15. 3Sum
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
ArrayExample 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
Approach
Solution
class Solution:
def threeSum(self, nums: list[int]) -> list[list[int]]:
nums.sort()
result = []
n = len(nums)
for i in range(n):
# skip all zero
if i > 0 and nums[i] == nums[i - 1]:
continue
# two pointers
left = i + 1
right = n - 1
target = -nums[i]
while left < right:
current = nums[left] + nums[right]
if current == target:
result.append([nums[i], nums[left], nums[right]])
# skip duplicates
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1
# shift pointers
left += 1
right -= 1
# since sorted, if current < target, then move left
elif current < target:
left += 1
# since sorted, if current > target, then move right
else:
right -= 1
return result