<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]]:
        num_to_index = {}
        known = set()
        for i, n in enumerate(nums):
            if n in num_to_index:
                num_to_index[n].append(i)
            else:
                num_to_index[n] = [i]
        ret = []
        unique_indexes = [num_to_index[i][0] for i in num_to_index.keys()]
        for i in unique_indexes:
            for y in range(i+1, len(nums)):
                # we need x such that x + nums[i] + nums[y] = 0
                searching = -nums[i]-nums[y]
                if searching in num_to_index and len(num_to_index[searching]) > 0:
                    k = 0
                    while k < len(num_to_index[searching]):
                        index = num_to_index[searching][k]
                        if index != i and index != y:
                            break
                        k+=1
                    if k == len(num_to_index[searching]):
                        continue
                    triple = [nums[i], nums[y], searching]
                    triple.sort()
                    string = str(triple[0]) + "," + str(triple[1]) + "," + str(triple[2])
                    if not string in known:
                        ret.append(triple)
                        known.add(string)
        return ret
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