Leetcode 18 Solution

This article provides solution to leetcode question 18 (4sum)
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https://leetcode.com/problems/4sum

Thinking Process

Similar to the question 3sum, we fix the first two numbers, and then apply the sorted two sum algorithm.

Time & Space Complexity

Assuming N is the size of the array

  • Time complexity: O(N^3)
  • Space complexity: O(N)

Solution

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());



        for (int i = 0; i < nums.size(); i++)
        {
            if (i > 0 && nums[i] == nums[i - 1])
                continue;



            for (int j = i + 1; j < nums.size(); j++)
            {
                if (j > i + 1 && nums[j] == nums[j - 1])
                    continue;



                int l = j + 1;
                int r = nums.size() - 1;



                while (l < r)
                {
                    int s = nums[i] + nums[j] + nums[l] + nums[r];



                    if (s < target)
                        l++;
                    else if (s > target)
                        r--;
                    else
                    {
                        vector<int> a;
                        a.push_back(nums[i]);
                        a.push_back(nums[j]);
                        a.push_back(nums[l]);
                        a.push_back(nums[r]);



                        res.push_back(a);



                        do
                        {
                            l++;
                        } while (nums[l] == nums[l - 1] && l < nums.size());



                        do
                        {
                            r--;
                        } while (nums[r] == nums[r + 1] && r >= j + 1);
                    }
                }
            }
        }



        return res;
    }
};