<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) {
        sort(nums.begin(), nums.end()); //NlogN

        int len = nums.size();

        set<vector<int>> ans;

        unordered_map<int, int> m; 
        for(auto &x: nums) m[x]++; 

        for(int i = 0; i < len - 2; i++) {
            if(nums[i] > 0) break;
            if(nums[i] == 0) {
                if(m[0] >= 3) ans.insert({0, 0, 0});
                break;
            }
            //Reduce used index count:
            if(m[nums[i]] == 1) m.erase(nums[i]);
            else m[nums[i]]--;
            
            for(int j = i + 1; j < len - 1; j++) {
                int toFind = -1 * (nums[i] + nums[j]);

                if(m[nums[j]] == 1) m.erase(nums[j]);
                else m[nums[j]]--;

                if(toFind >= nums[j] && m.find(toFind) != m.end()) {
                    ans.insert({nums[i], nums[j], toFind});
                }

                m[nums[j]]++;
            }
        }

        vector<vector<int>> finalAns(ans.begin(), ans.end());

        return finalAns; //O(N^2.logN)
    }
};
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