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

medium

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

Problem

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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

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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

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

Solution

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

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;
  }
};