L1 · FoundationsNumbers & loops~3 min · 5 tests

Count the Digits in a Number

Count how many digits an integer has using a loop with // 10, then see the O(1) logarithm shortcut and why it can be wrong. Live Python tests.

The problem

Return how many digits the integer n has. 0 has one digit; ignore the sign.

Examples

Example 1
Input
```
count_digits(12345)
```
Expected output
```
5
```
Example 2
Input
```
count_digits(0)
```
Expected output
```
1
```

+ 3 hidden tests on Submit — 20 nines.

Edge cases to ask about

Zero
Negative numbers
Very large numbers

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · count_digits

def count_digits(n):
    """Number of digits in n (0 has 1 digit)."""
    # Write your solution here
    pass

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Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and read why.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
Divide by 10 until zero	O(log n)	O(1)
len(str(abs(n)))	O(log n)	O(log n)	Simple and exact.
bestfloor(log10(n)) + 1	O(1)	O(1)	Floating-point error makes it wrong for very large n.

Walkthrough of the optimal approach (try it yourself first)

Divide by 10 until the number is 0, counting the steps — one per digit. Zero needs a special case because the loop would not run at all.

math.log10 looks O(1), but floating-point rounding makes it give wrong answers for numbers like 10**15 - 1. The hidden test checks a 20-digit number.

Complexity: O(log n) time, O(1) space. Each division removes one digit.

Reveal the reference solution

def count_digits(n):
    n = abs(n)
    if n == 0:
        return 1
    count = 0
    while n:
        n //= 10
        count += 1
    return count

Follow-ups interviewers ask

Why can the log10 version be wrong?

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Count the Digits in a Number in Python?

The optimal solution runs in O(log n) time and O(1) auxiliary space. Each division removes one digit.

What is the brute-force approach, and how do you optimise it?

Divide by 10 until zero: O(log n) time, O(1) space. len(str(abs(n))): O(log n) time, O(log n) space. Simple and exact. floor(log10(n)) + 1: O(1) time, O(1) space. Floating-point error makes it wrong for very large n.

What follow-up questions do interviewers ask about Count the Digits in a Number?

Why can the log10 version be wrong?