<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>

void f(int &i,const int &n,const vector<int>& nums)
{
    while(i<n&&nums[i]==nums[i-1])
        i++;
}
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>>niz;
        int i=0,j,k=3,n,s,n1;
        n=nums.size();
        pair<int,int>x;
        n=nums.size();   
        sort(nums.begin(),nums.end());
        n1=upper_bound(nums.begin(),nums.end(),0)-nums.begin();
        map<int,vector<int>>mapa;
        for(int i=0;i<n1;i++,f(i,n,nums))
            for(int j=i+1;j<n;j++,f(j,n,nums))
            {
                 s=-nums[i]-nums[j];
                 mapa[s].push_back(i);
                 mapa[s].push_back(j);
            }
        if(nums[0]!=0||nums[n-1]!=0||n<3000)     
        for(int i=max(n1-3,0);i<n;i++,f(i,n,nums))
        {
                s=nums[i];
                while(i<n-1&&nums[i]==nums[i+1])
                    i++;
                for(int j=0;j<mapa[nums[i]].size();j+=2)
                if(i>mapa[s][j]&&i>mapa[s][j+1])
                    niz.push_back({nums[mapa[s][j]],nums[mapa[s][j+1]],nums[i]});
        }
        else
        niz.push_back({0,0,0});
        return niz;
    }
};
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