<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) {
        
        if(nums.size()<3) return {};
        
        sort(nums.begin(), nums.end());
        int n = nums.size();
        
        if(nums.front() == 0 && nums.back() ==0) return vector<vector<int>>(1, vector<int>(3,0));
        
        if(nums[0] > 0) return {};
        vector<vector<int>>ans;
        
        unordered_map<int, int>mp;
        set<vector<int>>st;
        
        for(int i=0; i<n; i++)
        {
            mp[nums[i]] = i;
        }
        
        for(int i=0; i<n-2; i++)
        {
            int sum=0;
            sum+=nums[i];
            // vector<int>v1;
            
            for(int j=i+1; j<n-1; j++)
            {
                vector<int>v1;
                sum+=nums[j];
                int neg_sum = 0-sum;
                if(mp.find(neg_sum) != mp.end() && mp[neg_sum] > j)
                {
                    v1.push_back(nums[i]);
                    v1.push_back(nums[j]);
                    v1.push_back(neg_sum);
                    
                    if(st.find(v1) == st.end()) 
                    {
                        ans.push_back(v1);
                        st.insert(v1);
                    }
                    // break;
                }
                
                sum-=nums[j];
            }
        }
        
        
        return ans;
    }
};
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