<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) {
        set<int> s;vector<vector<int>> ans;
     for(auto it : nums){
        s.insert(it);
     }
     if(s.size()==1 && nums[0]==0){
        vector<int> v;
        v={0,0,0};
        ans.push_back(v);
        return ans;
     }
     map<int,int> mp;
     sort(nums.begin(),nums.end());
     for(auto it: nums){
        mp[it]++;
     }
     set<vector<int>> st;
     for(int i=0;i<nums.size();i++){
        for(int j=i+1;j<nums.size();j++){
           int a=-(nums[i]+nums[j]);
           if(a<nums[i]){
              break;
           }
           mp[nums[i]]--;mp[nums[j]]--;
           if(mp.find(a)!=mp.end() && mp[a]>0){
            vector<int> v;
             v={nums[i],nums[j],a};
             sort(v.begin(),v.end());
             st.insert(v);
           }
           mp[nums[i]]++;mp[nums[j]]++;
        }
     }
     for(auto it : st){
        ans.push_back(it);
     }
     return ans;
    }
}; 
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