[Python] Optimized approach for Counting odd Numbers in an interval range [low,high] in python...

Given two non-negative integers low and high. Return the count of odd numbers between low and high (inclusive).
I solved it multiple times using loops but those methods were not optimal enough to get my answer submitted is there much better alternative?

So The Thing is we can easily solve it using loops but is there any much more optimal way to do this faster ?

class solution:
def countOdds(self,low:int,high:int) -> int:
count = 0
for i in range (low,high-1):
if i % 2 != 0:
count+=1
return count


This was the most optimal answer that i was able to get is there much more optimal answer for this ?
Specifically in Python.

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