<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<pair<int, int>> twoSum(vector<int> &nums, int target, int start, map<int, int> &mp)
    {
        vector<pair<int, int>> ans;
        for(int i=start; i<nums.size(); i++)
        {
            if(mp.find(target - nums[i]) != mp.end() && mp[target - nums[i]] > i)
            {
                ans.push_back({i, mp[target - nums[i]]});
            }
        }
        return ans;
    }

    vector<vector<int>> threeSum(vector<int>& nums) {
        map<int, int> mp;
        for(int i=0; i<nums.size(); i++)
        {
            mp[nums[i]] = i;
        }
        set<vector<int>> ans;
        for(int i=0; i<nums.size(); i++)
        {
            vector<pair<int, int>> v = twoSum(nums, -nums[i], i+1, mp);
            vector<int> x;
            for(auto elem: v)
            {
                x = {nums[i], nums[elem.first], nums[elem.second]};
                sort(x.begin(), x.end());
                ans.insert(x);
            }
        }
        vector<vector<int>> ret;
        for(auto elem: ans)
        {
            ret.push_back(elem);
        }
        return ret;
    }
};
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