# Python Solution (108 ms)

• import math
import functools
import operator
class Solution(object):
def rangeBitwiseAnd(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
if m == 0:
return 0

if m == n:
return n

if int(math.log(m,2)) < int(math.log(n,2)):
return 0

return functools.reduce(operator.and_, range(m,n+1))

Explanation:

1.m==0

Anything ANDed with 0 is always 0.

2.m==n

If the first and last number are the same, any number ANDed with itself is always itself.

1. int(math.log(m,2)) < int(math.log(n,2))

The number of digits in a binary number is int(math.log(DECIMAL_NUMBER,2)) + 1. If the starting number, m, has less digits in its binary representation than ending number, n, does, that means that somewhere in that range there exists a number with a binary representation with 1 followed by int(math.log(NUMBER,2)) number of 0's, and m will be some number starting with 0 and followed by int(math.log(NUMBER,2)) digits

e.g. With (m=5, n=9), 5=="0101", 6=="0110", 7=="0111", 8=="1000", 5&6&7&8 == 0.

1. return functools.reduce(operator.and_, range(m,n+1))

If the integer range has passed all previous checks, return the AND of all the numbers.

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