[LeetCode] 357. Count Numbers with Unique Digits

 Problem : https://leetcode.com/problems/count-numbers-with-unique-digits/

The input number N means the maximum number of digits we can have for the wanted numbers.

Let F(N) = valid number of unique digits combination for N digits

F(0) = 0. Wanted number = 0. Total number = 1

F(1) = 1. Wanted number = [0 ... 9].   Total number = 10

F(2) =  9 * 9 + F (1) = 91.   The first digit has 9 options from 1 to 9. The second digit has 9 options from 0 to 9 ( excluding the digit used by the first digits)

F(3) = 9 * 9 * 8 + F(2) + F(1)

class Solution:
    def countNumbersWithUniqueDigits(self, n: int) -> int:
        if n == 0:
            return 1
        
        if n == 1:
            return 10
        
        result = 9
        digits = 9
        
        # calculate valid combiations of N digits
        for _ in range(1,n):
            result *= digits
            digits -= 1
        
        return result + self.countNumbersWithUniqueDigits(n-1)