'''
Given an integers a and b, return a count of numbers between a and b (inclusive) that have no repeated digits.

Example 1
Input:
a = 100; b = 1000
Output:
648
'''

class Solution(object):
    def solve(self,a:int, b:int) ->int:
        print(f"a: {a}, b: {b}")
        count = 0
        for num in range(a, b+1):               # Iterate through each number in the range
            if self.has_unique_digits(num):
                count += 1                      # increment count if the number has unique digits
        return count

    def has_unique_digits(self, num:int) -> bool:
        digits = set()
        while num > 0:
            digit = num % 10        # Looks at the last digit of num
            if digit in digits:      # If the digit is already in the set, return false
                return False
            digits.add(digit)       # Add the digit to the set of digits
            num //= 10              # Remove the last digit of num
        return True

s = Solution()
print(s.solve(100, 1000))  # Output: 648
print(s.solve(1000, 10000))  # Output: 5520
print(s.solve(10001, 100000))  # Output: 45360
print(s.solve(500, 600))  # Output: 403200