<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());
        vector<vector<int>> ans;
        set <vector<int>> triplets;
        int n = nums.size();
        unordered_map <int, int> freq;
        for(int i = 0;i<n;i++)
        {
            freq[nums[i]]++;
        }
        if(nums[0] == 0 && freq[0]>=3) return {{0,0,0}};
        if(nums[0] > 0) return {};
        int f, s, t;
        for(int i = 0;i<n;i++)
        {
            f = nums[i];
            freq[f]--;
            for(int j = i + 1;j<n;j++)
            {
                s = nums[j];
                freq[s]--;
                t = -f -s;
                if(t>=s && freq[t] > 0)
                {
                    vector<int> temp = {f,s,t};
                    //sort(temp.begin(), temp.end());
                    triplets.insert(temp);
                }
                freq[s]++;
            }
            freq[f]++;
        }
        for(auto it: triplets)
        {
            ans.push_back(it);
        }
        return ans;
    }
};
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