<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) {
        vector<vector<int>> answer;
        std::map<int, int> locations;
        std::map<vector<int>, int> test;
        std::sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); i++) {
            locations[nums[i]] = i;
        }
        if (locations.size() == 1 && nums[0] == 0) {
            vector<int> triplet = {0, 0, 0};            
            answer.emplace_back(triplet); 
            return answer;
        }
        for (int i = 0; i < nums.size() - 2; i++) {
            for (int j = i + 1; j < nums.size() - 1; j++) {
                const int pre_sum = -(nums[i] + nums[j]);
                const auto it = locations.find(pre_sum);
                if (it != locations.end()) {
                    const auto index = it->second;
                    if (index == i || index == j) {
                        continue;
                    }
                    vector<int> triplet;
                    if (pre_sum <= nums[i]) {
                         triplet = {pre_sum, nums[i], nums[j]};
                    } else if (pre_sum >= nums[j]) {
                         triplet = {nums[i], nums[j], pre_sum};                        
                    } else {
                         triplet = {nums[i], pre_sum, nums[j]};                                                
                    }
                        
                    if (test.count({triplet[0], triplet[1]}) == 0) {
                        answer.emplace_back(triplet);                                            
                        test[{triplet[0], triplet[1]}] = 0;
                    }
                }                
            }
        }
        
        return answer;
    }
};
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