<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{
private:
    void check_if_duplicate(int a, int b, int c, vector<vector<int>>& ans, unordered_map<string, int>& ans_map)
    {
        vector<int> vec = {a, b, c};
        sort(vec.begin(), vec.end());
        string s = to_string(vec[0]) + "/" + to_string(vec[1]) + "/" + to_string(vec[2]);

        if(ans_map.find(s) == ans_map.end())
        {
            ans.push_back(vec);
            ans_map[s] = 0;
        }
    }
public:
    vector<vector<int>> threeSum(vector<int>& nums){
        vector<vector<int>> ans;
        unordered_map<string, int> ans_map;
        unordered_map<int, vector<int>> nums_map;
        int x, left = 0, right = nums.size()-1;

        sort(nums.begin(), nums.end());

        for(int i = 0; i < nums.size(); i++)
            nums_map[nums[i]].push_back(i);

        while(left < nums.size() && nums[left] < 0)
        {
            while(nums[right] > 0)
            {
                if(nums[right] + nums[left] == 0)
                    x = 0;
                else
                    x = -(nums[right] + nums[left]);
                
                if(nums_map.find(x) != nums_map.end())
                    for(int k: nums_map[x])
                        if(k != left && k != right)
                        {
                            check_if_duplicate(nums[left], nums[right], x, ans, ans_map);
                            break;
                        }
                right--;
            }
            right = nums.size()-1;
            left++;
        }

        if(left < nums.size() && left+1 < nums.size() && left+2 < nums.size())
            if(nums[left] == 0 && nums[left+1] == 0 && nums[left+2] == 0)
                {
                    vector<int> vec = {0,0,0};
                    ans.push_back(vec);
                }
        return ans;
    }
};
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