'''
Given a positive integer, check whether it has alternating bits: namely, if two adjacent bits will always have different values.

Example 1:
Input: 5
Output: True
Explanation:
The binary representation of 5 is: 101
Example 2:
Input: 7
Output: False
Explanation:
The binary representation of 7 is: 111.
Example 3:
Input: 11
Output: False
Explanation:
The binary representation of 11 is: 1011.
Example 4:
Input: 10
Output: True
Explanation:
The binary representation of 10 is: 1010.

References: 1) divmod(a,b) divied a/b and gives the quotient and remainder.
            2) n//2 will ignore the remainder if its not 0 and returns the quotient
            3) all returns true, if for every check in an array is true. if not false.

Other way to do this is - 
def hasAlternatingBits(self, n):
        bits = bin(n)
        return all(bits[i] != bits[i+1]
                   for i in xrange(len(bits) - 1))

'''

class Solution:
    def hasAlternatingBits(self, n):
        """
        :type n: int
        :rtype: bool
        """
        n, cur = divmod(n, 2)
        while n:
            if cur == n % 2: return False
            n, cur = divmod(n, 2)
        return True