15. 3Sum

Problem Statement

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.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation: 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints:

3 <= nums.length <= 3000
-105 <= nums[i] <= 105

Intuition

Approach:

Fix one element and two sum for other two 
Use map to fix one, iterate in n^2 over other two

Links

https://leetcode.com/problems/3sum/description/

Video Links

Approach 1:

Map and Set

C++

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

        for(int i=0; i<n; i++)
            mp[nums[i]] = i;

        for(int i=0; i<n; i++){
            for(int j=i+1; j<n; j++){
                int k = -1 * (nums[i] + nums[j]);

                if(mp.find(k) != mp.end() and mp[k] > i and mp[k] > j){
                    s.insert({nums[i],nums[j],k});
                }
            }
        }

        for(auto &it: s)
            ans.push_back(it);

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++