Count Numbers Containing Specific Digits

import java.util.*;

class Solution {
    public int countValid(int n, int[] arr) {
        if (n == 1) {
            Arrays.sort(arr);
            if (arr[0] == 0)
                return arr.length - 1;
            else
                return arr.length;
        }
        long totalNumbers = 9; // Use long to prevent overflow for larger n
        for (int i = 0; i < n - 1; i++)
            totalNumbers *= 10;

        Set<Integer> arrSet = new HashSet<>();
        for (int digit : arr)
            arrSet.add(digit);

        int numUnavailableDigits = arr.length;
        int numAvailableNonArrDigits = 10 - numUnavailableDigits;

        long numbersWithoutArrDigits = 1;

        int firstDigitChoices = 0;
        for (int i = 1; i <= 9; i++) { // Iterate through 1-9
            if (!arrSet.contains(i))
                firstDigitChoices++;
        }
        numbersWithoutArrDigits = firstDigitChoices;

        for (int i = 0; i < n - 1; i++)
            numbersWithoutArrDigits *= numAvailableNonArrDigits;

        return (int) (totalNumbers - numbersWithoutArrDigits);
    }
}