<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<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
         vector<vector<int>> final;
         set<vector<int>> s;
         for(int i=0 ; i< nums.size(); ++i){
                
            int val = -nums[i];
            cout<<val;
            int j=i+1;
            int k=nums.size()-1;
                    while( j < k ){
                        
                       if(nums[j] + nums[k] == val){
                            vector<int> a = {nums[i],nums[j],nums[k]};
                            sort(a.begin(),a.end());
                            if(s.empty()){
                                s.insert(a);
                                 final.push_back(a);
                            }
                             else if( s.find(a)==s.end()){
                                s.insert(a);
                                final.push_back(a);
                             }
                             --k;
                             ++j;
                        }
                        else if(nums[j] + nums[k]  > val){
                            --k;
                        }
                        else if(nums[j] + nums[k] < val){
                            ++j;
                        }
                }
            
         }
         return final;
    }
};
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