<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 itertools import product
class Solution:

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

        nums.sort()

        size = len(nums)
        max_index = size - 1

        if size < 3:
            return []

        retval = []
        left = 0
        while left <= max_index - 2:
            middle, right = left + 1, max_index
            while middle < right:
                tup = nums[left], nums[middle], nums[right]
                total = sum(tup)
                if total == 0:
                    retval.append(tup)
                    middle = increment(nums, middle)
                elif total > 0:
                    right = decrement(nums, right)
                else:
                    middle = increment(nums, middle)
            left = increment(nums, left)

        return retval


def increment(nums, curr):
    index_max = len(nums) - 1
    steps = 1
    compare = nums[curr]
    while curr + steps < index_max:
        if nums[curr + steps] != compare:
            return curr + steps
        steps += 1
    return index_max

def decrement(nums, curr):
    index_min = 0
    steps = -1
    compare = nums[curr]
    while curr + steps > index_min:
        if nums[curr + steps] != compare:
            return curr + steps
        steps -= 1
    return index_min
19300