L1 · FoundationsNumbers & loops~3 min · 6 tests

Check If a Number Is a Palindrome

Check whether an integer reads the same backwards, like 121, without converting it to a string. Negative numbers are never palindromes. Run tests live.

The problem

Return True if the integer n reads the same forwards and backwards. Negative numbers are not palindromes. Avoid str().

Examples

Example 1
Input
```
is_palindrome_number(121)
```
Expected output
```
True
```
Example 2
Input
```
is_palindrome_number(123)
```
Expected output
```
False
```

+ 4 hidden tests on Submit.

Edge cases to ask about

Negative numbers
Zero
Trailing zero (10)

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 · is_palindrome_number

def is_palindrome_number(n):
    """True if the digits of n read the same both ways."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

Your code runs in real CPython inside your browser — nothing is sent anywhere. The first run downloads the interpreter (about 6 MB, once). Your code is saved on this device as you type.

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
str(n) == str(n)[::-1]	O(d)	O(d)
bestReverse arithmetically and compare	O(d)	O(1)

Walkthrough of the optimal approach (try it yourself first)

Reverse the number with the % 10 / // 10 loop and compare with the original. Negative numbers fail immediately because the - has no mirror.

Complexity: O(log n) time, O(1) space. Reversing touches each of the ~log₁₀(n) digits once.

Reveal the reference solution

def is_palindrome_number(n):
    if n < 0:
        return False
    original, rev = n, 0
    while n:
        rev = rev * 10 + n % 10
        n //= 10
    return rev == original

Follow-ups interviewers ask

Reverse only half the digits to stop early.

Frequently asked interview questions

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

What is the time complexity of Check If a Number Is a Palindrome in Python?

The optimal solution runs in O(log n) time and O(1) auxiliary space. Reversing touches each of the ~log₁₀(n) digits once.

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

str(n) == str(n)[::-1]: O(d) time, O(d) space. Reverse arithmetically and compare: O(d) time, O(1) space.

What follow-up questions do interviewers ask about Check If a Number Is a Palindrome?

Reverse only half the digits to stop early.