Title: Solving the 4-Sum Problem in JavaScript: A Step-by-Step Guide

When it comes to mastering Data Structures and Algorithms (DSA), solving complex problems like 4-Sum is essential. This problem involves finding all unique quadruplets (four numbers) in an array that sum up to a given target. It's an extension of the 2-Sum and 3-Sum problems, but with added complexity due to the additional combinations to consider.

In this article, we'll break down the 4-Sum problem and walk through an efficient solution using JavaScript. We’ll also explain the code in detail so that you can implement it with ease.

Problem Statement:

You are given an array nums of integers and a target integer. Your goal is to return all unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:

• a,b,c,da, b, c, da,b,c,d are distinct indices in the array.
• nums[a]+nums[b]+nums[c]+nums[d]=targetnums[a] + nums[b] + nums[c] + nums[d] = \text{target}nums[a]+nums[b]+nums[c]+nums[d]=target.

Example:

Input:

nums = [1, 0, -1, 0, -2, 2], target = 0

Output:

[[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]

Input:

nums = [2, 2, 2, 2, 2], target = 8

Output:

[[2, 2, 2, 2]]

Approach to Solve the Problem:

The 4-Sum problem can be tackled with a similar strategy to 3-Sum, but with one extra loop to account for the fourth number. Here’s how you can approach it step-by-step:

Step 1: Sort the Array

Sorting the array helps in avoiding duplicates and also allows us to efficiently use the two-pointer technique later.

Step 2: Use Nested Loops

We will use two loops to iterate over the array and fix two numbers at a time. After fixing two numbers, we will use two pointers (left and right) to find the remaining two numbers whose sum matches the target.

Step 3: Skip Duplicates

Since we are asked to find unique quadruplets, we need to ensure that we skip over duplicate numbers in the array.

Step 4: Optimize with Two Pointers

By using two pointers (one starting from the left and one from the right), we can reduce the time complexity, making our solution more efficient.

JavaScript Solution:

var fourSum = function(nums, target) {

// Step 1: Sort the array

nums.sort((a, b) => a - b);

let result = [];

// Step 2: First two loops to fix the first two numbers

for (let i = 0; i < nums.length - 3; i++) {

// Skip duplicate numbers for the first element

if (i > 0 && nums[i] === nums[i - 1]) continue;

for (let j = i + 1; j < nums.length - 2; j++) {

// Skip duplicate numbers for the second element

if (j > i + 1 && nums[j] === nums[j - 1]) continue;

// Step 3: Use two pointers to find the other two numbers

let left = j + 1;

let right = nums.length - 1;

while (left < right) {

let sum = nums[i] + nums[j] + nums[left] + nums[right];

// If the sum is less than the target, move the left pointer to the right

if (sum < target) {

left++;

}

// If the sum is greater than the target, move the right pointer to the left

else if (sum > target) {

right--;

}

// If the sum equals the target, we have found a valid quadruplet

else {

result.push([nums[i], nums[j], nums[left], nums[right]]);

// Skip duplicate numbers for the third and fourth elements

while (left < right && nums[left] === nums[left + 1]) left++;

while (left < right && nums[right] === nums[right - 1]) right--;

// Move the pointers to continue searching for other pairs

left++;

right--;

}

}

}

}

// Step 4: Return the result array containing all unique quadruplets

return result;

};

Explanation:

• Sorting the Array: Sorting ensures that we can easily skip over duplicates and simplifies the two-pointer technique.
• Two Loops for Fixing Two Numbers: The first two loops fix the first two numbers of the quadruplet. We use these loops to iterate over the sorted array while ensuring we skip duplicates.
• Two Pointers to Find the Remaining Two Numbers: After fixing two numbers, we use a two-pointer approach to find the remaining two numbers. This optimizes the search for the target sum by reducing the number of operations compared to a brute-force approach.
• Avoiding Duplicates: The checks nums[i] === nums[i - 1] and nums[j] === nums[j - 1] are crucial for skipping duplicate combinations. Similarly, the while loops for left and right pointers ensure that we don't count the same quadruplet multiple times.

Time and Space Complexity:

• Time Complexity: Sorting the array takes O(nlogn)O(n \log n)O(nlogn), and we have three nested loops, each of which runs in linear time after sorting. Therefore, the overall time complexity is O(n3)O(n^3)O(n3), where nnn is the length of the array.
• Space Complexity: The space complexity is O(k)O(k)O(k), where kkk is the number of quadruplets in the result. Apart from the result array, we use constant extra space for variables.

Conclusion:

The 4-Sum problem is a great exercise in improving your understanding of both nested loops and the two-pointer technique. By breaking the problem down into steps, using sorting, and efficiently skipping duplicates, we arrive at an optimal solution. Practicing problems like this on platforms like LeetCode will sharpen your problem-solving skills and make you more confident in tackling algorithmic challenges.

Call to Action: Have you solved any variations of the 4-Sum problem? Feel free to share your experiences or alternative solutions in the comments below!