81. Search in Rotated Sorted Array II

Description

There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).

Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].

Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.

You must decrease the overall operation steps as much as possible.

Example 1:

• Input: nums = [2,5,6,0,0,1,2], target = 0
• Output: true

Example 2:

• Input: nums = [2,5,6,0,0,1,2], target = 3
• Output: false

Constraints:

• 1 <= nums.length <= 5000
• -10⁴ <= nums[i] <= 10⁴
• nums is guaranteed to be rotated at some pivot.
• -10⁴ <= target <= 10⁴

Follow up: This problem is similar to 33. Search in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

Submitted Code

class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        l, r = 0, len(nums)-1

        while l <= r:
            m = (l + r) // 2

            if nums[m] == target:
                return True

            if nums[l] == nums[m] == nums[r]:       # 양쪽에서 한 칸씩 제거
                l += 1
                r -= 1
            elif nums[l] <= nums[m]:                # 왼쪽 정렬
                if nums[l] <= target < nums[m]:
                    r = m - 1
                else:
                    l = m + 1
            else:                                   # 오른쪽 정렬
                if nums[m] < target <= nums[r]:
                    l = m + 1
                else:
                    r = m - 1

        return False

Runtime: 0 ms | Beats 100.00%
Memory: 19.65 MB | Beats 32.32%

33번 문제와 달리 중복 숫자가 있어서 만약 [1,0,1,1,1] 처럼 왼쪽, 가운데, 오른쪽이 모두 같은 숫자일 경우 어느 쪽이 정렬되었는지 판단할 수 없게 된다. 이 조건만 처리하면 나머지 부분은 33번과 동일하게 진행할 수 있다. 다만 배열의 거의 모든 숫자가 중복인 최악의 경우 선형 탐색처럼 될 수 있다.

Other Solutions

1st

class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        left, right = 0, len(nums) - 1
        
        while left <= right:
            mid = (left + right) // 2
            
            if nums[mid] == target:
                return True
            
            if nums[mid] == nums[left]:
                left += 1
                continue

            if nums[left] <= nums[mid]:
                if nums[left] <= target < nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            else:
                if nums[mid] < target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1
        
        return False

time complexity: 𝑂(𝑛)
space complexity: 𝑂(1)