 # LeetCode Minimum Swaps to Group All 1's Together Solution

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 + nums + nums = (-1) + 0 + 1 = 0`.
`nums + nums + nums = 0 + 1 + (-1) = 0`.
`nums + nums + nums = (-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`

## Solution

Since we have learned how to find 2 Sum given a sorted array, 3 Sum is just one more number added onto the 2 sum. We can wrap the two pointer 2 sum implementation by a for loop that chooses the first element of the 3 sum for us. Using the deduplication patterns from Deduplication, we can easily find unique triplet that sums to 0.

• We avoid duplicate first element by checking whether `nums[i] == nums[i-1]` in every iteration.
• We avoid duplicating the second element by checking whether `nums[l] == nums[l-1]` in the selection of the left pointer.
• The value of the third element is based on the first and second elements, thus the triplets will be unique.

#### Implementation

``````1def threeSum(self, nums: List[int]) -> List[List[int]]:
2    nums.sort()
3    res = []
4    for i in range(len(nums)):
5        if nums[i] > 0 or i > 0 and nums[i] == nums[i-1]: continue
6        l, r = i+1, len(nums)-1
7        while l < r:
8            total = nums[i] + nums[l] + nums[r]
9            if total == 0:
10                res.append([nums[i], nums[l], nums[r]])
11                l, r = l+1, r-1
12                while l < len(nums) -1 and nums[l] == nums[l-1]:
13                    l += 1
14            elif total > 0:
15                r -= 1
16            else:
17                l += 1
18    return res``````