class Solution(object):
def findMin(self, nums):
if nums[0] <= nums[1]:
return nums[0]
for i in range(len(nums)1):
if nums[i+1] < nums[i]:
return nums[i+1]
We don't need binary search for this problem, do we?

@wonderfuly Using
Binary Search
the time complexity will beO(logn)
while your solution will give usO(n)
. Just imagine this case where you need tofrequently
retrieve theminimum
in a set of largescale different arrays. The difference will be huge. You should try to optimise it instead of just working it out, since you're here to improve yourself, right?
