# [Solved] 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.

## 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`
• `-105 <= nums[i] <= 105`

## 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```