4 Sum LeetCode

 4 Sum

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
• -109 <= nums[i] <= 109
• -109 <= target <= 109

Solution To the Above Question

class Solution

{

public:

    vector<vector<int>> fourSum(vector<int> &nums, int target)

    {

        sort(nums.begin(), nums.end());

        int n = nums.size();

        // for(int i=0; i<n; i++){

        //     cout<<nums[i]<<" ";

        // }

        cout << endl;

        vector<vector<int>> ans;

        vector<int> v;

        if (nums.size() < 4)

        {

            return ans;

        }

        for (int i = 0; i < n - 3; i++)

        {

            for (int x = i + 1; x < n - 2; x++)

            {

                int j = x + 1;

                int k = n - 1;

                while (j < k)

                {

                    long long int current = nums[i];

                    current += nums[x];

                    current += nums[j];

                    current += nums[k];

                    // cout<<current<<" ";

                    if (current == target)

                    {

                        v.push_back(nums[i]);

                        v.push_back(nums[x]);

                        v.push_back(nums[j]);

                        v.push_back(nums[k]);

                        bool flag = false;

                        for (int m = 0; m < ans.size(); m++)

                        {

                            if (ans[m] == v)

                            {

                                flag = true;

                            }

                        }

                        if (!flag)

                        {

                            ans.push_back(v);

                        }

                        v.clear();

                        j++;

                    }

                    else if (current > target)

                    {

                        k--;

                    }

                    else if (current < target)

                    {

                        j++;

                    }

                }

            }

        }

        return ans;

    }

};