<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(unordered_map<int, int> mp){
        for(pair<int, int> p : mp){
            cout << p.first << " " << p.second << endl;
        }
    }
    vector<vector<int>> threeSum(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int> > ans;
        sort(nums.begin(), nums.end());
        unordered_map<int, int> counts;
        for(int i : nums){
            counts[i]++;
        }
        vector<int> unique;
        for(pair<int, int> p : counts){
            unique.push_back(p.first);
        }
        n = unique.size();
        set<vector<int> > st;
        for(int i = 0; i < n; i++){
            for(int j = i; j < n; j++){
                int num1 = unique[i];
                int num2 = unique[j];
                counts[num1]--;
                if(counts[num2] <= 0){
                    counts[num1]++;
                    continue;
                }
                counts[num2]--;
                int num3 = -(num1 + num2);
                if(counts.find(num3) != counts.end() && counts[num3] > 0){
                    vector<int> arr = {num1, num2, num3};
                    sort(arr.begin(), arr.end());
                    st.insert(arr);
                }
                counts[num1]++;
                counts[num2]++;
            }
        }
        for(vector<int> v : st){
            ans.push_back(v);
        }
        return ans;
    }
};
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