Minimum Swaps to Group All 1's Together
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
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
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