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easyBasicsSometimes asked

Remove Duplicate Characters

Basics · 6 min

Given a string s, return a new string keeping only the first occurrence of each character, preserving order.

In plain English

Track which characters you've already emitted in a set; append a character only the first time you see it.

Examples

Example 1

Input: s = "banana"
Output: "ban"
Note: First b,a,n kept.

Example 2

Input: s = "aabbcc"
Output: "abc"
Note: One of each.

Example 3

Input: s = "abc"
Output: "abc"
Note: Already unique.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions1First question: Keep first or last occurrence?Explore questions

Q1.Keep first or last occurrence?
First, preserving order; last-occurrence is a small variant.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Keep a 'seen' set; for each character, if it isn't in the set, append it and add it. O(n) time, O(k) space.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you combine a seen-set with an order-preserving build.

Where candidates slip

Reordering output, or O(n²) concatenation.

How to stand out

Use the `set.add` boolean idiom to filter and record in one step.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Seen set build

TimeO(n)SpaceO(k)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

What structure tells you 'have I already kept this char?'

2Hint 2Explore hint

Append only on first sighting to preserve order.

3Hint 3Explore hint

Use a StringBuilder; `set.add` returns true only when newly added.

A · Core idea

Keep only the first occurrence of each character: track a seen set and append a character to the output only the first time you encounter it. Combines the seen-set membership pattern with a filtered build.

Why a set preserves order. You scan left to right and append on first sighting, so the kept characters appear in their original order — no sorting. Later repeats fail the membership test and are skipped.

The Java idiom `seen.add(ch)`. Set.add returns true only when the element was newly added, so if (seen.add(ch)) sb.append(ch); both records and filters in one line.

Performance. Use a StringBuilder, not +=, to keep the build O(n). The set gives O(1) membership, so the whole thing is O(n) time, O(k) space.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

s = "banana"

new

Stateb

Final result"ban"

C · Code with teaching comments

solution.javaJava

Seen set build — optimal

1 teaching note

1
public String removeDuplicates(String s) {
2
    StringBuilder sb = new StringBuilder();
3
    java.util.Set<Character> seen = new java.util.HashSet<>();
4
    for (char ch : s.toCharArray()) {
5
        // add returns true only the first time — append exactly once per char.
6
        if (seen.add(ch)) sb.append(ch);
7
    }
8
    return sb.toString();
9
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.9 lines

D · Why it works and complexityO(n) timeO(k) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(k) space — O(1) set lookup and append per character; the set holds k ≤ alphabet-size keys.

E · Edge cases and pitfalls5s = "aabbcc" — → "abc".Explore edge cases

Edge cases handled

s = "aabbcc"→ "abc".
s = "abc"Already unique → "abc".
s = ""Empty → "".

Pitfalls

Not preserving original order (e.g. sorting).
Using String += in the loop.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic String removeDuplicates(String s) { StringBuilder sb = new StringBuilder(); java.util.Set<Character> seen = new java.util.HashSet<>(); for (char ch : s.toCharArray()) { if (seen.add(ch)) sb.append(ch);Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Not preserving original order (e.g., sorting).
Using String += in the loop.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Seen-set filtered build

Code pattern to memorize

seen=set(); for ch: if ch not in seen: seen.add(ch); out+=ch

Rules to remember

1Append on first sighting.
2Set tracks seen chars.
3Order preserved.

Use this pattern when

First-occurrence dedup with order

Interview takeaway

First-occurrence-only = a seen set plus a builder.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Seen-set filtered build. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Same pattern

easyBasicsSometimes asked

Remove Duplicate Characters

Basics · 6 min

Given a string s, return a new string keeping only the first occurrence of each character, preserving order.

In plain English

Track which characters you've already emitted in a set; append a character only the first time you see it.

Examples

Example 1

Input: s = "banana"
Output: "ban"
Note: First b,a,n kept.

Example 2

Input: s = "aabbcc"
Output: "abc"
Note: One of each.

Example 3

Input: s = "abc"
Output: "abc"
Note: Already unique.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions1First question: Keep first or last occurrence?Explore questions

Q1.Keep first or last occurrence?
First, preserving order; last-occurrence is a small variant.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Keep a 'seen' set; for each character, if it isn't in the set, append it and add it. O(n) time, O(k) space.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you combine a seen-set with an order-preserving build.

Where candidates slip

Reordering output, or O(n²) concatenation.

How to stand out

Use the `set.add` boolean idiom to filter and record in one step.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Seen set build

TimeO(n)SpaceO(k)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

What structure tells you 'have I already kept this char?'

2Hint 2Explore hint

Append only on first sighting to preserve order.

3Hint 3Explore hint

Use a StringBuilder; `set.add` returns true only when newly added.

A · Core idea

The Java idiom `seen.add(ch)`. Set.add returns true only when the element was newly added, so if (seen.add(ch)) sb.append(ch); both records and filters in one line.

Performance. Use a StringBuilder, not +=, to keep the build O(n). The set gives O(1) membership, so the whole thing is O(n) time, O(k) space.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

s = "banana"

new

Stateb

Final result"ban"

C · Code with teaching comments

solution.javaJava

Seen set build — optimal

1 teaching note

1
public String removeDuplicates(String s) {
2
    StringBuilder sb = new StringBuilder();
3
    java.util.Set<Character> seen = new java.util.HashSet<>();
4
    for (char ch : s.toCharArray()) {
5
        // add returns true only the first time — append exactly once per char.
6
        if (seen.add(ch)) sb.append(ch);
7
    }
8
    return sb.toString();
9
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.9 lines

D · Why it works and complexityO(n) timeO(k) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(k) space — O(1) set lookup and append per character; the set holds k ≤ alphabet-size keys.

E · Edge cases and pitfalls5s = "aabbcc" — → "abc".Explore edge cases

Edge cases handled

s = "aabbcc"→ "abc".
s = "abc"Already unique → "abc".
s = ""Empty → "".

Pitfalls

Not preserving original order (e.g. sorting).
Using String += in the loop.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Not preserving original order (e.g., sorting).
Using String += in the loop.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Seen-set filtered build

Code pattern to memorize

seen=set(); for ch: if ch not in seen: seen.add(ch); out+=ch

Rules to remember

1Append on first sighting.
2Set tracks seen chars.
3Order preserved.

Use this pattern when

First-occurrence dedup with order

Interview takeaway

First-occurrence-only = a seen set plus a builder.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Seen-set filtered build. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Same pattern