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Binary search.py
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Binary search.py
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"""
Question one
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:
[4,5,6,7,0,1,2] if it was rotated 4 times.
[0,1,2,4,5,6,7] if it was rotated 7 times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
Given the sorted rotated array nums of unique elements, return the minimum element of this array.
You must write an algorithm that runs in O(log n) time.
Ref: LeetCode
"""
class Solution:
def findMin(self, nums: List[int]) -> int:
min_aixs, max_aixs = 0, len(nums) - 1
while max_aixs > min_aixs:
mid_aixs = (min_aixs + max_aixs) // 2
if nums[mid_aixs] < nums[max_aixs]:
max_aixs = mid_aixs
else:
min_aixs = mid_aixs + 1
return nums[min_aixs]
"""
Question two
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:
[4,5,6,7,0,1,4] if it was rotated 4 times.
[0,1,4,4,5,6,7] if it was rotated 7 times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array.
You must decrease the overall operation steps as much as possible.
Ref: LeetCode
"""
class Solution:
def findMin(self, nums: List[int]) -> int:
min_aixs, max_aixs = 0, len(nums) - 1
while max_aixs > min_aixs:
mid_aixs = (min_aixs + max_aixs) // 2
if nums[mid_aixs] < nums[max_aixs]:
max_aixs = mid_aixs
elif nums[mid_aixs] == nums[max_aixs]:
max_aixs -= 1
else:
min_aixs = mid_aixs + 1
return nums[min_aixs]