Easy

Binary Number with Alternating BitsPython

Full explanation · Time O(1) · Space O(1)

# Time:  O(1)
# Space: O(1)

class Solution(object):
    def hasAlternatingBits(self, n):
        """
        :type n: int
        :rtype: bool
        """
        n, curr = divmod(n, 2)
        while n > 0:
            if curr == n % 2:
                return False
            n, curr = divmod(n, 2)
        return True