L2 · Working engineerStrings~6 min · 6 tests#18

Check If Two Strings Are Anagrams

Decide whether two strings are anagrams with a character count in O(n), and see why sorting both is O(n log n). Run Python tests in your browser.

The problem

Return True if s1 and s2 are anagrams — the same characters with the same counts, in any order. Comparison is case-sensitive.

Examples

Example 1
Input
```
is_anagram('listen', 'silent')
```
Expected output
```
True
```
Example 2
Input
```
is_anagram('rat', 'car')
```
Expected output
```
False
```

+ 4 hidden tests on Submit — same letters, different counts, different length, case-sensitive.

Edge cases to ask about

Empty strings
Different lengths
Case differences
Unicode

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_anagram

def is_anagram(s1, s2):
    """True if s1 and s2 contain the same characters with the same counts."""
    # 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 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
Sort both	O(n log n)	O(n)	Two sorted copies compared.
bestCount characters	O(n)	O(k)	Increment for s1, decrement for s2; any shortfall means no.

Walkthrough of the optimal approach (try it yourself first)

Count characters in s1, then decrement for each character in s2. If a count would go below zero, the strings differ. A quick length check first makes the "same counts" logic complete.

sorted(s1) == sorted(s2) is fine to say out loud — then point out it is O(n log n) and the count version is O(n).

Complexity: O(n) time, O(k) space. Each string is walked once; the count table has one slot per distinct character.

Reveal the reference solution

def is_anagram(s1, s2):
    if len(s1) != len(s2):
        return False
    counts = {}
    for ch in s1:
        counts[ch] = counts.get(ch, 0) + 1
    for ch in s2:
        if counts.get(ch, 0) == 0:
            return False
        counts[ch] -= 1
    return True

The brute force, for comparison

def is_anagram(s1, s2):
    return sorted(s1) == sorted(s2)

Follow-ups interviewers ask

Ignore spaces and case.
Group a list of words into anagram groups (question 28).

Frequently asked interview questions

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

What is the time complexity of Check If Two Strings Are Anagrams in Python?

The optimal solution runs in O(n) time and O(k) auxiliary space. Each string is walked once; the count table has one slot per distinct character.

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

Sort both: O(n log n) time, O(n) space. Two sorted copies compared. Count characters: O(n) time, O(k) space. Increment for s1, decrement for s2; any shortfall means no.

What follow-up questions do interviewers ask about Check If Two Strings Are Anagrams?

Ignore spaces and case. Group a list of words into anagram groups (question 28).