## Question

Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]`

such that `i != j`

, `i != k`

, and `j != k`

, and `nums[i] + nums[j] + nums[k] == 0`

.

Notice that the solution set must not contain duplicate triplets.

**Example 1:**

Input:nums = [-1,0,1,2,-1,-4]Output:[[-1,-1,2],[-1,0,1]]Explanation: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.

**Example 2:**

Input:nums = [0,1,1]Output:[]Explanation:The only possible triplet does not sum up to 0.

**Example 3:**

Input:nums = [0,0,0]Output:[[0,0,0]]Explanation:The only possible triplet sums up to 0.

**Constraints:**

`3 <= nums.length <= 3000`

`-10`

^{5}<= nums[i] <= 10^{5}

## Python Solution

class Solution: def threeSum(self, nums: List[int]) -> List[List[int]]: nums=sorted(nums) l=[] for j in range(len(nums)): target = 0-nums[j] d={} for i in range(j+1,len(nums)): if nums[i] in d: l.append((nums[j],nums[d[nums[i]]],nums[i])) else: d[target-nums[i]] = i l=set(l) l=[list(a) for a in l] return l