L2 · Working engineerStrings~6 min · 5 tests#16

First Non-Repeating Character

Find the first character that appears exactly once in a Python string using a frequency count — two passes, O(n). Run tests and measure it live.

The problem

Return the first character in s that appears exactly once. Return None if every character repeats.

Examples

Example 1
Input
```
first_unique_char('aabbcdde')
```
Expected output
```
'c'
```
Example 2
Input
```
first_unique_char('swiss')
```
Expected output
```
'w'
```

+ 3 hidden tests on Submit — none unique, empty.

Edge cases to ask about

Empty string
All repeating
Single character

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 · first_unique_char

def first_unique_char(s):
    """First character that appears exactly once, else None."""
    # 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
s.count() per character	O(n²)	O(1)	Recounts the whole string for each character.
bestCount, then scan	O(n)	O(k)	One pass to count, one pass to find the first count of 1. k = alphabet size.

Walkthrough of the optimal approach (try it yourself first)

Two passes. First build a frequency map. Then walk the string again and return the first character whose count is 1 — walking the string, not the dict, is what guarantees "first".

Space is O(k) where k is the number of distinct characters — for lowercase English that is a constant 26, which is why some interviewers call it O(1).

Complexity: O(n) time, O(k) space. Two linear passes; the count dict holds at most one entry per distinct character (k ≤ alphabet size).

Reveal the reference solution

def first_unique_char(s):
    counts = {}
    for ch in s:
        counts[ch] = counts.get(ch, 0) + 1
    for ch in s:
        if counts[ch] == 1:
            return ch
    return None

The brute force, for comparison

def first_unique_char(s):
    for ch in s:
        if s.count(ch) == 1:
            return ch
    return None

Follow-ups interviewers ask

Return the index instead.
Characters arrive as a stream — answer after every character.

Frequently asked interview questions

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

What is the time complexity of First Non-Repeating Character in Python?

The optimal solution runs in O(n) time and O(k) auxiliary space. Two linear passes; the count dict holds at most one entry per distinct character (k ≤ alphabet size).

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

s.count() per character: O(n²) time, O(1) space. Recounts the whole string for each character. Count, then scan: O(n) time, O(k) space. One pass to count, one pass to find the first count of 1. k = alphabet size.

What follow-up questions do interviewers ask about First Non-Repeating Character?

Return the index instead. Characters arrive as a stream — answer after every character.