<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>> out;
        sort(nums.begin(), nums.end());

        for (int i = nums.size()-1; i >= 2; i--)
        {
            vector<vector<int>> ret = twoSum(nums, i);
            out.insert(out.end(), ret.begin(), ret.end());
        }

        return out;
    }

private:
    set<vector<int>> hmap;

    vector<vector<int>> twoSum(vector<int>& numbers, int size)
    {
        vector<vector<int>> out = {};
        int left = 0, right = size-1;
        int target = - numbers[size];

        while (left < right)
        {
            int sum = numbers[left] + numbers[right];

            if (sum == target)
            {
                vector<int> triplet = {numbers[left], numbers[right], numbers[size]};
                if (hmap.find(triplet) == hmap.end())
                {
                    out.push_back(triplet);
                    hmap.insert(triplet);
                }
                left++;
                right--;
            }
            else if (sum > target)
                right--;
            else
                left++;
        }

        return out;
    }
};
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