<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>> res;

        sort(nums.begin(), nums.end());
        int n= nums.size();
        int prev = INT_MIN ;
        unordered_map<string,int> mp; 
        for(int i=0; i<n; i++){
            if(prev == nums[i]){
                continue;
            }
           if(prev == INT_MIN){
               prev = nums[i];
           }

           for(int j = i+1, k = n-1; j<k; ){
               int sum = nums[i]+ nums[j]+nums[k];

               if(sum == 0){
                  vector<int> a;
                  a.push_back(nums[i]);
                  a.push_back(nums[j]);
                  a.push_back(nums[k]);
                  
                  stringstream oss;
                    for (int i : a) {
                        oss << i << " ";
                    }
                    std::string vectorStr = oss.str();
                    vectorStr.pop_back();  // Remove the trailing space
                  if(mp.find(vectorStr) == mp.end()){
                    mp[vectorStr] = 0;
                    res.push_back(a);
                  }

                  k--;
                  j++;
               }
               else if(sum>0){
                   k--;
               }
               else if(sum<0){
                   j++;
               }
           }
        }

        return res;

        
    }
};
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