<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> > fin;
    set<vector<int> > res;
    if(nums.size()<3)
        return fin;

    sort(nums.begin(), nums.end());
    int p=0;
    int q=1;
    int r=nums.size()-1;


    int posl=r;

    while(p+2<nums.size())
    {
        if(q>=r)
        {
            p++;
            q=p+1;
            r=nums.size()-1;
            if(p> r || p+2>=nums.size()) break;
        }

        int sum=nums[p]+nums[q]+nums[r];
        if(!sum)
        {
            vector<int> v;
            v.push_back(nums[p]);
            v.push_back(nums[q]);
            v.push_back(nums[r]);
            res.insert(v);
            q++;
            posl=r;
            if(nums[p]==nums[q] && nums[q]==nums[r]) break;
            continue;


        }
        if(sum>0)
            r--;
        else
        {
            q++;
        }




    }


    for(const auto& i : res)
        fin.push_back(i);
   
    
    return fin;

}
};
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