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easyBasicsSometimes asked

Count a Character

Basics · 6 min

Given a string s and a character ch, return how many times ch appears in s.

In plain English

One pass with a counter incremented on each match.

Examples

Example 1

Input: s = "banana", ch = 'a'
Output: 3
Note: Three a's.

Example 2

Input: s = "hello", ch = 'z'
Output: 0
Note: Absent.

Example 3

Input: s = "aaa", ch = 'a'
Output: 3
Note: All match.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions1First question: Case-sensitive match?Explore questions

Q1.Case-sensitive match?
I'll confirm; lower-casing both makes it case-insensitive.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Scan the string, incrementing a counter whenever the current character equals `ch`.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you tally all matches without early exit.

Where candidates slip

Stopping at the first occurrence.

How to stand out

Mention the frequency-map preprocessing for many queries.

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

Count matches

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Tally every match — should you stop early?

2Hint 2Explore hint

One counter, bump on equality, visit all characters.

3Hint 3Explore hint

Many targets? Build a frequency map once.

A · Core idea

Count how many times a target character appears: one pass, a counter bumped on each match, no early exit. It's count-occurrences for arrays applied to characters.

Why no early exit. You want the total, so unlike a search you must visit every character. Returning on the first match is the classic copy-from-search bug.

Comparing chars, not strings. Compare with == on char values (a value comparison) — fine. The Java == reference trap only bites when comparing String objects, not primitive chars.

Repeated queries. For many different target characters, a single frequency map answers each in O(1) after one O(n) build.

B · Visual dry run

Interactive dry run

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

s = "banana", ch = 'a'

b,n

Statecount=0

Final result3

C · Code with teaching comments

solution.javaJava

Count matches — optimal

1
public int count(String s, char ch) {
2
    int c = 0;
3
    for (int i = 0; i < s.length(); i++) {
4
        if (s.charAt(i) == ch) c++;
5
    }
6
    return c;
7
}

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

ComplexityO(n) time, O(1) space — every character is compared once; no early termination since all matches count.

E · Edge cases and pitfalls5s = "hello", ch = 'z' — Absent → 0.Explore edge cases

Edge cases handled

s = "hello", ch = 'z'Absent → 0.
s = "aaa", ch = 'a'All match → 3.
s = "", ch = 'a'Empty → 0.

Pitfalls

Returning after the first match.
Comparing String objects with == when you meant char equality.

Practice

Try the solution yourself

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

Editable code previewpublic int count(String s, char ch) { int c = 0; for (int i = 0; i < s.length(); i++) { if (s.charAt(i) == ch) c++; }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.

Returning after the first match.
Comparing strings with == in Java instead of char equality (here we use char, so fine).

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Count matches

Code pattern to memorize

c=0; for ch in s: if ch==target: c++

Rules to remember

1Counter from 0.
2Bump on equality.
3Visit every char.

Use this pattern when

Single-character tally

Interview takeaway

Character counting is a one-pass tally.

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 matches. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Character Frequency
Count all characters at once.

easyBasicsSometimes asked

Count a Character

Basics · 6 min

Given a string s and a character ch, return how many times ch appears in s.

In plain English

One pass with a counter incremented on each match.

Examples

Example 1

Input: s = "banana", ch = 'a'
Output: 3
Note: Three a's.

Example 2

Input: s = "hello", ch = 'z'
Output: 0
Note: Absent.

Example 3

Input: s = "aaa", ch = 'a'
Output: 3
Note: All match.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions1First question: Case-sensitive match?Explore questions

Q1.Case-sensitive match?
I'll confirm; lower-casing both makes it case-insensitive.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Scan the string, incrementing a counter whenever the current character equals `ch`.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you tally all matches without early exit.

Where candidates slip

Stopping at the first occurrence.

How to stand out

Mention the frequency-map preprocessing for many queries.

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

Count matches

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Tally every match — should you stop early?

2Hint 2Explore hint

One counter, bump on equality, visit all characters.

3Hint 3Explore hint

Many targets? Build a frequency map once.

A · Core idea

Count how many times a target character appears: one pass, a counter bumped on each match, no early exit. It's count-occurrences for arrays applied to characters.

Why no early exit. You want the total, so unlike a search you must visit every character. Returning on the first match is the classic copy-from-search bug.

Comparing chars, not strings. Compare with == on char values (a value comparison) — fine. The Java == reference trap only bites when comparing String objects, not primitive chars.

Repeated queries. For many different target characters, a single frequency map answers each in O(1) after one O(n) build.

B · Visual dry run

Interactive dry run

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

s = "banana", ch = 'a'

b,n

Statecount=0

Final result3

C · Code with teaching comments

solution.javaJava

Count matches — optimal

1
public int count(String s, char ch) {
2
    int c = 0;
3
    for (int i = 0; i < s.length(); i++) {
4
        if (s.charAt(i) == ch) c++;
5
    }
6
    return c;
7
}

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

ComplexityO(n) time, O(1) space — every character is compared once; no early termination since all matches count.

E · Edge cases and pitfalls5s = "hello", ch = 'z' — Absent → 0.Explore edge cases

Edge cases handled

s = "hello", ch = 'z'Absent → 0.
s = "aaa", ch = 'a'All match → 3.
s = "", ch = 'a'Empty → 0.

Pitfalls

Returning after the first match.
Comparing String objects with == when you meant char equality.

Practice

Try the solution yourself

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

Editable code previewpublic int count(String s, char ch) { int c = 0; for (int i = 0; i < s.length(); i++) { if (s.charAt(i) == ch) c++; }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.

Returning after the first match.
Comparing strings with == in Java instead of char equality (here we use char, so fine).

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Count matches

Code pattern to memorize

c=0; for ch in s: if ch==target: c++

Rules to remember

1Counter from 0.
2Bump on equality.
3Visit every char.

Use this pattern when

Single-character tally

Interview takeaway

Character counting is a one-pass tally.

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 matches. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Character Frequency
Count all characters at once.