<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>& vec) {
        
        // sort(vec.begin(),vec.end());
        set<int>s;
        unordered_map<int,int>mp;
        vector<int>temp;
        set<multiset<int>> ans;
        vector<vector<int>> ansf;
        for(auto xy:vec)
        {
            if(mp[xy]==0)
            temp.push_back(xy);
            mp[xy]++;
        }
        sort(temp.begin(),temp.end());
        int n=temp.size();
        for(int i=0;i<n;i++)
        {
            for(int j=i;j<n;j++)
            {
                if(i==j && mp[temp[i]]<2)
                continue;
                int t=-1*(temp[i]+temp[j]);
                if(mp[t]>0)
                {
                    if(t==temp[i] && t==temp[j] )
                    {
                    if(mp[t]>=3)
                    ans.insert({temp[i],temp[j],t,});
                    }
                    else if(temp[i]==temp[j])
                    ans.insert({temp[i],temp[j],t});
                    else if((t==temp[i] || t==temp[j]))
                    {
                    if(mp[t]>=2)
                    ans.insert({temp[i],temp[j],t});
                    }
                    else 
                    ans.insert({temp[i],temp[j],t});
                }
            }
        }
        int c=0;
        for(auto xy:ans)
        {
            ansf.push_back({});
            for(auto yz:xy)
            {
                // cout<<yz<<" ";
            ansf[c].push_back(yz);
            }
            // cout<<"\n";
            c++;
        }
        return ansf;
    }
};
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