<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>

#include <bits/stdc++.h>
template <typename T> using vt = vector<T>;
typedef uint32_t u32;

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vt<vt<int>> ans;
    
        sort(nums.begin(), nums.end());
        
        u32 n = nums.size();

        u32 count = 1;
        int current = 0;
        for (u32 i = 0; i < n; i++) {
            if (nums[i] != current) {
                count = 1;
                current = nums[i];
            } else if (count > 3) {
                nums.erase(nums.begin() + i);
                n--;
            } else {
                count++;
            }
        }
    
        for (u32 cp = 0; cp < n; cp++) {
            int c = nums[cp];
            u32 ap = 0, bp = n - 1;
    
            for (; ap < bp && ap < cp && cp < bp;) {
                if (abs((double) ap-bp) <= 1) {
                    break;
                }

                int absum = nums[ap] + nums[bp];

                if (absum < -c) {
                    ap++;
                } else if (absum > -c) {
                    bp--;
                } else {
                    vector<int> subans =
                            {nums[ap], nums[bp], nums[cp]};
                    sort(subans.begin(), subans.end());
                    ans.push_back(subans);
                    ap++;
                    bp--;
                }
            }
        }

        sort(ans.begin(), ans.end());
        ans.erase(unique(ans.begin(), ans.end()), ans.end());

        return ans;
    }
};

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