The challenge
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):
the numbers of interest are
1, 4, 9, 10, 11, 12, 13, 14, 19, 21 which squared 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.
The solution in Java code
Option 1:
public class CountDig {
public static int nbDig(int n, int d) {
int sum = 0;
for (int i = 0; i <= n; i++) {
sum += countOfDigit((int) Math.pow(i, 2), d);
}
return sum;
}
public static int countOfDigit(int n, int d) {
int count = 0;
do {
if (n % 10 == d)
count ++;
n /= 10;
} while (n > 0);
return count;
}
}
Option 2:
import java.util.stream.IntStream;
public class CountDig {
public static int nbDig(int n, int d) {
return (int) IntStream
.rangeClosed(0, n)
.map(i -> i * i)
.flatMap(i -> String.valueOf(i).chars())
.mapToObj(i -> (char)i)
.mapToInt(Character::getNumericValue)
.filter(i -> i == d)
.count();
}
}
Option 3:
public class CountDig {
public static int nbDig(int n, int d) {
int needle = String.valueOf(d).charAt(0);
int count = 0;
for (int i = 0; i <= n; i++) {
String haystack = String.valueOf(i * i);
count += (int) haystack.chars().filter(ch -> ch == needle).count();
}
return count;
}
}
Test cases to validate our solution
import static org.junit.Assert.*;
import org.junit.Test;
public class CountDigTest {
private static void testing(int actual, int expected) {
assertEquals(expected, actual);
}
@Test
public void test() {
System.out.println("Fixed Tests nbDig");
testing(CountDig.nbDig(5750, 0), 4700);
testing(CountDig.nbDig(11011, 2), 9481);
testing(CountDig.nbDig(12224, 8), 7733);
testing(CountDig.nbDig(11549, 1), 11905);
}
}
