<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) {
        ios_base::sync_with_stdio(false);
        cout.tie(NULL);
        cin.tie(NULL);

        int n = nums.size();
        set<vector<int>> set_store;
        unordered_map<int, int> count;
        for(auto i:nums){
            count[i]++;
        }
        if(count[0]>=3){
            vector<int> t = {0,0,0};
            set_store.insert(t);
        }
        for(int i = 0; i<n; i++){
            for(int j = i+1; j<n; j++){
                while(j<n && nums[j]==0){
                    j++;
                }
                if(j==n){
                    continue;
                }
                int comp = 0-(nums[i]+nums[j]);
                if(count[comp]>0){
                    if((comp==nums[i] || comp==nums[j]) && count[comp]==1){
                        continue;
                    }else {

                        vector<int> temp = {nums[i],nums[j],comp};
                        if(temp[0] == 0 && temp[1] == 0 && temp[2] == 0 && count[0]<3){
                            continue;
                        }
                        sort(temp.begin(),temp.end());
                        set_store.insert(temp);
                    }
                }

            }
        }
        vector<vector<int>> ans(set_store.begin(),set_store.end());
        return ans;

    }
};
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