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easyBasicsSometimes asked

First Non-Repeating Character

Basics · 6 min

Given a string s, return the first character that appears exactly once, or a space ' ' if there is none.

In plain English

Count every character, then scan again and return the first whose count is 1.

Examples

Example 1

Input: s = "leetcode"
Output: 'l'
Note: l appears once and is first.

Example 2

Input: s = "aabb"
Output: ' '
Note: No unique char.

Example 3

Input: s = "abca"
Output: 'b'
Note: b is the first unique.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions2First question: Return the char or its index?Explore questions

Q1.Return the char or its index?
I'll return the character; index is a trivial change.
Q2.What if none is unique?
Return the agreed sentinel (space here).

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Two passes: first build a frequency map, then walk the string and return the first character with count 1. O(n) time.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you use a count-then-scan two-pass and preserve order for 'first'.

Where candidates slip

Iterating the map and losing positional order.

How to stand out

Mention LinkedHashMap / storing first-index as one-structure alternatives.

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

Frequency then scan

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Can you decide uniqueness before seeing the whole string?

2Hint 2Explore hint

Pass 1: count all characters. Pass 2: find the first with count 1.

3Hint 3Explore hint

Re-scan the string (ordered), not the map, for 'first'.

A · Core idea

Two passes: first count every character's frequency, then re-scan the string in order and return the first character whose count is exactly 1. O(n) time.

Why two passes are necessary. You cannot know a character is unique until you've seen the entire string, so the count must be complete before you judge any character. Pass one builds that knowledge; pass two applies it.

Why re-scan the string, not the map. 'First' means first by position in the original string. A hash map has no positional order, so iterating the map could return the wrong character. Re-scanning the input preserves order. (A LinkedHashMap or storing first-index would also work.)

Return convention. When no unique character exists, return the agreed sentinel (a space here, or -1 if returning an index).

B · Visual dry run

Interactive dry run

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

s = "abca"

count

a:2, b:1, c:1

Statetable

Final result'b'

C · Code with teaching comments

solution.javaJava

Frequency then scan — optimal

2 teaching notes

1
public char firstUnique(String s) {
2
    int[] count = new int[256];
3
    for (int i = 0; i < s.length(); i++) count[s.charAt(i)]++;
4
    for (int i = 0; i < s.length(); i++) {
5
        if (count[s.charAt(i)] == 1) return s.charAt(i);
6
    }
7
    return ' ';
8
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.8 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

The 'count first, then re-scan in order' two-pass shape recurs in many string problems.

ComplexityO(n) time, O(1) space — two linear passes (count, then scan) with a fixed-size count table.

E · Edge cases and pitfalls5s = "leetcode" — 'l' is first with count 1.Explore edge cases

Edge cases handled

s = "leetcode"'l' is first with count 1.
s = "aabb"No unique char → ' '.
s = ""Empty → ' '.

Pitfalls

Scanning the (unordered) map instead of the string for 'first'.
Returning an index when a character is expected (or vice versa).

Practice

Try the solution yourself

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

Editable code previewpublic char firstUnique(String s) { int[] count = new int[256]; for (int i = 0; i < s.length(); i++) count[s.charAt(i)]++; for (int i = 0; i < s.length(); i++) { if (count[s.charAt(i)] == 1) return s.charAt(i);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.

Scanning the map (unordered) instead of the string for the 'first' requirement.
Returning an index instead of the character.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Count then re-scan in order

Code pattern to memorize

count all; then for ch in s: if count[ch]==1 return ch

Rules to remember

1Pass 1: frequencies.
2Pass 2: first count == 1.
3Re-scan the string for order.

Use this pattern when

'First/earliest with property' over counts

Interview takeaway

When 'first' matters, re-scan the input, not the map.

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: Count then re-scan in order. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

First Unique Character in a String
Coming soon
Return the index instead of the char.
Character Frequency
The count table this relies on.

easyBasicsSometimes asked

First Non-Repeating Character

Basics · 6 min

Given a string s, return the first character that appears exactly once, or a space ' ' if there is none.

In plain English

Count every character, then scan again and return the first whose count is 1.

Examples

Example 1

Input: s = "leetcode"
Output: 'l'
Note: l appears once and is first.

Example 2

Input: s = "aabb"
Output: ' '
Note: No unique char.

Example 3

Input: s = "abca"
Output: 'b'
Note: b is the first unique.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions2First question: Return the char or its index?Explore questions

Q1.Return the char or its index?
I'll return the character; index is a trivial change.
Q2.What if none is unique?
Return the agreed sentinel (space here).

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Two passes: first build a frequency map, then walk the string and return the first character with count 1. O(n) time.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you use a count-then-scan two-pass and preserve order for 'first'.

Where candidates slip

Iterating the map and losing positional order.

How to stand out

Mention LinkedHashMap / storing first-index as one-structure alternatives.

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

Frequency then scan

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Can you decide uniqueness before seeing the whole string?

2Hint 2Explore hint

Pass 1: count all characters. Pass 2: find the first with count 1.

3Hint 3Explore hint

Re-scan the string (ordered), not the map, for 'first'.

A · Core idea

Two passes: first count every character's frequency, then re-scan the string in order and return the first character whose count is exactly 1. O(n) time.

Return convention. When no unique character exists, return the agreed sentinel (a space here, or -1 if returning an index).

B · Visual dry run

Interactive dry run

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

s = "abca"

count

a:2, b:1, c:1

Statetable

Final result'b'

C · Code with teaching comments

solution.javaJava

Frequency then scan — optimal

2 teaching notes

1
public char firstUnique(String s) {
2
    int[] count = new int[256];
3
    for (int i = 0; i < s.length(); i++) count[s.charAt(i)]++;
4
    for (int i = 0; i < s.length(); i++) {
5
        if (count[s.charAt(i)] == 1) return s.charAt(i);
6
    }
7
    return ' ';
8
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.8 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

The 'count first, then re-scan in order' two-pass shape recurs in many string problems.

ComplexityO(n) time, O(1) space — two linear passes (count, then scan) with a fixed-size count table.

E · Edge cases and pitfalls5s = "leetcode" — 'l' is first with count 1.Explore edge cases

Edge cases handled

s = "leetcode"'l' is first with count 1.
s = "aabb"No unique char → ' '.
s = ""Empty → ' '.

Pitfalls

Scanning the (unordered) map instead of the string for 'first'.
Returning an index when a character is expected (or vice versa).

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.

Scanning the map (unordered) instead of the string for the 'first' requirement.
Returning an index instead of the character.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Count then re-scan in order

Code pattern to memorize

count all; then for ch in s: if count[ch]==1 return ch

Rules to remember

1Pass 1: frequencies.
2Pass 2: first count == 1.
3Re-scan the string for order.

Use this pattern when

'First/earliest with property' over counts

Interview takeaway

When 'first' matters, re-scan the input, not the map.

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: Count then re-scan in order. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

First Unique Character in a String
Coming soon
Return the index instead of the char.
Character Frequency
The count table this relies on.