There is 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 pivot index k (1 <= 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, [0,1,2,4,5,6,7] might be rotated at pivot index 3 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 in nums.
You must write an algorithm with O(log n) runtime complexity.
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 -
-104 <= nums[i] <= 104 - All values of
numsare unique. -
numsis an ascending array that is possibly rotated. -
-104 <= target <= 104
SOLUTION:
class Solution:
def getPivot(self, nums):
n = len(nums)
beg = 0
end = n
while beg <= end:
mid = (beg + end) // 2
if mid < n - 1 and nums[mid] > nums[mid + 1]:
return mid + 1
if beg >= end - 1:
return 0
elif nums[mid] > nums[0]:
beg = mid
else:
end = mid
def binarySearch(self, nums, beg, end, target):
while beg <= end:
mid = (beg + end) // 2
if mid >= end:
break
elif nums[mid] == target:
return mid
elif beg == end:
break
elif nums[mid] > target:
end = mid
else:
beg = mid + 1
return -1
def search(self, nums: List[int], target: int) -> int:
n = len(nums)
k = self.getPivot(nums)
left = self.binarySearch(nums, 0, k, target)
if left != -1:
return left
right = self.binarySearch(nums, k, n, target)
if right != -1:
return right
return -1
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