15. 3Sum.cpp

/*
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
*/

// Two Pointers Approach and Skipping duplicates to reduce number of comparisions
class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> output;
        int n = nums.size();
        for (int i = 0; i < n - 2; i++){
            if (i > 0 && nums[i] == nums[i - 1])
                continue;
            int j = i + 1, k = n - 1;
            while (j < k){
                int sum = nums[i] + nums[j] + nums[k];
                if (sum == 0){
                    output.push_back({nums[i], nums[j], nums[k]});
                    j++;
                    k--;
                    while (j < k && nums[j] == nums[j - 1])
                        j++;
                    while (j < k && nums[k] == nums[k + 1])
                        k--;
                }
                else if (sum < 0)
                    j++;
                else
                    k--;
            }
        }
        return output;
    }
};