Skip to content

Create find_previous_power_of_two.py #10971

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Closed
wants to merge 2 commits into from
Closed
Changes from 1 commit
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
45 changes: 45 additions & 0 deletions bit_manipulation/find_previous_power_of_two.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,45 @@
# https://stackoverflow.com/questions/1322510/given-an-integer-how-do-i-find-the-next-largest-power-of-two-using-bit-twiddlin

def find_previous_power_of_two(n: int) -> int:

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Please provide descriptive name for the parameter: n

"""
Find the largest power of two that is less than or equal to a given integer.

>>> find_previous_power_of_two(10)
8
>>> find_previous_power_of_two(16)
16
>>> find_previous_power_of_two(5)
4
>>> find_previous_power_of_two(1)
1
>>> find_previous_power_of_two(0)
0
>>> find_previous_power_of_two(-5)
Traceback (most recent call last):
...
ValueError: Input must be a non-negative integer
>>> find_previous_power_of_two(10.5)
Traceback (most recent call last):
...
TypeError: Input must be an integer
"""

if not isinstance(n, int):
raise TypeError("Input must be an integer")

if n < 0:
raise ValueError("Input must be a non-negative integer")

if n == 0:
return 0

power = 1
while power <= n:
power <<= 1 # Equivalent to multiplying by 2

return power >> 1

if __name__ == "__main__":
import doctest

doctest.testmod()