Write a Java program to count the number of digits in a number.

public class NumberOfDigits {
    public static void main(String[] args) {
        int count = 0;
        int num   = 3452;

        while (num != 0) {
            num /= 10;   // 3452 → 345 → 34 → 3 → 0
            ++count;     // increment digit count each iteration
        }

        System.out.println("Number of digits: " + count);
    }
}

Number of digits: 4

num=3452, count=0
Iteration 1: num = 3452/10 = 345,  count = 1
Iteration 2: num = 345/10  = 34,   count = 2
Iteration 3: num = 34/10   = 3,    count = 3
Iteration 4: num = 3/10    = 0,    count = 4
Loop ends (num == 0)

Result: 4 digits

int num = 3452;
int digits = String.valueOf(Math.abs(num)).length();
System.out.println("Digits: " + digits);  // 4

public static int countDigits(int num) {
    if (num == 0) return 1;            // special case: 0 has 1 digit
    num = Math.abs(num);               // handle negatives: -345 → 345
    int count = 0;
    while (num != 0) {
        num /= 10;
        count++;
    }
    return count;
}

System.out.println(countDigits(0));       // 1
System.out.println(countDigits(12345));   // 5
System.out.println(countDigits(-999));    // 3
System.out.println(countDigits(1000000)); // 7

int num = 3452;
int digits = (int) Math.log10(num) + 1;
System.out.println("Digits: " + digits);  // 4
// log10(3452) ≈ 3.537 → (int)3.537 = 3 → 3+1 = 4

// Validate phone number digit count
String phone = driver.findElement(By.id("phone")).getAttribute("value");
int digitCount = phone.replaceAll("[^0-9]", "").length();
assertEquals(digitCount, 10, "Phone number must be 10 digits");