<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) {
        set<tuple<int, int, int>> res;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            int j = i + 1;
            int k = n - 1;
            int target = -nums[i];
            while (j < k) {
                int sum = nums[j] + nums[k];
                if (sum == target) {
                    int tup[3] = {nums[i], nums[j], nums[k]};
                    sort(tup, tup + 3);
                    res.emplace(tup[0], tup[1], tup[2]);
                    ++j;
                    --k;
                } else if (sum < target) {
                    ++j;
                } else {
                    --k;
                }
            }
        }
        vector<vector<int>> vec;
        for (auto &&t : res) {
            vec.push_back(vector<int>{std::get<0>(t),std::get<1>(t),std::get<2>(t)});
        }
        return vec;
    }
};
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