L2 · Working engineerStrings~5 min · 6 tests#19

Check If a String Is a Palindrome

Check whether a Python string reads the same backwards using two pointers in O(n) time and O(1) space, then compare with the slicing one-liner.

The problem

Return True if s reads the same forwards and backwards. Compare characters exactly (case-sensitive), and use two pointers rather than s[::-1].

Examples

Example 1
Input
```
is_palindrome('madam')
```
Expected output
```
True
```
Example 2
Input
```
is_palindrome('python')
```
Expected output
```
False
```

+ 4 hidden tests on Submit — even length, case-sensitive.

Edge cases to ask about

Empty string
Single character
Even vs odd length

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

def is_palindrome(s):
    """True if s reads the same in both directions."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

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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 measure your code against the optimal one at growing input sizes.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

Measure it

Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.

From brute force to optimal

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

Approach	Time	Space	Idea
Reverse and compare	O(n)	O(n)	s[::-1] builds a full reversed copy.
bestTwo pointers	O(n)	O(1)	Compare from both ends inward; stop at the first mismatch.

Walkthrough of the optimal approach (try it yourself first)

Put i at the start and j at the end, compare, and walk them toward each other. The first mismatch means False; meeting in the middle means True.

s == s[::-1] is also O(n) time, but it allocates a full reversed copy (O(n) space) and cannot stop early.

Complexity: O(n) time, O(1) space. At most n/2 comparisons and two integer indices — no copy of the string.

Reveal the reference solution

def is_palindrome(s):
    i, j = 0, len(s) - 1
    while i < j:
        if s[i] != s[j]:
            return False
        i += 1
        j -= 1
    return True

The brute force, for comparison

def is_palindrome(s):
    return s == s[::-1]

Follow-ups interviewers ask

Ignore non-alphanumeric characters and case ('A man, a plan…').
Valid if you may delete at most one character.

Frequently asked interview questions

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

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

The optimal solution runs in O(n) time and O(1) auxiliary space. At most n/2 comparisons and two integer indices — no copy of the string.

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

Reverse and compare: O(n) time, O(n) space. s[::-1] builds a full reversed copy. Two pointers: O(n) time, O(1) space. Compare from both ends inward; stop at the first mismatch.

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

Ignore non-alphanumeric characters and case ('A man, a plan…'). Valid if you may delete at most one character.