<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) {
        int size = nums.size();
        sort(nums.begin(), nums.end());

        if(size<3)
            return {{}};
        
        if(nums[0] == nums[size-1] && nums[0] == 0)
            return {{nums[0], nums[0], nums[0]}};

        set<vector<int>> ans;
        for(int first = 0; first < size; first++) {
            int second = 0;
            int third = size-1;

            int requiredSum = -1*nums[first];
            while(second<third) {
                if(second==first)
                    second++;
                else if(third==first)
                    third--;
                else {
                    if(requiredSum == nums[second] + nums[third]) {
                        vector<int> temp = {nums[first], nums[second], nums[third]};
                        sort(temp.begin(), temp.end());
                        ans.insert(temp);
                        second++;
                    }
                    else if(requiredSum > nums[second] + nums[third])
                        second++;
                    else 
                        third--;
                }
            }
            
        }

        vector<vector<int>> answer;
        for(auto a: ans)
            answer.push_back(a);
        
        return answer;
    }
};
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