4Sum
Post
Cancel

# LeetCode #18: 4Sum (C/C++).

medium

source: https://leetcode.com/problems/4sum/
C/C++ Solution to LeetCode problem 18. 4Sum.

## Problem

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.

## Examples

### 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

## Solution

This is an extension of the 3Sum problem, that is an extension of the 2Sum problem.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 class Solution { public: vector<vector<int>> fourSum(vector<int>& nums, int target) { if (nums.size() < 4) return {}; vector<vector<int>> result; sort(nums.begin(), nums.end()); for (int i=0; i<nums.size() - 3; i+=1) { if (i>0 && nums[i] == nums[i-1]) continue; for (int j=i+1; j<nums.size() - 2; j+=1) { if (j>i+1 && nums[j] == nums[j-1]) continue; int l = j+1; int r = nums.size() - 1; long sum = nums[i] + nums[j]; while (l < r) { long s = sum + nums[l] + nums[r]; if (s == target) { result.push_back({nums[i], nums[j], nums[l], nums[r]}); l += 1; r -= 1; while (l < r && nums[l] == nums[l-1]) l += 1; while (l < r && nums[r] == nums[r+1]) r -= 1; } else if (s > target) r -= 1; else l += 1; } } } return result; } };