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

<p>Notice that the solution set must not contain duplicate triplets.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
<strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> [[0,0,0]]
<strong>Explanation:</strong> The only possible triplet sums up to 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= nums.length &lt;= 3000</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    // vector<int> twosum(int i,vector<int>&nums){
    //     int sum = -nums[i];
    //     for(;i<s;i++){

    //     }
    // }

    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        set<vector<int>>gabbu;
        int n = nums.size();
        sort(nums.begin(),nums.end());
        if(nums[0] == nums[n-1] && nums[0] == 0){
            vector<int> arr= {0,0,0};
            res.push_back(arr);
            return res;
        }
        for(int i=0;i<n-2;i++){
            int sum = nums[i];
            sum = -1*sum;
            int j = i+1;
            int k = n-1;
            while(k>j){
                if(nums[j] + nums[k]<sum){
                    j++;
                    continue;
                }
                if(nums[j]+nums[k] >sum){
                    k--;
                    continue;
                }
                if(nums[j] +nums[k] == sum){
                    vector<int > arr;
                    arr.push_back(-1*sum);
                    arr.push_back(nums[j]);
                    arr.push_back(nums[k]);
                    gabbu.insert(arr);
                    j++;
                    continue;
                }
            }
        }
        for(auto k:gabbu){
            res.push_back(k);
        }
        return res;
    }
};
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