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]]
Example 2:
Input: nums = [] Output: []
Example 3:
Input: nums = [0] Output: []
Constraints:
0 <= nums.length <= 3000
-105 <= nums[i] <= 105
Solution:
Sort based algorithm
- a+b = -c. 3SUM reduces to 2SUM problem.
Handling Duplicates in 2SUM
- Say index s and e are forming a solution in a sorted array. Now givens nums[s], there is a unique nums[e] such that nums[s]+nums[e]=Target. Therefore, if nums[s+1] is the same as nums[s], then searching in range s+1 to e will give us a duplicate solution. Thus we must move s till nums[s] != nums[s-1] to avoid getting duplicates.
while s<e and nums[s] == nums[s-1]:
s = s+1
Handling Duplicates in 3SUM
- Imagine we are at index i and we have invoked the 2SUM problem from index i+1 to end of the array. Now once the 2SUM terminates, we will have a list of all triplets which include nums[i]. To avoid duplicates, we must skip all nums[i] where nums[i] == nums[i-1].
if i > 0 and nums[i] == nums[i-1]:
continue
Code
class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
nums.sort()
N, result = len(nums), []
for i in range(N):
if i > 0 and nums[i] == nums[i-1]:
continue
target = nums[i]*-1
s,e = i+1, N-1
while s<e:
if nums[s]+nums[e] == target:
result.append([nums[i], nums[s], nums[e]])
s = s+1
while s<e and nums[s] == nums[s-1]:
s = s+1
elif nums[s] + nums[e] < target:
s = s+1
else:
e = e-1
return result