4Sum

Approach

The problem asks us to find all unique quadruplets in an array that sum up to a given target. Pattern: Generalized k-Sum (Sorting + Two Pointers + Recursion/Loops).

Similar to 3Sum, a brute force approach of four nested loops would yield $O(N^4)$ time, which is unacceptable. We can extend the optimal logic of 3Sum:

Sort the array.
Fix one number, reducing the problem to 3Sum.
Fix a second number, reducing the problem to Two Sum II (Two Pointers).

This gives an $O(N^3)$ solution for 4Sum. For a scalable approach, we can write a generalized kSum function. If $k=2$ , we use the two-pointer approach. If $k>2$ , we iterate through the array, fix the current element, and recursively call kSum with $k-1$ and a reduced target.

Code

Brute-Force Approach

Pseudocode:

Four nested loops i, j, k, l.
Check if the sum equals target.
Store in a set to avoid duplicates.

// Unacceptably slow, purely for theoretical understanding
class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        set<vector<int>> res;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        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 long)nums[i] + nums[j] + nums[k] + nums[l] == target) {
                            res.insert({nums[i], nums[j], nums[k], nums[l]});
                        }
                    }
                }
            }
        }
        return vector<vector<int>>(res.begin(), res.end());
    }
};

Optimal Approach (Sorting + Generalized k-Sum)

Pseudocode:

Sort nums.
Define kSum(start_index, target, k).
Base Cases for kSum:
- If start_index is out of bounds, return [].
- If the smallest possible sum (k * smallest element) > target, return [].
- If the largest possible sum (k * largest element) < target, return [].
If $k == 2$ : perform standard Two Sum using two pointers.
If $k > 2$ $k > 2$ :
- Loop i from start_index to n-1.
- Skip duplicates: if i > start_index and nums[i] == nums[i-1].
- Call kSum(i + 1, target - nums[i], k - 1).
- Append nums[i] to each result returned and collect them.

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        return kSum(nums, target, 0, 4);
    }
 
private:
    vector<vector<int>> kSum(vector<int>& nums, long long target, int start, int k) {
        vector<vector<int>> res;
        
        // Base cases to early terminate
        if (start == nums.size()) {
            return res;
        }
        
        long long average_value = target / k;
        // We cannot obtain a sum of target if the smallest value is greater than the average
        // or the largest value is smaller than the average.
        if (nums[start] > average_value || average_value > nums.back()) {
            return res;
        }
        
        if (k == 2) {
            return twoSum(nums, target, start);
        }
        
        for (int i = start; i < nums.size(); ++i) {
            // Skip duplicates
            if (i == start || nums[i - 1] != nums[i]) {
                for (vector<int>& subset : kSum(nums, target - nums[i], i + 1, k - 1)) {
                    res.push_back({nums[i]});
                    res.back().insert(res.back().end(), subset.begin(), subset.end());
                }
            }
        }
        
        return res;
    }
 
    vector<vector<int>> twoSum(vector<int>& nums, long long target, int start) {
        vector<vector<int>> res;
        int left = start;
        int right = nums.size() - 1;
        
        while (left < right) {
            long long total = (long long)nums[left] + nums[right];
            
            if (total < target) {
                left++;
            } else if (total > target) {
                right--;
            } else {
                res.push_back({nums[left], nums[right]});
                left++;
                right--;
                while (left < right && nums[left] == nums[left - 1]) {
                    left++;
                }
            }
        }
        
        return res;
    }
};

Complexity

Time: $O(N^{k-1})$ . For 4Sum, $k=4$ , so the time complexity is $O(N^3)$ . The recursive kSum function nests loops up to $k-2$ times, and the base case (Two Sum) takes $O(N)$ . Thus, $O(N^{k-2} \times N) = O(N^{k-1})$ .
Space: $O(k)$ for the recursion stack. For 4Sum, this is $O(1)$ auxiliary space. Sorting in C++ generally takes $O(\log N)$ space.

Edge Cases

Large Target/Elements: Overflows can happen if numbers are huge. Using long long for targets and sums handles arbitrarily large integers automatically. The early termination nums[start] > average_value || average_value > nums.back() handles mathematically impossible targets instantly.
Duplicates everywhere: [2, 2, 2, 2, 2] with target 8. Handled cleanly by the i == start || nums[i - 1] != nums[i] deduplication logic.

Notes

Thought Process & Recognition: When scaling from 3Sum to 4Sum, you should immediately recognize that writing 3 nested loops (with a two-pointer base) is brittle. Hardcoding n nested loops is terrible practice. Instead, recognize the recursive nature: $k$ -Sum is just a loop that picks an element and calls $(k-1)$ -Sum. The real magic is in the early termination. Calculating the average needed target / k and checking if the smallest or largest elements can even support that average prunes massive branches of the recursion tree, turning a slow $O(N^3)$ into an extremely fast execution in practice.

4Sum

published

modified

length

source