L2 · Working engineerStrings~4 min · 5 tests#23

Remove Duplicate Characters from a String

Remove repeated characters from a Python string, keeping first occurrences in order, with a seen-set and join in O(n). Run hidden tests in your browser.

The problem

Return s with every repeated character removed, keeping the first occurrence of each character in order.

Examples

Example 1

Input

remove_duplicate_chars('programming')

Expected output

'progamin'

Example 2
Input
```
remove_duplicate_chars('aabbcc')
```
Expected output
```
'abc'
```

+ 3 hidden tests on Submit — space is a character.

Edge cases to ask about

Empty string
No duplicates
Spaces

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 · remove_duplicate_chars

def remove_duplicate_chars(s):
    """Keep the first occurrence of each character, in order."""
    # 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 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
`ch not in out` on a string	O(n·k)	O(k)	Each check scans the result.
bestSeen set + list + join	O(n)	O(k)	Or simply ''.join(dict.fromkeys(s)).

Walkthrough of the optimal approach (try it yourself first)

Same pattern as question 2: a seen set for O(1) membership and a list to keep order, then one "".join().

"".join(dict.fromkeys(s)) works because dicts preserve insertion order.

Complexity: O(n) time, O(k) space. One pass with O(1) set checks; extra memory is one entry per distinct character.

Reveal the reference solution

def remove_duplicate_chars(s):
    seen = set()
    out = []
    for ch in s:
        if ch not in seen:
            seen.add(ch)
            out.append(ch)
    return "".join(out)

The brute force, for comparison

def remove_duplicate_chars(s):
    out = ""
    for ch in s:
        if ch not in out:
            out += ch
    return out

Follow-ups interviewers ask

Remove only adjacent duplicates ('aabba' → 'aba').
Case-insensitive version.

Frequently asked interview questions

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

What is the time complexity of Remove Duplicate Characters from a String in Python?

The optimal solution runs in O(n) time and O(k) auxiliary space. One pass with O(1) set checks; extra memory is one entry per distinct character.

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

`ch not in out` on a string: O(n·k) time, O(k) space. Each check scans the result. Seen set + list + join: O(n) time, O(k) space. Or simply ''.join(dict.fromkeys(s)).

What follow-up questions do interviewers ask about Remove Duplicate Characters from a String?

Remove only adjacent duplicates ('aabba' → 'aba'). Case-insensitive version.