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MEDIUM

4Sum

Given an array nums of n integers, return all 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

Input:

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

Output:

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

Explanation:
There are three unique quadruplets whose sum is 0.

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Constraints

• 1 ≤ nums.length ≤ 200
• -10⁹ ≤ nums[i] ≤ 10⁹
• -10⁹ ≤ target ≤ 10⁹

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Solution: Two Pointers

• Time Complexity: O(n³)
• Space Complexity: O(1) (excluding the space for the answer)

C++

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        for (int i = 0; i < n - 3; ++i) {
            if (i > 0 && nums[i] == nums[i-1]) continue;
            for (int j = i + 1; j < n - 2; ++j) {
                if (j > i + 1 && nums[j] == nums[j-1]) continue;
                int left = j + 1, right = n - 1;
                while (left < right) {
                    long long sum = (long long)nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        res.push_back({nums[i], nums[j], nums[left], nums[right]});
                        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 res;
    }
};