<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) {
         vector<vector<int>>ans;
         vector<int>v(3);
         sort(nums.begin(),nums.end());
         map<vector<int>,bool>b;
       //  cout<<nums[0];
  unordered_map<int,int>m;
 //check for 
 for(int i=0;i<nums.size();i++){

    m[nums[i]]++;
 }
int i=0;
int j=nums.size()-1;
for(int i=0;i<nums.size()-2;i++){
    for(int j=i+1;j<nums.size();j++){
//cout<<i<<" "<<j;
int p=-nums[i]-nums[j];
 v[0]=nums[i];
       v[1]=nums[j];
       v[2]=p;
//cout<<nums[i]<<" "<<nums[j]<<" "<<p<<" "<<"\n";
 
       sort(v.begin(),v.end());
if(m[p]>0){
if(nums[i]==p&&nums[i]!=0){cout<<"bue";
    if(m[nums[i]]>1){
     b[v]=true;
    }
}
else if(nums[j]==p&&nums[j]!=0){

 if(m[nums[j]]>1){
   b[v]=true;
    }

}
else if(nums[i]==nums[j]&&nums[i]==0){
//cout<<"gg";
if(m[0]>2){
   
      b[v]=true;
}}
else{ b[v]=true;
}


}

}
while(i!=nums.size()-2){
if(nums[i]==nums[i+1]){
i++;}
else{
    break;
}
}}


    map<vector<int>, bool>::iterator it = b.begin();

while(it!=b.end()){

    
        vector<int>temp(3);
temp[0]=it->first[0];
 temp[1]=it->first[1];
   temp[2]=it->first[2];
 ans.push_back(temp);

    
    it++;
}
return ans;
}
 
    
};
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