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]]

This is a bit tricky problem first we need to sort the array as now we have to find all combination of 4 numbers that makes the sum equal to the target so in this, we approach the problem with 2 for loops where first loop i.e i 0 to n-1 (where n is the size of the array) and j from i to n then we add the 2 numbers and subtract that from the target we the required number which we want to get than inside the j loop we do a binary search and make 2 pointers to j+1 and n and use the same approach we did in 3 sums but now we will also check for repeated numbers for optimal and non-repeating answers so we while loop to counter this problem of repeating answers.

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> res;
        sort(nums.begin(),nums.end());
        int n=nums.size()-1;
        for(int i=0;i<n-1;)
        {
            for(int j=i+1;j<n;)
            {
                int sum=nums[i]+nums[j];
                int req=target-sum;
                int lo=j+1,hi=n;
                while(lo<hi)
                {
                    int sum2=nums[lo]+nums[hi];
                    if(sum2==req)
                    {
                        res.push_back({nums[i],nums[j],nums[lo++],nums[hi--]});
                        while(lo<n and nums[lo]==nums[lo-1]) lo++;
                        while(hi>j and nums[hi]==nums[hi+1]) hi--;
                    }
                    else if(sum2>req) hi--;
                    else lo++;
                }
                do{j++;}while(j<n and nums[j]==nums[j-1]);
            }
            do{i++;}while(i<n and nums[i]==nums[i-1]);
        }
        return res;
    }
};