<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) {
        int n = nums.size();
        set<vector<int>> ans;

        sort(nums.begin(), nums.end());
        int zero_idx = lower_bound(nums.begin(), nums.end(), 0) - nums.begin();
        int pos_idx  = upper_bound(nums.begin(), nums.end(), 0) - nums.begin();

        // all zeros
        if (pos_idx - zero_idx >= 3) {
            ans.insert({0,0,0});
        }

        // one zero, one positive, one negative
        if (pos_idx - zero_idx >= 1) {
            unordered_set<int> negatives;
            for (int i = 0; i < zero_idx; i++) {
                negatives.insert(nums[i]);
            }
            for (int i = pos_idx; i < n; i++) {
                if (negatives.find(-nums[i]) != negatives.end()) {
                    ans.insert({-nums[i],0,nums[i]});
                }
            }
        }

        // two negatives, one positive
        unordered_map<int, vector<pair<int,int>>> negsums;
        for (int i = 0; i < zero_idx; i++) {
            for (int j = i+1; j < zero_idx; j++) {
                negsums[nums[i]+nums[j]].push_back({nums[i],nums[j]});
            }
        }
        for (int i = pos_idx; i < n; i++) {
            if (negsums.find(-nums[i])!=negsums.end()) {
                for (auto& p : negsums[-nums[i]]) {
                    ans.insert({p.first, p.second, nums[i]});
                }
            }
        }

        // two positives, one negative
        unordered_map<int, vector<pair<int,int>>> possums;
        for (int i = pos_idx; i < n; i++) {
            for (int j = i+1; j < n; j++) {
                possums[nums[i]+nums[j]].push_back({nums[i],nums[j]});
            }
        }
        for (int i = 0; i < zero_idx; i++) {
            if (possums.find(-nums[i])!=possums.end()) {
                for (auto& p : possums[-nums[i]]) {
                    ans.insert({p.first, p.second, nums[i]});
                }
            }
        }

        vector<vector<int>> vecans;
        for (auto& a : ans) {
            vecans.push_back(a);
        }
        return vecans;

    }
};
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