<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>> st;
        sort(nums.begin(),nums.end());
        //If even after sorting there is no -ve number then 0 sum cant be made
        if(nums[0]>0) return {};

        int n=nums.size();
        for(int k=0;k<n;k++)
        {
            int ele=-nums[k];
            for(int i=k+1,j=n-1;i<j;)
            {
                if(nums[i]+nums[j]==ele)
                { 
                    st.push_back({nums[i],nums[j],nums[k]});  
                    int x1=nums[j];
                    int x2=nums[k];
                    i++;  j--;
                    while(i<j&& nums[i]==x1) i++;
                    while(i<j&& nums[j]==x2) j--;
                }
                else if(nums[i]+nums[j]>ele) j--;
                else i++;
            }
        }
        set<vector<int>> set(st.begin(),st.end());
        return vector<vector<int>>(set.begin(),set.end());
    }
};
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