3sum

Problem Description

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Intuition

We can break down this problem by fixing one element and converting it into a 2Sum problem:

  1. Fix one element as the first number
  2. Solve for the remaining elements using the 2Sum approach

Time Complexity: $O(n^2)$
Space Complexity: $O(n)$ for sorting, $O(1)$ extra space

Algorithm

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class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int n = nums.size();
        set<vector<int>> s;
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());

        for(int i = 0; i < n - 2; i++){
            int target  = nums[i];
            int left = i + 1;
            int right = n - 1;
            while(left < right){
                if(nums[left] + nums[right] + target == 0) {
                    s.insert({target, nums[left], nums[right]});
                    left++;
                    right--; 
                }
                else if(nums[left] + nums[right] + target < 0) {
                    left++;
                }
                else{
                    right--;
                }
            }
        }

        for(auto vec:s){
            ans.push_back(vec);
        }

        return ans;
    }
};