LeetCode- Count Numbers with Unique Digits

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

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])

Analysis:

A number of unique digits is a number which is a combination of unrepeated digits. So, we can calculate the total number. for number with n digits, like 100-999 or 1000-9999, the total numbers with unique digits equals to 9*9*8...*(11-n). because the highest digit cannot be 0.

 

Solution:

public class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if (n==0) return 1;
        
        int res = 10;
        int cur = 9;
        for (int i=2;i<=Math.min(10,n);i++){
            cur = cur*(11-i);
            res += cur;
        }
        return res;
    }
}