<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) {
        sort(nums.begin(), nums.end());

        vector<vector<pair<int, int>>> sum12 (2e5+1);
        vector<int> sum3 (2e5+1, 0);
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            if (i > 0 && nums[i] == nums[i-1]) continue;
            for (int j = i+1; j < n; j++) {
                if (nums[j] == nums[j-1] && j != i+1) continue;
                if ((nums[i] <= 0 && nums[j] >= 0) || (nums[i] >= 0 && nums[j] <= 0)) {
                    continue;
                }
                if (nums[i] + nums[j] > 1e5 || nums[i] + nums[j] < -1e5) {
                    continue;
                } else {
                    sum12[nums[i] + nums[j] + 1e5].push_back({nums[i], nums[j]});
                }
            }
        }

        vector<vector<int>> out;

        for (int i = 0; i < n; i++) {
            sum3[nums[i] + 1e5]++;
        }
        for (int i = 0; i <= 2e5; i++) {
            for (pair<int, int> p : sum12[i]) {
                if (sum3[2e5-i] > 0) {
                    // cout << sum3[2e5-i];
                    vector<int> temp = {p.first, p.second, 100000-i};
                    out.push_back(temp);
                }
            }
        }

        if (sum3[0 + 1e5] >= 3) {
            out.push_back({0, 0, 0});
        }

        if (sum3[0 + 1e5] >= 1) {
            for (int i = 0; i < 1e5; i++) {
                if (sum3[i] > 0 && sum3[2e5-i] > 0) {
                    out.push_back({ i-(int)1e5, (int)1e5 - i, 0});
                }
            }
        }

        return out;
    }
};
356309