mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-18 17:20:16 +00:00
61 lines
1.3 KiB
Python
61 lines
1.3 KiB
Python
|
"""
|
||
|
Author : Naman Sharma
|
||
|
Date : October 2, 2023
|
||
|
|
||
|
Task:
|
||
|
To Find the largest power of 2 less than or equal to a given number.
|
||
|
|
||
|
Implementation notes: Use bit manipulation.
|
||
|
We start from 1 & left shift the set bit to check if (res<<1)<=number.
|
||
|
Each left bit shift represents a pow of 2.
|
||
|
|
||
|
For example:
|
||
|
number: 15
|
||
|
res: 1 0b1
|
||
|
2 0b10
|
||
|
4 0b100
|
||
|
8 0b1000
|
||
|
16 0b10000 (Exit)
|
||
|
"""
|
||
|
|
||
|
|
||
|
def largest_pow_of_two_le_num(number: int) -> int:
|
||
|
"""
|
||
|
Return the largest power of two less than or equal to a number.
|
||
|
|
||
|
>>> largest_pow_of_two_le_num(0)
|
||
|
0
|
||
|
>>> largest_pow_of_two_le_num(1)
|
||
|
1
|
||
|
>>> largest_pow_of_two_le_num(-1)
|
||
|
0
|
||
|
>>> largest_pow_of_two_le_num(3)
|
||
|
2
|
||
|
>>> largest_pow_of_two_le_num(15)
|
||
|
8
|
||
|
>>> largest_pow_of_two_le_num(99)
|
||
|
64
|
||
|
>>> largest_pow_of_two_le_num(178)
|
||
|
128
|
||
|
>>> largest_pow_of_two_le_num(999999)
|
||
|
524288
|
||
|
>>> largest_pow_of_two_le_num(99.9)
|
||
|
Traceback (most recent call last):
|
||
|
...
|
||
|
TypeError: Input value must be a 'int' type
|
||
|
"""
|
||
|
if isinstance(number, float):
|
||
|
raise TypeError("Input value must be a 'int' type")
|
||
|
if number <= 0:
|
||
|
return 0
|
||
|
res = 1
|
||
|
while (res << 1) <= number:
|
||
|
res <<= 1
|
||
|
return res
|
||
|
|
||
|
|
||
|
if __name__ == "__main__":
|
||
|
import doctest
|
||
|
|
||
|
doctest.testmod()
|