✅ C++ || WITHOUT USING SET || 100% FAST
INSTEAD OF USING SET FOR REMOVING DUPLICATES WE ARE USING SOME WHILE LOOPS AND IF CONDITIONS:\n\n\nclass Solution {\npublic:\n    #define ll long long\n    vect
# INSTEAD OF USING SET FOR REMOVING DUPLICATES WE ARE USING SOME WHILE LOOPS AND IF CONDITIONS:\n\n```\nclass Solution {\npublic:\n    #define ll long long\n    vector<vector<int>> fourSum(vector<int>& nums, int target) {\n        vector<vector<int>>ans;\n        sort(nums.begin(),nums.end());\n        for(ll i=0;i<nums.size();i++){\n            if(i>0 && nums[i]==nums[i-1])continue;       //REMOVES DUPLICATE IN 1ST INDEX\n            ll t=target-nums[i];\n            for(ll j=i+1;j<nums.size();j++){\n                if(j>i+1 && nums[j]==nums[j-1])continue;     //REMOVE DUPLICATE IN 2ND INDEX\n                ll s=t-nums[j];\n                ll left=j+1,right=nums.size()-1;\n                while(left<right){\n                    ll sum=nums[left] + nums[right];\n                    if(sum==s){\n                        ans.push_back({nums[i],nums[j],nums[left],nums[right]});\n                        while(left<nums.size()-1 && nums[left]==nums[left+1])left++;     //REMOVE DUPLICATE IN 3RD INDEX\n                        while(right>0 && nums[right]==nums[right-1])right--;       //REMOVE DUPLICATE IN 4TH INDEX\n                        left++;\n                        right--;\n                    }\n                    else if(sum<s)left++;\n                    else right--;\n                }\n            }\n        }\n        return ans;\n    }\n\n};\n```\n\n![image](https://assets.leetcode.com/users/images/10a8794f-c64c-4561-b4d9-2b7a2524753d_1654456657.7689903.png)\n\n\n\uD83D\uDE4C\uD83D\uDE4C HAPPY CODING !!
['Two Pointers', 'Binary Search', 'C', 'Binary Tree']