<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:

    void print(vector<int> &nums)
    {
        cout << "[ " << nums[0] << ", ";
        for (int i = 1 ; i < nums.size() ; i++)
        {
            cout << nums[i] << ", ";
        }

        cout << " ]" << endl;
    }
    vector<vector<int>> threeSum(vector<int>& nums) 
    {
        sort(nums.begin(), nums.end());
        int iter = 0;
        vector<vector<int>> out;
        print(nums);
        unordered_map<int,vector<unordered_set<int>>> seen;
        while (iter < nums.size() && nums[iter] <= 0)
        {
            if (seen.find(nums[iter]) == seen.end())
            {
                seen[nums[iter]] = vector(2, unordered_set<int>());
                vector<vector<int>> triplet = twoNums(nums, iter+1, -nums[iter], seen[nums[iter]]);
                if (!triplet.empty())
                    out.insert(out.end(), triplet.begin(), triplet.end());
            }
            iter++;
        }

        return out;
    }

    vector<vector<int>> twoNums(vector<int>& nums, int start, int target, vector<unordered_set<int>> &seen)
    {
        int n = nums.size()-1;
        vector<vector<int>> out;
        for (int i = start ; i < nums.size() - 1; i++)
        {
            int num1 = nums[i];
            int num2 = bs(i+1, n, nums, num1, target);
            vector<int> triplet = {-target, num1, num2};
            if (num1 + num2 == target && unique({num1, num2}, seen))
            {
                out.push_back(triplet);
            }
        }

        return out;
    }

    int bs(int start, int end, vector<int> &nums, int num1, int target)
    {
        int mid = start + (end - start)/2;
        //cout << "   Performing Binary Search : target = -" <<target<< ", num1 = " << num1 <<endl;
        while (start < end)
        {
            mid = start + (end - start)/2;
            if (nums[mid] + num1 > target)
            {
                end = mid - 1;
            }

            else if (nums[mid] + num1 < target)
            {
                start = mid + 1;
            }

            else
            {   
                //cout << "   - num2 = " << nums[mid] << endl;
                return nums[mid];
            }

        }

        //cout << "   - num2 = " << nums[start] << endl;
        return nums[start];
    }

    bool unique(vector<int> triplet, vector<unordered_set<int>> &seen)
    {
        bool ret = false;
        for (int i = 0 ; i < seen.size() ; i++)
        {
            if (seen[i].find(triplet[i]) == seen[i].end())
            {
                seen[i].insert(triplet[i]);
                ret = true;
            }
        }

        return ret;
    }
};
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