<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>> ans;
        sort(nums.begin(),nums.end());
        unordered_map<int,int> left,right;
        left[nums[0]]++;
        for(int i=2;i<nums.size();i++){
            right[nums[i]]++;
        }
        for(int i=1;i<nums.size()-1;i++){
            int sum=-nums[i];
            /*cout<<i<<endl;
            for(auto it:left){
                cout<<it.first<<" "<<it.second<<endl;
            }
            cout<<"*"<<endl;
            for(auto it:right){
                cout<<it.first<<" "<<it.second<<endl;
            }
            cout<<"sum:"<<sum<<endl;*/
            for(auto it:left){
                //cout<<it.first<<" "<<it.second<<" "<<sum-it.first<<endl;
                //cout<<right[sum-it.first]<<endl;
                if(it.second>0 and right[sum-it.first]>0){
                    vector<int> vec={it.first,nums[i],sum-it.first};
                    /*for(auto j:vec){
                        cout<<j<<" ";
                    }cout<<endl;*/
                    ans.push_back(vec);
                }
            }
            left[nums[i]]++;
            right[nums[i+1]]--;
        }
        sort(ans.begin(), ans.end());
        ans.erase(unique(ans.begin(), ans.end()), ans.end());
        return ans;
    }
};
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