<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>> outside;
        int low = 0;
        map<int, int> mp;
        for (int i = 0; i < nums.size(); i++) {
            mp[nums[i]]++;
        }
        vector<int> v1={0,0,0};
        vector<vector<int>> outside1={v1};
        if(mp[0]==nums.size())return outside1; 
        sort(nums.begin(), nums.end());
        map<vector<int>, int> Duplicate;
        int high = nums.size() - 1;
        while (nums[low]<0 and low<=high) {
            if(nums[low]==nums[low+1]){
                low++;
                continue;
            }
            high=nums.size() - 1;
            while(nums[high]>0 and high>=low)
            {
                int sum = 0;
                vector<int> v;
                sum += (nums[low] + nums[high]);
                mp[nums[low]]--;
                mp[nums[high]]--;
                int oppSign = sum * (-1); 
                if (mp[oppSign]) {
                    v.push_back(oppSign);
                    v.push_back(nums[low]);
                    v.push_back(nums[high]);
                    sort(v.begin(),v.end());
                    if(Duplicate[v]==0){
                        outside.push_back(v);
                        Duplicate[v]++;
                    }
                }
                mp[nums[low]]++;
                mp[nums[high]]++;
                high--;
            }
            low++;    
        }
        
        if(mp[0]>=3)outside.push_back(v1);
        return outside;
    }
};
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