4Sum

//Optimized Approach using two pointer - O(n^3) time and O(n) space
class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        set<vector<int>> s;
        vector<vector<int>> output;
        for (int i = 0; i < nums.size(); i++){
            for(int j = i+1; j < nums.size(); j++){
                int k = j + 1;
                int l = nums.size() - 1;
                while (k < l) {
                    //by writing below 4 statement this way it will not give runtime error
                    long long int sum = nums[i];
                    sum += nums[j];
                    sum += nums[k];
                    sum += nums[l];
                    if (sum == target) {
                        s.insert({nums[i], nums[j], nums[k], nums[l]});
                        k++;
                        l--;
                    } else if (sum < target) {
                        k++;
                    } else {
                        l--;
                    }
                }
            }
        }
        for(auto quadruplets : s)
            output.push_back(quadruplets);
        return output;
    }
};