<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) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> ans;
        int mid=0;
        while (mid<nums.size()&&nums[mid]<0)mid++;
        if((mid+2)<nums.size()&&nums[mid]==0&&nums[mid+1]==0&&nums[mid+2]==0) ans.push_back({0,0,0});
        set<int> positive;
        set<int> negative;
        for (int i =0;i<mid;i++) {
            negative.insert(nums[i]);
        }
        for (int i =mid;i<nums.size();i++) {
            positive.insert(nums[i]);
        }
        set<pair<int,int>> check;
        for (int i =0;i<mid;i++) {
            for (int j=i+1;j<mid;j++) {
                if (check.contains({nums[i],nums[j]})) continue;
                check.insert({nums[i],nums[j]});
                if (positive.contains((nums[i]+nums[j])*-1)) {
                    ans.push_back({nums[i],nums[j],(nums[i]+nums[j])*-1});
                }
            }
        }
        for (int i =mid;i<nums.size();i++) {
            for (int j=i+1;j<nums.size();j++) {
                if (check.contains({nums[i],nums[j]})) continue;
                check.insert({nums[i],nums[j]});
                if (negative.contains((nums[i]+nums[j])*-1)) {
                    ans.push_back({nums[i],nums[j],(nums[i]+nums[j])*-1});
                }
            }
        }
        return ans;
    }
};
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