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

      unordered_map<int,int>mp;

      sort(nums.begin(),nums.end());

      for(int i = 0;i<n;i++)
      {
        if(i>0 and nums[i] == nums[i-1]) continue;
         int sum = -1 * (nums[i]);
         for(int j = i+1;j<n;j++)
         {
            if(mp.find(nums[j])!=mp.end()) {
                if(mp[nums[j]] == 1)
                {  
                    ans.push_back({nums[i],sum - nums[j],nums[j]});
                    mp[nums[j]]++;
                }
                 
            }
            else if(mp[sum - nums[j]] == 0) mp[sum - nums[j]]++;
             
         }  
         //so that at each iteration new pairs are formed and no old pairs interfere
         mp.clear();


      }
        return ans;
    }
};
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