Find Minimum in Rotated Sorted Array
Given two sorted arrays nums1 and nums2 of size `m` and `n` respectively, return the median of the two sorted arrays
Problem Statement
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[n1]]
1 time results in the array [a[n1], a[0], a[1], a[2], ..., a[n2]]
.
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.
Example 1:
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3
 Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:
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3
 Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:
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3
 Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

Constraints:
n
== nums.length
 1 <=
n
<= 5000  5000 <=
nums[i]
<= 5000  All the integers of nums are unique.
 nums is sorted and rotated between 1 and
n
times.
See the problem
Solution
Python3 Solution:
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class Solution:
def rotation_point(self, nums):
n = len(nums)
start = 0
end = n  1
if n == 1:
return 0
while True:
mid = (start + end) // 2
if (
nums[mid] <= nums[(mid + 1 + n) % n]
and nums[mid] < nums[(mid  1 + n) % n]
):
return mid
elif nums[mid] > nums[end]:
start = mid + 1
else:
end = mid  1
def findMin(self, nums: List[int]) > int:
i = self.rotation_point(nums)
return nums[i]

Test Cases
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 [3,4,5,1,2]
[3,4,5,6, 7, 0,1, 2]
[1]
[1,2, 3, 4]
[5,1,2,3,4]
[2,3,4,5,1]
[1,2]
[2,1]
[1,2,3]
[3,1,2]
[2,3,1]
