<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>

//first come to twoSum

class Solution {
public:

    vector<vector<int>> threeSum(vector<int>& n) 
    {
        auto print = [&]()
        {
            for(auto nn : n)
                cout << nn << " ";
            cout << endl;
        };

        std::sort(n.begin(), n.end());
        
        //value to indicies map for fast lookups in twoSum
        unordered_map<int, vector<int>> m;
        for(int i = 0; i < n.size(); i++)
            m[n[i]].push_back(i);
        

        auto twoSum = [&](int skip,  int target) -> set<pair<int, int>>
        {
            set<pair<int, int>> r;
            for(int i = skip+1; i < n.size(); i++)
            {
                auto x = -(target + n[i]);
                if(auto it = m.find(x); it != m.end())
                {
                    auto [_, jj] = *it;
                    for(auto j : jj)
                        if(j > i)
                            r.insert({n[i], n[j]});
                }
            }
            return r;
            
        };

        if(n.size() < 2)
            return {};

       
        set<int> visited;
        vector<vector<int>> r;
        for(int i = 0; i < (int)n.size()-2; i++)
        {
            //target
            auto t = n[i];
            
           
            if(m[t].size() > 1)
            {
                 //avoid duplication for the same target at different index
                if(visited.find(t) != visited.end())
                continue;
                visited.insert(t);
            }

            for(auto [j, k] : twoSum(i, t))
            {
                r.emplace_back(vector<int>{t, j, k});
            }
        }
            
        return r;
    }
};
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