357-count-numbers-with-unique-digits

Question

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

Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10^n.

Example:

Given n = 2, return 91. 
(The answer should be the total numbers in the range of 0 ≤ x < 100, 
excluding [11,22,33,44,55,66,77,88,99])

Thought Process

1. DP
   1. Start from n = 1, we have 10 choices
   2. For n = 2, we can only have 9 choice for the first digit (excluding 0), and 9 choices for the second digit (avoiding the same number), 9 * 9 unique 2 digits number in additional to the 10 before
   3. For n = 3, we have 9, 9, and 8 choice for each digit, 9 *9 * 8 in additional to 91 before
   4. Time complexity O(n) or O(10) at most
   5. Space complexity O(1)

Solution

class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        int res = 1;
        int choice = 9, available = 9;
        while (n-- >= 1 && available > 0) {
            res += choice;
            choice *= available;
            available--;
        }
        return res;
    }
}

Additional