<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:
    def threeSum(self, nums: List[int]) -> List[List[int]]:

        res = set()

        #1. Split nums into three lists: negative numbers, positive numbers, and zeros
        n, p, z = [], [], []
        for num in nums:
            if num > 0:
                p.append(num)
            elif num < 0: 
                n.append(num)
            else:
                z.append(num)

        #2. Create a separate set for negatives and positives for O(1) look-up times
        N, P = set(n), set(p)

        #3. If there is at least 1 zero in the list, add all cases where -num exists in N and num exists in P
        #   i.e. (-3, 0, 3) = 0
        if z:
            for num in P:
                if -1*num in N:
                    res.add((-1*num, 0, num))

        #3. If there are at least 3 zeros in the list then also include (0, 0, 0) = 0
        if len(z) >= 3:
            res.add((0,0,0))

        #4. For all pairs of negative numbers (-3, -1), check to see if their complement (4)
        #   exists in the positive number set
        for i in range(len(n)):
            for j in range(i+1,len(n)):
                target = -1*(n[i]+n[j])
                if target in P:
                    res.add(tuple(sorted([n[i],n[j],target])))

        #5. For all pairs of positive numbers (1, 1), check to see if their complement (-2)
        #   exists in the negative number set
        for i in range(len(p)):
            for j in range(i+1,len(p)):
                target = -1*(p[i]+p[j])
                if target in N:
                    res.add(tuple(sorted([p[i],p[j],target])))

        return res
            
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