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0015 3Sum.py
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0015 3Sum.py
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class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
"""
https://leetcode.com/problems/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.
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.
https://youtu.be/jzZsG8n2R9A
"""
nums.sort() # sort the numbers
triplets = [] # initialize triplets
for i in range(0, len(nums) - 2):
if i > 0 and nums[i-1] == nums[i]: # checking from 2nd element onwards
continue # skips duplicate number
l = i + 1 # `l` starts from the second element,
r = len(nums) - 1 # `r` start from the last element
while l < r:
triplet = (nums[i], nums[l], nums[r]) # define `triplet`
current_sum = sum(triplet) # `current_sum`
if current_sum < 0: # if `current_sum` < 0,
l += 1 # increase `l`
elif current_sum > 0: # if `current_sum` > 0,
r -= 1 # decrease `r`
else: # if `sum` = 0
triplets.append(triplet) # Add triplet to triplets
r -= 1 # decrease `r`, keep moving
while l < r and nums[r] == nums[r+1]: # skips duplicate number
r -= 1 # decrease `r`, keep moving
return triplets