153. Find Minimum in Rotated Sorted Array
# O(log(n)) time || O(1) space
def find_min(self, nums: List[int]) -> int:
low, high = 0, len(nums) - 1
while low < high:
mid = low + (high - low) // 2
if nums[mid] > nums[high]:
low = mid + 1
else:
high = mid
return nums[high]
array might be offset but still is sorted.
use binary search, take in consider in which part to have a search.
if mid element bigger than last one, low border is mid + 1