<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:
    void twoSum(vector<vector<int>> & ans, vector<int> & v, int r, int tgt) {
        if (tgt < v[0] + v[1] || v[r - 1] + v[r] < tgt) return;
        unordered_map<int, int> mp;
        for (int i = r; i >= 0; --i) {
            mp[v[i]] = i;
        }
        int pre = INT_MAX;
        for (int i = r; i >= 0; --i) {
            if (v[i] == pre) continue;
            int x = tgt - v[i];
            if (mp.find(x) != mp.end() && mp[x] < i) {
                ans.push_back({x, v[i], v[r + 1]});
            }
            pre = v[i];
        }
        return;
    }
    
    vector<vector<int>> threeSum(vector<int>& v) {
        vector<vector<int>> ans;
        int n = v.size();
        sort(v.begin(), v.end());
        int pre = INT_MAX;
        for (int i = n - 1; i >= 2; --i) {
            if (v[i] == pre) continue;
            twoSum(ans, v, i - 1, -v[i]);
            pre = v[i];
        }
        return ans;
    }
};
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