Leetcode Count Numbers with Unique Digits problem solution

In this Leetcode Count Numbers with Unique Digits problem solution you have given an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10n.

Problem solution in Python.

class Solution:
    def countNumbersWithUniqueDigits(self, n):
        if n == 0:
            return 1
        
        if n == 1:
            return 10
        
        if n == 2:
            return 91
        
        i = 3
        total = 91
        k = 9*9
        
        while i <= n:
            total += k * (11 - i)
            k *= (11 - i)
            i += 1
            
            if i == 11:
                break
        
        return total

Problem solution in Java.

class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if(n==0){
            return 1;
        }
        int sum = 0;
        for(int k=1;k<=n;k++){
            sum += count(k);
        }
        return sum+1;
    }
    
    public int count(int n){
        int product = 9;        
        for(int i=0;i<n-1;i++){
            product = product * (9-i);
        }
        return product;
    }
}

Problem solution in C++.

int countNumbersWithUniqueDigits(int n) {
        int s = 0;
        int p = 1;
        for (int i=1; i<=min(10, n); i++) {
            s += p;
            p *= 10 - i;
        }
        return s * 9 + 1;
    }

Problem solution in C.

int fac(int a, int b){
    int result = 1;
    while(a > b){
        result *= a;
        a--;
    }
    return result;
}
int countNumbersWithUniqueDigits(int n) {
    if(n == 1 || n == 0) return pow(10, n);
    if(n > 10) return countNumbersWithUniqueDigits(10);
    
    int result = 0;
    result = countNumbersWithUniqueDigits(n - 1) + fac(9, 9-n) + fac(9, 9 - n + 1) * (n - 1);
    return result;
}