<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>> GetTwoSum(vector<int> v, int index)
    {
        int t = 0 - (v[index]);
        int front = index + 1;
        int back = v.size() - 1;
        vector<vector<int>> res;
        while (front < back)
        {
            if (v[front] + v[back] == t)
            {
                vector<int> r;
                r.push_back (v[front]);
                r.push_back (v[back]);
                res.push_back (r);
                ++front;
                --back;
            }
            else if (v[front] + v[back] < t)
            {
                ++front;
            }
            else
            {
                --back;
            }
        }
        return res;
    }

    vector<int> RemoveDups(vector<int>& nums)
    {
        unordered_map <int, int> m;
        for (auto n : nums) 
        {
            m[n]++;
        }
        nums.clear();
        for (auto e : m)
        {
            if (e.second > 3)
            {
                nums.push_back (e.first);
                nums.push_back (e.first);
                nums.push_back (e.first);
            }
            else
            {
                for (auto i = 0; i < e.second; ++i)
                {
                    nums.push_back (e.first);
                }
            }
        }
        return nums;
    }

    vector<vector<int>> RemoveDups2(vector<vector<int>>& res)
    {
        set<vector<int>> s;
        for (auto r : res)
        {
            s.insert(r);
        }
        res.clear();
        for (auto e : s)
        {
            res.push_back (e);
        }
        return res;
    }

    vector<vector<int>> threeSum(vector<int>& nums) 
    {
        vector<vector<int>> res;
        nums = RemoveDups(nums);
    
        sort (nums.begin(), nums.end());
        for (auto i = 0; i < nums.size() - 2; ++i)
        {
            vector<vector<int>> ts = GetTwoSum (nums, i);
            if (ts.size() != 0)
            {
                for (auto v : ts)
                {
                    vector<int> r;
                    r.push_back (nums[i]);
                    r.push_back (v[0]);
                    r.push_back (v[1]);
                    res.push_back (r);
                }
            }
        }
        res = RemoveDups2(res);
        return res;
    }
};
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