Intuition
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
Approach
- Sort
- TwoSum2
- Consider duplicate
ref:https://www.youtube.com/watch?v=jzZsG8n2R9A&list=PLot-Xpze53lfOdF3KwpMSFEyfE77zIwiP&ab_channel=NeetCode
Complexity
-
Time complexity: O(nlogn) + O(n^2) = O(n^2)
-
Space complexity: O(1) or O(n) depends on sorting
Code
class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
res = []
nums.sort()
for index, value in enumerate(nums):
if index > 0 and value == nums[index-1]:
continue
# continue next iteration of the loop
#TwoSum2
left, right = index+1, len(nums)-1
while left < right:
threeSum = value + nums[left] + nums[right]
if threeSum > 0:
right -= 1
elif threeSum < 0:
left += 1
else:
res.append([value, nums[left], nums[right]])
# Need to update but what if [-2, -2, 0, 0, 2, 2]
# Need to skip one number from left
# No Need to skip from right since Sum is not right.
left += 1
while nums[left] == nums[left-1] and left < right:
# Need to keep updated if its same
left += 1
return res