xample: Input: 5 Output: True because the binary representation of 5 is: 101. Input: 7 Output: False because the binary representation of 7 is: 111. Input: 11 Output: False because the binary representation of 11 is: 1011. Input: 10 Output: True because The binary representation of 10 is: 1010. """
Given a positive integer, check whether it has alternating bits: namely,
if two adjacent bits will always have different values.
For example:
Input: 5
Output: True because the binary representation of 5 is: 101.
Input: 7
Output: False because the binary representation of 7 is: 111.
Input: 11
Output: False because the binary representation of 11 is: 1011.
Input: 10
Output: True because The binary representation of 10 is: 1010.
"""
# Time Complexity - O(number of bits in n)
def has_alternative_bit(n):
first_bit = 0
second_bit = 0
while n:
first_bit = n & 1
if n >> 1:
second_bit = (n >> 1) & 1
if not first_bit ^ second_bit:
return False
else:
return True
n = n >> 1
return True
# Time Complexity - O(1)
def has_alternative_bit_fast(n):
mask1 = int('aaaaaaaa', 16) # for bits ending with zero (...1010)
mask2 = int('55555555',.
Step by step
Solved in 3 steps with 1 images
