[LeetCode] 3Sum

yoonene·2023년 1월 23일
0

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난이도 : ⭐️⭐️

첫 번째 제출 (시간초과)

itertools의 combinations 사용한 풀이
combinations 시간복잡도 : n! / r! / (n - r)!
조합의 시간복잡도는 위와 같이 크기 때문에 n이 커질수록 복잡도가 커진다. 따라서 n이 큰 입력의 경우 시간초과가 발생한다.

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        lst = []
        for a,b,c in combinations(nums, 3):
            if a+b+c == 0 and sorted([a,b,c]) not in lst:
                lst.append(sorted([a,b,c]))
        
        return lst

두 번째 제출 (정답)

투 포인터를 사용하는 방법을 통해 O(n^3)을 O(n^2)으로 줄일 수 있다.

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        result = list()
        for i in range(len(nums)-2):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            left, right = i + 1, len(nums)-1
            while left < right:
                sum = nums[i] + nums[left] + nums[right]
                if sum > 0:
                    right -= 1
                elif sum < 0:
                    left += 1
                else:
                    result.append([nums[i], nums[left], nums[right]])
                    while left < right and nums[left] == nums[left+1]:
                        left += 1
                    while left < right and nums[right] == nums[right-1]:
                        right -= 1
                    left += 1
                    right -= 1
        return result

세 번째 제출 (시간초과)

두 번째 제출에서 합이 0일 때의 값과 다음 지정될 값이 같을 때 continue로 넘겨주지 않으면 시간초과가 날 지 궁금해서 제출해봤다. 예상대로 시간초과가 났다.

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        nums.sort()
        result = list()
        for i in range(len(nums)-2):
            left, right = i + 1, len(nums)-1
            while left < right:
                sum = nums[i] + nums[left] + nums[right]
                if sum > 0:
                    right -= 1
                elif sum < 0:
                    left += 1
                else:
                    if [nums[i], nums[left], nums[right]] not in result:
                        result.append([nums[i], nums[left], nums[right]])
                    left += 1
                    right -= 1

        return result

INPUT 예시

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같은 입력이 저렇게 많을 때마다 연산을 하느라 timeout이 난다.
합이 0일 때의 값이 이후에 지정될 값과 같다면 굳이 같은 연산을 할 필요가 없기 때문에 continue로 넘겨버려야 한다.

+)

  • 투 포인트를 쓸 때는 당연히 정렬 후 쓰는 경우가 많다.
  • sum > 0 일 땐 right 포인터를 왼쪽으로 옮겨 값을 작게, 반대일 땐 left를 오른쪽으로 옮겨줘야 한다는 생각을 해야 했다.
  • 시간초과가 날 수 있는 예외값들을 고려해서 같은 값이 나오면 넘어가도록 설정해야 했다.
  • 항상 최악의 입력에 대해 예외처리를 해줘야 한다는 것을 잊지 말자.
    (최대 입력, 최소 입력 등)
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