# Time:  O(nlogr) = O(n)
# Space: O(logr) = O(1)

# A self-dividing number is a number that is divisible by every digit it contains.
#
# For example, 128 is a self-dividing number because 128 % 1 == 0,
# 128 % 2 == 0, and 128 % 8 == 0.
#
# Also, a self-dividing number is not allowed to contain the digit zero.
#
# Given a lower and upper number bound, output a list of every possible self dividing number,
# including the bounds if possible.
#
# Example 1:
# Input:
# left = 1, right = 22
# Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]
#
# Note:
# - The boundaries of each input argument are 1 <= left <= right <= 10000.

class Solution(object):
    def selfDividingNumbers(self, left, right):
        """
        :type left: int
        :type right: int
        :rtype: List[int]
        """
        def isDividingNumber(num):
            n = num
            while n > 0:
                n, r = divmod(n, 10)
                if r == 0 or (num%r) != 0:
                    return False
            return True
        
        return [num for num in xrange(left, right+1) if isDividingNumber(num)]


# Time:  O(nlogr) = O(n)
# Space: O(logr) = O(1)
import itertools


class Solution2(object):
    def selfDividingNumbers(self, left, right):
        """
        :type left: int
        :type right: int
        :rtype: List[int]
        """
        return [num for num in xrange(left, right+1) \
                if not any(itertools.imap(lambda x: int(x) == 0 or num%int(x) != 0, str(num)))]

# Time:  O(nlogr) = O(n)
# Space: O(logr) = O(1)

# A self-dividing number is a number that is divisible by every digit it contains.
#
# For example, 128 is a self-dividing number because 128 % 1 == 0,
# 128 % 2 == 0, and 128 % 8 == 0.
#
# Also, a self-dividing number is not allowed to contain the digit zero.
#
# Given a lower and upper number bound, output a list of every possible self dividing number,
# including the bounds if possible.
#
# Example 1:
# Input:
# left = 1, right = 22
# Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]
#
# Note:
# - The boundaries of each input argument are 1 <= left <= right <= 10000.

class Solution(object):
    def selfDividingNumbers(self, left, right):
        """
        :type left: int
        :type right: int
        :rtype: List[int]
        """
        def isDividingNumber(num):
            n = num
            while n > 0:
                n, r = divmod(n, 10)
                if r == 0 or (num%r) != 0:
                    return False
            return True
        
        return [num for num in xrange(left, right+1) if isDividingNumber(num)]


# Time:  O(nlogr) = O(n)
# Space: O(logr) = O(1)
import itertools


class Solution2(object):
    def selfDividingNumbers(self, left, right):
        """
        :type left: int
        :type right: int
        :rtype: List[int]
        """
        return [num for num in xrange(left, right+1) \
                if not any(itertools.imap(lambda x: int(x) == 0 or num%int(x) != 0, str(num)))]

# Time:  O(nlogr) = O(n) # Space: O(logr) = O(1) # A self-dividing number is a number that is divisible by every digit it contains. # # For example, 128 is a self-dividing number because 128 % 1 == 0, # 128 % 2 == 0, and 128 % 8 == 0. # # Also, a self-dividing number is not allowed to contain the digit zero. # # Given a lower and upper number bound, output a list of every possible self dividing number, # including the bounds if possible. # # Example 1: # Input: # left = 1, right = 22 # Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22] # # Note: # - The boundaries of each input argument are 1 <= left <= right <= 10000. :type left: int         :type right: int         :rtype: List[int] # Time:  O(nlogr) = O(n) # Space: O(logr) = O(1) :type left: int         :type right: int         :rtype: List[int]