<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>

from typing import List, Set, Tuple

class Solution:
    # my twoSum assumes sorted nums array and returns unique set of pairs that sum to target target
    def twoSum(self, sortedNums: List[int], left, right, target: int) -> Set[Tuple[int, int]]:
        if len(sortedNums) < 2:
            return set()

        resultSet = set()
        
        while left < right:
            currSum = sortedNums[left] + sortedNums[right]
            if currSum == target:
                resultSet.add((sortedNums[left], sortedNums[right]))
                left += 1
                right -= 1
            elif currSum < target:
                left += 1
            else:
                right -= 1
        return resultSet

    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        resultSet = set()

        listLen = len(nums)
        for i in range(listLen - 2):
            firstElement = nums[i]
            rightArrayPairSet = self.twoSum(nums, left=(i+1), right=(listLen-1), target=(0 - firstElement))
            
            for pair in rightArrayPairSet:
                resultSet.add((firstElement, pair[0], pair[1]))

        return resultSet

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