<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>> output;
        map<int,int> mid; //required number
        unordered_map<long long,bool> end;
        unordered_map<int,int> exists; //number of occurences
        unordered_map<int,int> start; //number of occurences
        
        for(int i=0;i<nums.size();i++){
            exists[nums[i]]++;
            start[nums[i]]++;
        }
        for(auto it=start.begin();it!=start.end();++it){
            int num = -it->first;
            int testNum;
            for(int i=0;i<nums.size();i++){
                testNum = num-nums[i];
                if(exists[num-nums[i]]>0){
                    vector<int> t = {-num,nums[i],num-nums[i]};
                    bool push = false;
                    if(-num==nums[i]&&num-nums[i]==-num){
                        if(exists[nums[i]]>2){
                            push=true;
                        }
                    }
                    else if(num-nums[i]==nums[i]){
                        if(exists[nums[i]]>1){
                            push=true;
                        }
                    }else if(-num == nums[i]){
                        if(exists[nums[i]]>1){
                            push=true;
                        }
                    }else if(-num == num-nums[i]){
                        if(exists[-num]>1){
                            push=true;
                        }
                    }else{
                        push=true;
                    }
                    if(push){
                        sort(t.begin(),t.end());
                        if(!end[(long long)(t[0])*100000+t[1]]){
                            output.push_back(t);
                            end[(long long)(t[0])*100000+t[1]] = true;
                        }
                    }
                }
            }
        }
        return output;
    }
};
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