<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) {
        vector<vector<int>> res;
        set<vector<int>> seen;

        sort(nums.begin(), nums.end());

        for (int i = 0; i < nums.size() - 2; i++){
            int left = i + 1;
            int right = nums.size() - 1;

            int l = nums[i];
            int m = nums[left];
            int r = nums[right];
            int s = l + m + r;
            while (left < right){
                m = nums[left];
                r = nums[right];
                s = l + m + r;
                if (s > 0){
                    right -= 1;
                } else if (s < 0){
                    left += 1;
                }
                if (s == 0 && left < right){
                    vector<int> temp_res;
                    temp_res.push_back(l);
                    temp_res.push_back(m);
                    temp_res.push_back(r);
                    left += 1;
                    right -= 1;
                    if (!seen.contains(temp_res)){
                        res.push_back(temp_res);
                        seen.insert(temp_res);
                        while (left < right && m == nums[left]){
                            left +=1;
                        }
                        while (left < right && r == nums[right]){
                            right -=1;
                        }
                    }
                }
                if (l == r){
                    break;
                }
            }
        }
        return res;
    }
};
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