations because it explores different possibilities by making choices and undoing them if they don't lead to a valid solution.
Insight 2: Swapping elements in the array is an efficient way to explore different orderings during backtracking. It avoids the need for creating copies of the array repeatedly.
Insight 3: The base case for the recursive backtracking is when a complete permutation is formed (i.e., when the recursion reaches the end of the array). At this point, the permutation is added to the result.

Complexity Analysis

Time Complexity: O(n!)

Space Complexity: O(n*n!)

Topics

This problem involves: Array, Backtracking.

Companies

Asked at: Adobe, Amazon, American Express, Apple, Barclays, Bloomberg, Booking.com, Cisco, Citadel, Goldman Sachs, J.P. Morgan, LinkedIn, Meta, Microsoft, Microstrategy, Oracle, Qualcomm, ServiceNow, TikTok, Uber, Workday, Yahoo, Zoho, eBay.

Permutations | LeetCode 46

Medium

Array Backtracking

Time: O(n!)Space: O(n*n!)

Solution Code

Available in:

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

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

        backtrack(nums, 0, result);
        return result;
    }

    private void backtrack(int[] nums, int start_index, List<List<Integer>> result) {
        if (start_index == nums.length) {
            List<Integer> permutation = new ArrayList<>();
            for (int num : nums) {
                permutation.add(num);
            }
            result.add(permutation);
            return;
        }

        for (int i = start_index; i < nums.length; i++) {
            swap(nums, start_index, i);
            backtrack(nums, start_index + 1, result);
            swap(nums, start_index, i);
        }
    }

    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

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Problem Info

Description

Problem 46: Permutations - Detailed explanation and solution approaches.

Topics

Array Backtracking

Companies

AdobeAmazonAmerican ExpressAppleBarclaysBloombergBooking.comCisco

Stats

Difficulty:Medium

Languages:4

Companies:24

Problem #:46

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

10 steps

1
Step 1: Initialize an empty list `result` to store the permutations.
2
Step 2: Define a recursive function `backtrack(start_index)` to generate permutations starting from the given `start_index`.
3
Step 3: Base Case: If `start_index` is equal to the length of the input array `n`, it means a complete permutation is formed. Add a copy of the current `nums` array to the `result` list and return.
4
Step 4: Iterate through the elements from `start_index` to `n-1` using a `for` loop.
5
Step 5: Swap the element at `start_index` with the element at the current index `i` in the loop.
6
Step 6: Recursively call `backtrack(start_index + 1)` to explore permutations for the rest of the array.
7
Step 7: After the recursive call returns, swap the elements back to their original positions (undo the swap) to explore other possibilities. This is the 'backtracking' step.
8
Step 8: After the `for` loop completes, all permutations starting from `start_index` have been explored. The `backtrack` function returns.
9
Step 9: Call `backtrack(0)` to start the permutation generation process from the beginning of the array.
10
Step 10: Return the `result` list containing all the generated permutations.

💡

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!)

💾

Space Complexity

O(n*n!)

📝

Why?

The time complexity is O(n!) because the algorithm explores all possible n! permutations. The space complexity includes the space to store the result, which is O(n*n!) because each permutation is of length n, and there are n! permutations. The recursion depth is at most n, contributing O(n) to the space complexity of the call stack. Thus, in total, the space complexity is O(n*n!).

📖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

Iterative Approach using Heap's Algorithm

Heap's algorithm provides an iterative approach to generate permutations. It systematically swaps elements to produce all permutations without recursion.

Time: O(n!)Space: O(n)

Trade-offs: The iterative approach avoids the overhead of recursion and might be slightly faster in practice. However, the backtracking approach is generally easier to understand and implement.

✨

Key Insights

1Insight 1: Backtracking is a suitable approach for generating permutations because it explores different possibilities by making choices and undoing them if they don't lead to a valid solution.
2Insight 2: Swapping elements in the array is an efficient way to explore different orderings during backtracking. It avoids the need for creating copies of the array repeatedly.
3Insight 3: The base case for the recursive backtracking is when a complete permutation is formed (i.e., when the recursion reaches the end of the array). At this point, the permutation is added to the result.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: Empty input array. The provided code does not explicitly handle this case, but it will correctly return an empty list of permutations.
Edge case 2: Input array with a single element. The algorithm will still work correctly, returning a list containing a single permutation which is the input array itself.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not creating a copy of the `nums` array before adding it to the `result` list. Without creating a copy, all permutations in the `result` list will point to the same modified `nums` array, resulting in incorrect output. This is especially critical in the Python version.

✗Common mistake 2: Forgetting to undo the swap after the recursive call. This is crucial for exploring all possible permutations. If the swap is not undone, the algorithm will only generate a subset of the permutations.

✅

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: Password generation: Generating all possible password combinations within a given character set and length.

🏦Real-world scenario 2: Route optimization: Finding all possible routes for a traveling salesperson to visit a set of cities, although this quickly becomes intractable for larger sets of cities.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: July 17, 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

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

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What is the difference between backtracking and brute force?

Backtracking prunes invalid paths early using constraints, while brute force explores all possibilities. In backtracking, you "choose" an option, "explore" it recursively, then "unchoose" (backtrack) if it doesn't lead to a solution. This eliminates entire branches of the search tree.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Permutations"?

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 backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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