<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>& a) {
        int n = a.size();
        sort(a.begin(), a.end());
        auto two_pointers = [&](int l, int r, int target) {
            vector<tuple<int,int>> v;
            while (r > l) {
                if (a[l] + a[r] == target) {
                    v.push_back(make_tuple(a[l], a[r]));
                    l++, r--;
                } 
                if (a[l] + a[r] < target) {
                    l++;
                }
                if (a[l] + a[r] > target) {
                    r--;
                }
            }
            return v;
        };

        set<tuple<int,int,int>> s;
        for (int i = 0; i < n; i++) {
            auto v = two_pointers(i+1, n-1, 0-a[i]);
            for (auto [x, y] : v) {
                s.insert({a[i], x, y});
            }
        }
        vector<vector<int>> ans;
        for (auto [x, y, z] : s) {
            ans.push_back({x, y, z});
        }
        return ans;
    }
};
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