Hard
Count Binary Palindromic Numbers — Python
Full explanation · Time O(logn) · Space O(1)
# Time: O(logn)
# Space: O(1)
# bitmasks, combinatorics
class Solution(object):
def countBinaryPalindromes(self, n):
"""
:type n: int
:rtype: int
"""
def length(n):
result = 0
while n:
result += 1
n >>= 1
return result
def reverse(n, l):
result = 0
for i in xrange(l):
if n&(1<<i):
result |= 1<<((l-1)-i)
return result
l = length(n)//2
return ((1<<l)-1)+(n>>l)+int(((n>>l)<<l)|reverse(n>>(length(n)-l), l) <= n)
# Time: O(logn)
# Space: O(logn)
# bitmasks, combinatorics
class Solution2(object):
def countBinaryPalindromes(self, n):
"""
:type n: int
:rtype: int
"""
s = map(int, bin(n)[2:])
l = len(s)//2
return ((1<<l)-1)+(n>>l)+int(s[:len(s)-l]+s[:l][::-1] <= s)