<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<int> tmp;
        set<vector<int>> s;
        unordered_map<int, vector<int>> m;
        int j = 0;
        for(int i = 0; i < nums.size(); i++) {
            // we don't need to store more than 3 of the same number
            // as duplicate triplets are not allowed
            if(m[nums[i]].size() > 3) continue;

            m[nums[i]].push_back(j);
            j++;
            tmp.push_back(nums[i]);
        }
        nums = tmp;
        int missing;
        for(int i = 0; i < nums.size() - 2; i++) {
            for(int j = i+1; j < nums.size() - 1; j++) {
                missing = -1 * (nums[i] + nums[j]);
                if(m.find(missing)==m.end()) {
                    continue;
                }

                for(auto& k: m[missing]) {
                    if(k != i && k!= j) {
                        // cout << i << ", " << j << ", " << k << " with " << nums[i] << nums[j] << nums[k] << endl;
                        vector<int> v = {nums[i], nums[j], missing};
                        sort(v.begin(), v.end());
                        s.insert(v);
                    }
                }
            }
        }

        vector<vector<int>> ans;
        for(auto& it: s) {
            ans.push_back(it);
        }

        return ans;
    }
};
81279