4 Sum

Description

4 Sum
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
• -109 <= nums[i] <= 109
• -109 <= target <= 109

For 4Sum, the brute force method uses four nested loops, while the optimal solution sorts the array and combines two fixed loops with two pointers to achieve $O(N^3)$ time complexity.

1. Brute Force Approach ($O(N^4)$)

Intuition
Check all possible quadruplets using four nested loops and insert sorted quadruplets into a set to eliminate duplicates.

public List<List<Integer>> fourSum(int[] nums, int target) {
    Set<List<Integer>> resultSet = new HashSet<>();
    int n = nums.length;
 
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            for (int k = j + 1; k < n; k++) {
                for (int l = k + 1; l < n; l++) {
                    if ((long) nums[i] + nums[j] + nums[k] + nums[l] == target) {
                        List<Integer> quad = Arrays.asList(nums[i], nums[j], nums[k], nums[l]);
                        Collections.sort(quad);
                        resultSet.add(quad);
                    }
                }
            }
        }
    }
    return new ArrayList<>(resultSet);
}
 

• Time Complexity: $O(N^4)$
• Space Complexity: $O(N^4)$ worst-case to store quadruplets in the set.

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2. Optimized Approach ($O(N^3)$)

Intuition: Fixed Outer Loops + Two Pointers
Think of 4Sum as extending 3Sum: fix the first two numbers (nums[i] and nums[j]) using loops, then solve the remaining two numbers using a left and right two-pointer search. Sorting first allows skipping duplicates naturally without extra set overhead.

4-Step Formula to Remember:

1. Sort the array.
2. Loop i from 0 to n - 4. Skip duplicate nums[i].
3. Loop j from i + 1 to n - 3. Skip duplicate nums[j].
4. Set left = j + 1 and right = n - 1. Move left forward if sum < target, right backward if sum > target. When sum == target, record the answer and jump over duplicate values for both pointers.

public List<List<Integer>> fourSum(int[] nums, int target) {
    List<List<Integer>> result = new ArrayList<>();
    int n = nums.length;
    if (n < 4) return result;
 
    Arrays.sort(nums);
 
    for (int i = 0; i < n - 3; i++) {
        // Skip duplicates for 1st number
        if (i > 0 && nums[i] == nums[i - 1]) continue;
 
        for (int j = i + 1; j < n - 2; j++) {
            // Skip duplicates for 2nd number
            if (j > i + 1 && nums[j] == nums[j - 1]) continue;
 
            int left = j + 1;
            int right = n - 1;
 
            while (left < right) {
                // Cast to long to prevent integer overflow
                long sum = (long) nums[i] + nums[j] + nums[left] + nums[right];
 
                if (sum == target) {
                    result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
 
                    // Skip duplicates for 3rd and 4th numbers
                    while (left < right && nums[left] == nums[left + 1]) left++;
                    while (left < right && nums[right] == nums[right - 1]) right--;
 
                    left++;
                    right--;
                } else if (sum < target) {
                    left++;
                } else {
                    right--;
                }
            }
        }
    }
    return result;
}
 

• Time Complexity: $O(N^3)$ — $O(N\log N)$ for sorting + $O(N^3)$ for two nested loops with two pointers.
• Space Complexity: $O(1)$ auxiliary memory (excluding the output list).