15. 3Sum
MediumLeetCodeGiven 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
-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup>
How to solve the problem
Sorting the array
nums
first if we want to use two pointer approach.Once we find a triplet we need, we have to do
j++
andk--
in the mean time to find another one. And usingnums[j] == nums[j-1]
andnums[k] == nums[k+1]
to make sure there is no duplicate triplet.while j < k
is essential because we have already donej++
andk--
.Using
ans.append([nums[i], nums[j], nums[k]])
return all the triplets as an element instead of adding any single element intoans
.
Code
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
ans = []
for i in range(len(nums) - 2):
j = i + 1
k = len(nums) - 1
if i > 0 and nums[i] == nums[i - 1]:
continue
while j < k:
s = nums[i] + nums[j] + nums[k]
if s < 0:
j += 1
elif s > 0:
k -= 1
else:
ans.append([nums[i], nums[j], nums[k]])
j += 1
k -= 1
while j < k and nums[j] == nums[j - 1]:
j += 1
while j < k and nums[k] == nums[k + 1]:
k -= 1
return ans
Complexity
Time complexity: O(n²)
- Traversing
nums[i]
is O(n), two pointer is O(n), so in total is O(n²).
Space complexity: O(1)
- We ignore the memory required for the output. For the purpose of complexity analysis, we also ignore sorting algorithm.
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