<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<pair<int, int>> helper(vector<int>& nums, int start, int end, int target) {
        vector<pair<int, int>> pairs;
        while (start < end) {
            int sum = nums[start] + nums[end];
            if (sum < target) {
                start++;
            } else if (sum == target) {
                pairs.push_back({nums[start], nums[end]});
                start++;
                end--;
                // Skip duplicates for the second and third elements
                while (start < end && nums[start] == nums[start - 1]) start++;
                while (start < end && nums[end] == nums[end + 1]) end--;
            } else {
                end--;
            }
        }
        return pairs;
    }

    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> ans;
        sort(nums.begin(), nums.end());
        int n = nums.size();
        
        for (int i = 0; i < n - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;

            int sumToFind = -nums[i];
            auto pairs = helper(nums, i + 1, n - 1, sumToFind);
            for (auto p : pairs) {
                ans.push_back({nums[i], p.first, p.second});
            }
        }

        return ans;
    }
};

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