4Sum

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
• -1e⁹ <= nums[i] <= 1e⁹
• -1e⁹ <= target <= 1e⁹

Code

Python Code

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()
        n = len(nums)
        ans = set()
        for i in range(n):
            for j in range(i+1, n):
                l, r = j + 1, n - 1
                remain = target - nums[i] - nums[j]
                while l < r:
                    if nums[l] + nums[r] == remain:
                        ans.add((nums[i], nums[j], nums[l], nums[r]))
                        l += 1
                        r -= 1
                    elif nums[l] + nums[r] > remain:
                        r -= 1
                    else:
                        l += 1
        return ans

C++

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<vector<int>> ans;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                int l = j + 1, r = n - 1;
                int remain = target - nums[i] - nums[j];
                while (l < r) {
                    if (nums[l] + nums[r] == remain) {
                        ans.push_back({nums[i], nums[j], nums[l], nums[r]});
                        ++l; --r;
                        while (l < r && nums[l-1] == nums[l]) ++l; // Skip duplicate nums[l]
                    } else if (nums[l] + nums[r] > remain) {
                        --r;
                    } else {
                        ++l;
                    }
                }
                while (j+1 < n && nums[j] == nums[j+1]) ++j; // Skip duplicate nums[j]
            }
            while (i+1 < n && nums[i] == nums[i+1]) ++i; // Skip duplicate nums[i]
        }
        return ans;
    }
};