<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<pair<int, int>> new_pos{{0, 1}};
        new_pos.reserve(nums.size());
        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] == nums[i - 1]) {
                new_pos.back().second++;
            } else {
                new_pos.emplace_back(i, 1);
            }
        }

        unordered_set<int> nums_set(nums.begin(), nums.end());
        vector<vector<int>> result;
        int target = 0;
        for (int a = 0; a < new_pos.size(); a++) {
            const auto &[i, cnta] = new_pos[a];
            const int numsi = nums[i];
            if (numsi > 0) {
                break;
            }
            if (numsi == 0) {
                if (cnta >= 3) {
                    result.push_back({0, 0, 0});
                }
                break;
            }
            target = -2 * numsi;
            if (cnta >= 2 && nums_set.contains(target)) {
                result.push_back({numsi, numsi, target});
            }
            for (int b = a + 1; b < new_pos.size(); b++) {
                const auto &[j, cntb] = new_pos[b];
                const int numsj = nums[j];
                if (numsj > -numsi / 2) {
                    break;
                }
                if (numsi + 2 * numsj == 0) {
                    if (cntb >= 2) {
                        result.push_back({numsi, numsj, numsj});
                    }
                    break;
                }
                target = -numsi - numsj;
                if (nums_set.contains(target)) {
                    result.push_back({numsi, numsj, target});
                }
            }
        }

        return result;
    }
};
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