<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>

const int OFFSET = 1e5;

class Solution {
public:
    int cnt[200001];
    void add(int val) {
        ++cnt[val + OFFSET];
    }
    void remove(int val) {
        --cnt[val + OFFSET];
    }

    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(begin(nums), end(nums));
        vector<vector<int>> res;
        set<vector<int>> S;
        for (auto x: nums) add(x);

        for (int i = 0; i< nums.size(); ++i) {
            remove(nums[i]);
            for (int j = i + 1; j < nums.size(); ++j) {
                remove(nums[j]);
                if ((abs(-nums[i] - nums[j]) <= 1e5) && cnt[-nums[i] - nums[j] + OFFSET] > 0) {
                    vector<int> tmp = {nums[i], nums[j], -nums[i] - nums[j]};
                    sort(begin(tmp), end(tmp));
                    if (!S.count(tmp)) {
                        res.push_back(tmp);
                        S.insert(tmp);
                    }
                    if (tmp[0] == 0 && tmp[1] == 0 && tmp[2] == 0) {
                        return res;
                    }
                }
                add(nums[j]);
            }
            add(nums[i]);
        }
        return res;
    }
};
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