Binary Search
08. Search in Rotated Sorted Array
Given an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown index k (1 <= k < nums.length) such that the resulting array becomes:
[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed).
For example, [0,1,2,4,5,6,7] might be rotated by 3 positions and become [4,5,6,7,0,1,2].
Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not.
You must write an algorithm with O(log n) runtime complexity.
Examples
Example 1
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
Example 3
Input: nums = [1], target = 0
Output: -1
Constraints
- 1 <= nums.length <= 5000
- -10^4 <= nums[i] <= 10^4
- All values of nums are unique
- nums is an ascending array that is possibly rotated
- -10^4 <= target <= 10^4
from typing import List
nums: List[int] = [4,5,6,7,0,1,2]
target: int = 8
l: int = 0
r: int = len(nums) - 1
while l <= r:
mid: int = (l + r) // 2
if nums[mid] == target:
print(mid)
break
if nums[l] <= nums[mid]:
# Left half is sorted
if nums[l] <= target < nums[mid]:
r = mid - 1
else:
l = mid + 1
else:
# Right half is sorted
if nums[mid] < target <= nums[r]:
l = mid + 1
else:
r = mid - 1
else:
print(-1)
# <fix>
# The condition was nums[l] < nums[mid], but when l == mid (2-element window)
# nums[l] == nums[mid] and it fell into the wrong branch, returning -1 for valid targets.
# Fixed to nums[l] <= nums[mid].Keyboard shortcuts
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