Count the Digit

Take an integer n (n >= 0) and a digit d (0 <= d <= 9) as an integer.

Square all numbers k (0 <= k <= n) between 0 and n.

Count the numbers of digits d used in the writing of all the k**2.

Call nb_dig (or nbDig or ...) the function taking n and d as parameters and returning this count.

Examples:

n = 10, d = 1 
the k*k are 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
We are using the digit 1 in: 1, 16, 81, 100. The total count is then 4.

nb_dig(25, 1) returns 11 since
the k*k that contain the digit 1 are:
1, 16, 81, 100, 121, 144, 169, 196, 361, 441.
So there are 11 digits 1 for the squares of numbers between 0 and 25.

Note that 121 has twice the digit 1.

Solution

public class CountDig {
    public static int nbDig(int n, int d) {
        int count = 0;
        for (int i = 0; i <= n; i++)
          count += Integer.toString(i*i).chars().filter(ch -> ch == Character.forDigit(d, 10)).count();
        return count;
    }
}