What is the time complexity of 4Sum?

The optimal solution has a time complexity of O(n^3).

What data structures are used to solve 4Sum?

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

What are the key insights for solving 4Sum?

Insight 1: Sorting the array enables efficient searching for quadruplets and helps avoid duplicate solutions using the two-pointer technique. Insight 2: The two-pointer technique, applied after fixing the first two numbers of the quadruplet, allows finding the remaining two numbers that satisfy the target sum in O(n) time.

Which companies ask LeetCode 18. 4Sum in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Cloudflare and 15 more companies.

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4Sum - LeetCode 18 Solution

Q: How to solve 4Sum?

The provided solution employs a nested loop approach combined with the two-pointer technique to find the quadruplets. First, the input array is sorted. The outer two loops iterate through all possible pairs of numbers. Inside these loops, the two-pointer technique is used to find two additional numbers that, when added to the first two, sum up to the target. The code also includes several optimizations, such as skipping duplicate elements and early pruning to avoid unnecessary computations.

4Sum - Complete Solution Guide

4Sum is LeetCode problem 18, 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 n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that: 0 <= a, b, c, d < n a , b , c , and d are distinct . nums[a] + nums[b] + nums[c] + nums[d] == target You may return the answer in any order . Example 1: Input: nums = [1,0,-1,0,-2,2], target = 0 Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]] Example 2: Input: nums = [2,2,2,2,2], target = 8 Output: [[2,2,2,2]] Constraints: 1 <= nums.length <= 200 -10 9 <= nums[i] <= 10 9 -

Detailed Explanation

The problem requires finding all unique quadruplets (sets of four numbers) within a given array `nums` that sum up to a specified `target` value. The quadruplets must consist of distinct elements from the input array, and the order of elements within a quadruplet, as well as the order of quadruplets in the output, does not matter. The constraints include the size of the array (1 <= n <= 200) and the range of the numbers in the array and the target (-10^9 <= nums[i], target <= 10^9).

Solution Approach

The provided solution employs a nested loop approach combined with the two-pointer technique to find the quadruplets. First, the input array is sorted. The outer two loops iterate through all possible pairs of numbers. Inside these loops, the two-pointer technique is used to find two additional numbers that, when added to the first two, sum up to the target. The code also includes several optimizations, such as skipping duplicate elements and early pruning to avoid unnecessary computations.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order.
Step 2: Iterate through the array using the first pointer `i` from 0 to n-4.
Step 3: Inside the first loop, skip duplicate values of `nums[i]` to avoid duplicate quadruplets.
Step 4: Perform early pruning: check if the minimum possible sum (using `nums[i]` and the next three elements) is greater than the target or if the maximum possible sum (using `nums[i]` and the last three elements) is less than the target. If either condition is true, skip the current iteration.
Step 5: Iterate through the array using the second pointer `j` from `i+1` to n-3.
Step 6: Inside the second loop, skip duplicate values of `nums[j]` to avoid duplicate quadruplets.
Step 7: Perform early pruning similar to step 4, but considering `nums[i]` and `nums[j]`.
Step 8: Initialize two pointers, `left` (pointing to `j+1`) and `right` (pointing to `n-1`).
Step 9: While `left < right`, calculate the `current_sum` of `nums[i]`, `nums[j]`, `nums[left]`, and `nums[right]`.
Step 10: If `current_sum == target`, add the quadruplet to the result list, increment `left`, decrement `right`, and skip any duplicate values of `nums[left]` and `nums[right]` to avoid duplicate quadruplets.
Step 11: If `current_sum < target`, increment `left` to increase the sum.
Step 12: If `current_sum > target`, decrement `right` to decrease the sum.
Step 13: Return the list of unique quadruplets.

Key Insights

Insight 1: Sorting the array enables efficient searching for quadruplets and helps avoid duplicate solutions using the two-pointer technique.
Insight 2: The two-pointer technique, applied after fixing the first two numbers of the quadruplet, allows finding the remaining two numbers that satisfy the target sum in O(n) time.
Insight 3: Early pruning using checks on minimum and maximum possible sums involving the current elements can significantly reduce the search space.

Complexity Analysis

Time Complexity: O(n^3)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Sorting.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Cloudflare, DE Shaw, DoorDash, Google, Infosys, Meta, Microsoft, Rubrik, Samsung, ServiceNow, TikTok, Uber, Yahoo, Zoox, oyo, tcs.

4Sum | LeetCode 18

Medium

Array Two Pointers Sorting

Time: O(n^3)Space: O(1)

Solution Code

Available in:

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

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        int n = nums.length;
        if (n < 4) {
            return new ArrayList<>();
        }

        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();

        for (int i = 0; i < n - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }

            if ((long) nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {
                break;
            }

            if ((long) nums[i] + nums[n - 3] + nums[n - 2] + nums[n - 1] < target) {
                continue;
            }

            for (int j = i + 1; j < n - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }

                if ((long) nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {
                    break;
                }

                if ((long) nums[i] + nums[j] + nums[n - 1] + nums[n - 2] < target) {
                    continue;
                }

                int left = j + 1;
                int right = n - 1;
                while (left < right) {
                    long current_sum = (long) nums[i] + nums[j] + nums[left] + nums[right];
                    if (current_sum == target) {
                        result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
                        left += 1;
                        right -= 1;
                        while (left < right && nums[left] == nums[left - 1]) {
                            left += 1;
                        }
                        while (left < right && nums[right] == nums[right + 1]) {
                            right -= 1;
                        }
                    } else if (current_sum < target) {
                        left += 1;
                    } else {
                        right -= 1;
                    }
                }
            }
        }

        return result;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 18: 4Sum - Detailed explanation and solution approaches.

Topics

Array Two Pointers Sorting

Companies

AdobeAmazonAppleBloombergCloudflareDE ShawDoorDashGoogle

Stats

Difficulty:Medium

Languages:4

Companies:20

Problem #:18

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

13 steps

1
Step 1: Sort the input array `nums` in ascending order.
2
Step 2: Iterate through the array using the first pointer `i` from 0 to n-4.
3
Step 3: Inside the first loop, skip duplicate values of `nums[i]` to avoid duplicate quadruplets.
4
Step 4: Perform early pruning: check if the minimum possible sum (using `nums[i]` and the next three elements) is greater than the target or if the maximum possible sum (using `nums[i]` and the last three elements) is less than the target. If either condition is true, skip the current iteration.
5
Step 5: Iterate through the array using the second pointer `j` from `i+1` to n-3.
6
Step 6: Inside the second loop, skip duplicate values of `nums[j]` to avoid duplicate quadruplets.
7
Step 7: Perform early pruning similar to step 4, but considering `nums[i]` and `nums[j]`.
8
Step 8: Initialize two pointers, `left` (pointing to `j+1`) and `right` (pointing to `n-1`).
9
Step 9: While `left < right`, calculate the `current_sum` of `nums[i]`, `nums[j]`, `nums[left]`, and `nums[right]`.
10
Step 10: If `current_sum == target`, add the quadruplet to the result list, increment `left`, decrement `right`, and skip any duplicate values of `nums[left]` and `nums[right]` to avoid duplicate quadruplets.
11
Step 11: If `current_sum < target`, increment `left` to increase the sum.
12
Step 12: If `current_sum > target`, decrement `right` to decrease the sum.
13
Step 13: Return the list of unique quadruplets.

💡

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^3)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by the three nested loops. The outer loop runs n times, the second loop runs n times, and the two-pointer approach takes O(n) time. Therefore, the overall time complexity is O(n * n * n) = O(n^3). The space complexity is O(1) because the algorithm primarily uses a fixed amount of extra space, regardless of the input size. The space used to store the output is not considered for the O(1) space complexity since the question asks us to *return* the result. Note: if the space for the output is considered, then the space complexity would be O(K) where K is the number of quadruplets found, but it is generally not the norm on leetcode problems.

📖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

Hash Map Approach

An alternative approach could involve using a hash map to store the sums of all possible pairs of numbers. Then, iterate through the remaining pairs and check if the complement (target - sum of the pair) exists in the hash map. This approach can potentially improve the time complexity but often results in higher space complexity.

Time: O(n^2) average, O(n^4) worstSpace: O(n^2)

Trade-offs: The hash map approach might be faster in some cases but requires significantly more space and is susceptible to worst-case O(n^4) performance if hash collisions are frequent. The two-pointer approach provides a better balance between time and space complexity and is generally preferred.

✨

Key Insights

1Insight 1: Sorting the array enables efficient searching for quadruplets and helps avoid duplicate solutions using the two-pointer technique.
2Insight 2: The two-pointer technique, applied after fixing the first two numbers of the quadruplet, allows finding the remaining two numbers that satisfy the target sum in O(n) time.
3Insight 3: Early pruning using checks on minimum and maximum possible sums involving the current elements can significantly reduce the search space.

📚

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.

Sorting

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

Edge Cases

Edge case 1: Array size less than 4. The solution handles this by returning an empty list.
Edge case 2: Duplicate numbers in the input array. The solution uses checks (`if i > 0 and nums[i] == nums[i-1]: continue`) to skip duplicate values and ensure unique quadruplets in the result.
Edge case 3: Target sum is very large or very small causing overflow. Solution uses long long (or long in Java) to avoid integer overflow when calculating the sum.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling duplicate numbers correctly, leading to duplicate quadruplets in the output. Skipping duplicate `nums[i]`, `nums[j]`, `nums[left]`, and `nums[right]` is crucial.

✗Common mistake 2: Integer overflow when calculating the sum of four numbers. Using `long long` (C++) or `long` (Java) to avoid overflow is important.

✗Common mistake 3: Not sorting the array, which prevents using the efficient two-pointer technique and leads to a much slower and more complex solution.

✅

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: Data analysis: Identifying sets of four data points that satisfy a specific condition or target value in a large dataset.

🏦Real-world scenario 2: Financial modeling: Finding combinations of assets that meet a target portfolio return or risk profile.

🎮Real-world scenario 3: Image processing: Detecting patterns of four pixels with a specific combined intensity or color value.

💡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: June 19, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

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

@TechLeadJul 6

❓

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.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

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 "4Sum"?

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.

🤖 Need help with this problem?