<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:
    
    void merge(vector<int> &nums,int s, int mid, int e){
        vector<int> n1(mid-s+1);
        vector<int> n2(e-mid);
        for(int i=0;i<=mid-s;i++)
        {
            n1[i]=nums[s+i];
        }
        
        for(int j=0;j<e-mid;j++){
            n2[j]=nums[j+mid+1];
        }
        
        int count=0, i=0, j=0;
        while(i<(mid-s+1)&&j<(e-mid)){
            if(n1[i]>n2[j]){
                nums[count+s]=n2[j];
                j++;
            }
            else{
                nums[count+s]=n1[i];
                i++;
            }
            count++;
        }
        
        while(i<(mid-s+1)){
            nums[count+s]=n1[i];
            i++;
            count++;
        }
        
        while(j<(e-mid)){
            nums[count+s]=n2[j];
            j++;
            count++;
        }
    }
    
    void mergeSort(vector<int> &nums, int s, int e){
        if(s<e)
        {
            int mid=s+(e-s)/2;
            mergeSort(nums, s, mid);
            mergeSort(nums, mid+1, e);
            merge(nums, s, mid, e);
        }
    }
    
    
    vector<vector<int>> threeSum(vector<int>& nums) 
    {
        multimap<int, int> ma;
        for(int i=0;i<nums.size();i++){
            ma.insert({nums[i],i});
        }
        set<vector<int>> ans;
        mergeSort(nums,0,nums.size()-1);
        for(int i=0;i<nums.size()-2;i++)
        {
            //int j=i+1, k=nums.size()-1;
            
            for(int j=i+1, k=nums.size()-1;j<k;)
            {
                if(nums[j]+nums[k]+nums[i]<0)
                {
                    j++;
                }
                else if(nums[j]+nums[k]+nums[i]>0)
                {
                    k--;
                }
                else //if(nums[j]+nums[k]+nums[i]==0)
                {
                    vector<int> tm(3);
                    tm[0]=nums[i];
                    
                    tm[1]=nums[j];
                    
                    tm[2]=nums[k];
                    ans.insert(tm);
                    j++;
                    k--;
                }
            }
            
        }
        //cout<<ans.size()<<endl;
        vector<vector<int>> ans2;
     
        int level=0;
        
        for (vector<int> v: ans){
            ans2.push_back(v);       
        }
       
        return ans2;
    }
};
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