<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:
    # time complex: O(n**2)
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        # follows https://en.wikipedia.org/wiki/3SUM
        ans = set()
        nums.sort()
        len_ = len(nums)
        for i in range(len_ - 1):
            a = nums[i]
            if a > 0:
                # array is sorted, if we have bigger than 0 numbers already,
                # there won't be any more negative numbers further
                break

            start = i + 1
            end = len_ - 1
            target = -a
            while start < end:
                b = nums[start]
                c = nums[end]
                if b + c == target:
                    #if [a, b, c] not in ans:
                    ans.add((a, b, c))
                    start += 1
                    end -= 1

                    while nums[start] == nums[start-1] and start < end:
                        # if previous number is the same as current
                        # there is no point in checking 3sum
                        start += 1

                elif b + c > target:
                    end -= 1
                else:
                    start += 1
        return ans


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