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Count Digits

Java coding interview problem for Basic Number Programs: Count Digits.

Counting the number of digits in an integer is one of the most frequently asked Java coding interview questions. It helps interviewers evaluate your understanding of loops, arithmetic operations, integer manipulation, and problem-solving skills.

This problem serves as the foundation for many other number-based coding questions such as Armstrong Number, Palindrome Number, Reverse Integer, and Sum of Digits.

What is Count Digits?

Count Digits means determining how many digits are present in a given number.

For example,

Input

12345

Digits

1

2

3

4

5

Output

5

Example 1

Input

987654

Output

6

Example 2

Input

1000

Output

4

Example 3

Input

7

Output

1

Example 4

Input

0

Output

1

Note: Zero has exactly one digit.

Real Interview Question

Write a Java program to count the number of digits in an integer without converting it into a String.

Understanding the Logic

The idea is straightforward.

Every time we divide a number by 10, its last digit is removed.

For example,

12345

becomes

1234

↓

123

↓

12

↓

1

↓

0

The number becomes zero after 5 divisions.

Therefore,

Number of Digits = 5

Brute Force Approach

The simplest solution is

  • Initialize a counter.
  • Divide the number by 10 repeatedly.
  • Increase the counter after every division.
  • Continue until the number becomes zero.

Algorithm

Step 1

Read the input number.

Step 2

Handle the special case.

If number == 0

Return 1

Step 3

Initialize

count = 0;

Step 4

Repeat while

number != 0

Step 5

Increase the counter.

count++;

Step 6

Remove the last digit.

number = number / 10;

Step 7

Print the total count.

Dry Run

Input

12345
Iteration Number Count
1 12345 1
2 1234 2
3 123 3
4 12 4
5 1 5
End 0 5

Output

5

Another Dry Run

Input

567890
Iteration Number Count
1 567890 1
2 56789 2
3 5678 3
4 567 4
5 56 5
6 5 6
End 0 6

Output

6

Approach 1 — Using Arithmetic Operators

This is the most common and interview-preferred solution.

Complete Java Program

public class CountDigits {

    public static void main(String[] args) {

        int number = 12345;

        if (number == 0) {

            System.out.println("Number of Digits = 1");

            return;
        }

        int count = 0;

        while (number != 0) {

            count++;

            number = number / 10;

        }

        System.out.println("Number of Digits = " + count);

    }

}

Output

Number of Digits = 5

Step-by-Step Code Explanation

Step 1

Declare the input number.

int number = 12345;

Current value

12345

Step 2

Handle the special case.

if (number == 0) {

    System.out.println("Number of Digits = 1");

    return;

}

Since

0

contains one digit,

its answer should be

1

Step 3

Initialize the counter.

int count = 0;

Initially

Count = 0

Step 4

Run the loop.

while (number != 0)

The loop continues until all digits are removed.

Step 5

Increase the count.

count++;

Every iteration processes exactly one digit.

Step 6

Remove the last digit.

number = number / 10;

Example

12345

↓

1234

↓

123

↓

12

↓

1

↓

0

Each division removes one digit from the right.

Step 7

Print the answer.

System.out.println("Number of Digits = " + count);

Output

Number of Digits = 5

Why Does This Work?

The algorithm relies on the fact that integer division by 10 removes the last digit.

For example,

98765

becomes

9876

↓

987

↓

98

↓

9

↓

0

Each division removes one digit.

By counting the number of divisions until the number becomes zero, we obtain the total number of digits.

Advantages of This Approach

  • Easy to understand.
  • Uses only arithmetic operators.
  • No extra data structures are required.
  • Works efficiently for integers.
  • Preferred in Java coding interviews.
  • Uses constant extra memory.

Drawbacks

Although this solution is efficient and widely accepted, interviewers often ask follow-up questions such as:

  • Can you solve it using a reusable method?
  • Can you solve it using logarithms?
  • Can you solve it using String conversion?
  • How would you handle negative numbers?
  • Which approach is the fastest?

we'll cover:

  • Reusable method approach
  • Logarithmic (Math.log10()) approach
  • String-based approach
  • Time and space complexity
  • Comparison of all approaches
  • Common interview mistakes
  • Frequently asked interview questions
  • Related coding problems
  • Key takeaways
  • Interview tips

Approach 2 — Using a Reusable Method

Instead of writing the logic inside the main() method, we can create a reusable method.

This approach improves

  • Code reusability
  • Readability
  • Unit testing
  • Maintainability

Java Program

public class CountDigitsMethod {

    static int countDigits(int number) {

        if (number == 0) {
            return 1;
        }

        number = Math.abs(number);

        int count = 0;

        while (number != 0) {

            count++;

            number = number / 10;

        }

        return count;

    }

    public static void main(String[] args) {

        int number = 987654;

        System.out.println("Number of Digits = " + countDigits(number));

    }

}

Output

Number of Digits = 6

Approach 3 — Using Math.log10()

This approach uses a mathematical formula.

Formula

Digits = floor(log10(number)) + 1

Example

Number = 12345

log10(12345) = 4.09

floor(4.09) = 4

4 + 1 = 5

Java Program

public class CountDigitsLogarithm {

    public static void main(String[] args) {

        int number = 12345;

        if (number == 0) {

            System.out.println("Number of Digits = 1");

        } else {

            int digits = (int) Math.floor(Math.log10(Math.abs(number))) + 1;

            System.out.println("Number of Digits = " + digits);

        }

    }

}

Output

Number of Digits = 5

Approach 4 — Using String Conversion

Although interviewers usually expect an arithmetic solution,

Java also allows counting digits by converting the number into a String.

Java Program

public class CountDigitsString {

    public static void main(String[] args) {

        int number = -12345;

        int count = String.valueOf(Math.abs(number)).length();

        System.out.println("Number of Digits = " + count);

    }

}

Output

Number of Digits = 5

Time Complexity

Arithmetic Solution

Operation Complexity
Time O(log₁₀ n)
Space O(1)

Logarithmic Solution

Operation Complexity
Time O(1)
Space O(1)

String Solution

Operation Complexity
Time O(log₁₀ n)
Space O(log₁₀ n)

Comparison of All Approaches

Approach Time Space Recommended
Arithmetic Operators O(log n) O(1) ✅ Best for Interviews
Reusable Method O(log n) O(1) Reusable
Math.log10() O(1) O(1) Fastest
String Conversion O(log n) O(log n) Easy but Less Preferred

Common Mistakes

Mistake 1

Ignoring the special case

0

Wrong

int count = 0;

while(number != 0){

    count++;

}

Output

0

Correct Output

1

Mistake 2

Not handling negative numbers.

Example

-12345

Always use

number = Math.abs(number);

before processing.

Mistake 3

Using

Math.log10()

for zero.

Wrong

Math.log10(0)

This is undefined.

Always check

number == 0

first.

Mistake 4

Forgetting to divide the number.

Wrong

count++;

Correct

count++;

number = number / 10;

Otherwise,

the loop never terminates.

Mistake 5

Using String conversion when the interviewer specifically asks

"Without converting the number into a String."

Use arithmetic operators instead.

Interview Follow-up Questions

Q1. Count digits without using String.

Q2. Count digits recursively.

Q3. Count even digits.

Q4. Count odd digits.

Q5. Count occurrences of a particular digit.

Q6. Count zeros in a number.

Q7. Count digits in a very large number.

Q8. Explain the logarithmic approach.

Q9. Compare all approaches.

Q10. Which solution would you use in production?

Related Coding Problems

  • Sum of Digits
  • Reverse Integer
  • Armstrong Number
  • Palindrome Number
  • Product of Digits
  • Largest Digit
  • Smallest Digit
  • Digital Root

Key Takeaways

  • Count digits by repeatedly dividing the number by 10.
  • Integer division removes one digit during each iteration.
  • Handle 0 as a special case because it contains exactly one digit.
  • Use Math.abs() to correctly process negative numbers.
  • Math.log10() provides a constant-time mathematical solution but requires special handling for zero.
  • The arithmetic solution is the most common and interview-friendly approach.

Interview Tip

If an interviewer asks:

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

Start with the arithmetic solution because it demonstrates your understanding of integer manipulation using % and /. Mention the Math.log10() solution as an optimization and discuss its limitation with zero. Finally, explain the String-based approach as an alternative, noting that it is simpler but usually not preferred when the interviewer expects an arithmetic solution. Demonstrating multiple approaches and their trade-offs shows strong problem-solving ability and practical Java knowledge.