Find unique triplets in an array such that the triplet sum = 0

Author: Sreejit7

C++/Data-Structures/Array/threeSum.cpp

@@ -0,0 +1,55 @@
+/*
+ Given an array nums of n integers, find all unique triplets in the array which gives the sum of zero.
+*/
+#include<bits/stdc++.h>
+using namespace std;
+vector<vector<int>> threeSum(vector<int>& nums) {
+ int n = nums.size();
+ if(n < 3)
+ return {};
+ sort(nums.begin(),nums.end());
+ vector <vector<int>> result;
+ unordered_set<string> triplets;
+ int i = 0;
+ while(i < n -2){
+ int j = i+1, k = n-1;
+ while(j<k){
+ int sum = nums[i] + nums[j] + nums[k];
+ if(sum == 0){
+ string newTriplet = to_string(nums[i]) + ":"
+ + to_string(nums[j]) + ":" 
+ + to_string(nums[k]);
+ if(triplets.find(newTriplet) == triplets.end()){
+ triplets.insert(newTriplet);
+ result.push_back({nums[i],nums[j],nums[k]});
+ }
+ j++;
+ k--;
+ }
+ else if(sum > 0)
+ k--;
+ else if(sum < 0)
+ j++;
+ }
+ i++;
+ } 
+ return result;
+}
+
+int main(){
+ int n;
+ vector <int> nums;
+ cin>>n;
+ for(int i = 0; i < n; i++){
+ int input;
+ cin>>input;
+ nums.push_back(input);
+ }
+ vector <vector<int>> result = threeSum(nums);
+ for(int i = 0; i< result.size(); i++){
+ for(int j = 0; j< result[0].size(); j++)
+ cout<<result[i][j]<<" ";
+ cout<<"\n";
+ }
+}
+//Time complexity: O(n^2), Space complexity: O(n)