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.
solution
For any given m value, the pivot will sit on either the left or right. We should make our determination of where to split by using the side that doesn’t have the pivot (in other words, is properly sorted).
We can handle duplicates by special casing when nums[l] == nums[m] == nums[r]. In this case, we can’t determine which side is sorted, and we also know none of those values are target. So, we can move l and r inwards.
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
# we can't determine which side is sorted
if nums[l] == nums[m] == nums[r]:
l += 1
r -= 1
# [l:m] is sorted, check it
elif nums[l] <= nums[m]:
# it's in the left half, remove right half
if nums[l] <= target <= nums[m]:
r = m-1
else: # it's not in left half, look in right half
l = m+1
# [l:m] is rotated, binary search on [m:r]
else:
# target is in right half, omit left
if nums[m] <= target <= nums[r]:
l = m+1
else:
r = m-1
return False