<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;
        vector<int>mp(2000000);
        vector<int>v;
        for(auto i:nums) {
            mp[i + 200000]++;
            if(mp[i + 200000] == 1)
                v.push_back(i);
        }
        n = v.size();
        for(int i = 0;i < n;i++){
            int a = v[i];
            for(int j = i + 1;j < n;++j){
                int b = v[j];
                int c = 0 - (a + b);
                if ( mp[c + 200000] == 0 ) continue;
                else if ( (c == a || c == b) && mp[c + 200000] <= 1)
                    continue;  
                else if ( c == a && a == b && mp[c + 200000] <= 2) 
                    continue;
                else ans.push_back({a,b,c});
            }
        }
        if(mp[200000] >= 3) ans.push_back({0,0,0});
        for(auto &i:ans) 
            sort(i.begin(),i.end());
        sort(ans.begin(),ans.end());
        ans.resize(unique(ans.begin(),ans.end()) - ans.begin());
        return ans;
    }
};
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