<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]]:

        m = {i:1 for i in nums}
        ans = []
        nums.sort()
        negatives = [i for i in nums if i < 0]
        zeroes = [i for i in nums if i == 0]
        positives = [i for i in nums if i > 0]

        if len(nums) < 3:
            return []
        
        if len(zeroes) >= 3:
            ans.append([0,0,0])

        if len(zeroes) > 0:
            for i in set(negatives):
                if m.get(-i):
                    ans.append([i,0,-i])

        for ni in set(negatives):
            ts_ar = self.twoSum(positives, -ni)
            if len(ts_ar) > 0:
                for ts in ts_ar:
                    ans.append([ni,ts[0],ts[1]])

        for pi in set(positives):
            ts_ar = self.twoSum(negatives, -pi)
            if len(ts_ar) > 0:
                for ts in ts_ar:
                    ans.append([ts[0],ts[1],pi])

        return ans

    def twoSum(self, nums, target):
        d={}
        ans = []

        for i in range(len(nums)):
            if i>0 and nums[i]==nums[i-1]:
                if i>1 and nums[i-1]==nums[i-2]:
                    pass
                elif nums[i]+nums[i-1]==target:
                    ans.append([nums[i], nums[i-1]])
            elif d.get(target-nums[i]) is not None:
                ans.append([nums[i], nums[d.get(target-nums[i])]])
            d[nums[i]]=i       
        
        return ans
21000