<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:

    unordered_map<int,vector<int> > present;
    // map<vector<int>,int> check;

    vector<vector<int>> threeSum(vector<int>& nums) 
    {
        int n = nums.size();

        bool check=true;

        for(int x:nums)
        {
            if(x!=0){check=false; break;}
        }
        if(check) return vector<vector<int>>{{0,0,0}};

        for(int i=0; i<nums.size(); i++)
        {
            present[nums[i]].push_back(i);
        }

        set<vector<int> > ans;


        for(int i=0; i<n; i++)
        {
            for(int j=i+1; j<n; j++)
            {
                int a = nums[i];
                int b = nums[j];
                int c = 0-a-b;
                vector<int> &x = present[c];
                auto last = upper_bound(x.begin(),x.end(),j);
                if(last!=x.end())
                {
                    vector<int> ans1 = {a,b,c};
                    sort(ans1.begin(),ans1.end());
                    ans.insert(ans1);
                }
            }
        }

        vector<vector<int>> final;
        final.reserve(ans.size()*3);

        for(auto &v:ans) final.push_back(v);

        return final;
    }
};
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