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easyBasicsSometimes asked

Count Vowels and Consonants

Basics · 6 min

Given a string s of English letters, return how many vowels and how many consonants it contains, as [vowels, consonants].

In plain English

Lower-case each letter and check membership in the vowel set; everything else that's a letter is a consonant.

Examples

Example 1

Input: s = "hello"
Output: [2, 3]
Note: e,o vowels; h,l,l consonants.

Example 2

Input: s = "xyz"
Output: [0, 3]
Note: No vowels.

Example 3

Input: s = "aeiou"
Output: [5, 0]
Note: All vowels.

Constraints

0 ≤ s.length ≤ 10^4
Letters only

Clarifying questions2First question: Is 'y' a vowel?Explore questions

Q1.Is 'y' a vowel?
Usually no for this problem; easy to include if specified.
Q2.Are there non-letters?
If so I skip them rather than counting as consonants.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

For each letter, normalise case and test if it's one of a,e,i,o,u; increment vowels or consonants accordingly.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you use set membership and exclude non-letters.

Where candidates slip

Treating any non-vowel (including spaces/digits) as a consonant.

How to stand out

Note the design lets you add 'y' or other alphabets trivially.

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

Vowel-set membership

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

How do you test if a character is one of a,e,i,o,u cleanly?

2Hint 2Explore hint

Lower-case first so 'A' and 'a' share one check.

3Hint 3Explore hint

Only count consonants for characters that are actually letters.

A · Core idea

Classify each letter by set membership in the fixed vowel set {a,e,i,o,u}: a vowel if it's in the set, otherwise (if it's a letter) a consonant.

Why normalise case first. Lower-casing each character means one membership test handles both 'A' and 'a'. Without it you'd need to check ten characters instead of five.

Why guard with isLetter. Spaces, digits, and punctuation are neither vowels nor consonants. Counting them as consonants (the default 'else') is the common bug — gate the consonant branch on 'is this actually a letter?'.

Why a set, not a chain of ORs. Membership in a small fixed alphabet is clearest as a set lookup; it also generalises if the alphabet grows (e.g. treating 'y' as a vowel).

B · Visual dry run

Interactive dry run

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

s = "Hi!"

letter, not vowel

Statecons=1

Final result[1, 1]

C · Code with teaching comments

solution.javaJava

Vowel-set membership — optimal

1
public int[] countVowelsConsonants(String s) {
2
    int vowels = 0, consonants = 0;
3
    String set = "aeiou";
4
    for (char ch : s.toLowerCase().toCharArray()) {
5
        if (Character.isLetter(ch)) {
6
            if (set.indexOf(ch) >= 0) vowels++;
7
            else consonants++;
8
        }
9
    }
10
    return new int[]{vowels, consonants};
11
}

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

ComplexityO(n) time, O(1) space — one case-fold and one membership test per character.

E · Edge cases and pitfalls5s = "xyz" — No vowels → [0, 3].Explore edge cases

Edge cases handled

s = "xyz"No vowels → [0, 3].
s = "aeiou"All vowels → [5, 0].
s = "a1 b"Digit and space skipped → vowels 1, consonants 1.

Pitfalls

Counting uppercase vowels as consonants (normalise case).
Counting spaces/digits as consonants.

Practice

Try the solution yourself

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

Editable code previewpublic int[] countVowelsConsonants(String s) { int vowels = 0, consonants = 0; String set = "aeiou"; for (char ch : s.toLowerCase().toCharArray()) { if (Character.isLetter(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.

Counting uppercase vowels as consonants (normalise case first).
Counting spaces or digits as consonants.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Set membership classify

Code pattern to memorize

if ch in 'aeiou' vow++ elif isLetter(ch) cons++

Rules to remember

1Lower-case first.
2Vowel ⟺ in {a,e,i,o,u}.
3Other letters → consonant.

Use this pattern when

Classifying over a small fixed alphabet

Interview takeaway

Fixed small alphabets are best tested with set membership.

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: Set membership classify. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Count Upper and Lower
Same classify-and-count with ASCII ranges.

easyBasicsSometimes asked

Count Vowels and Consonants

Basics · 6 min

Given a string s of English letters, return how many vowels and how many consonants it contains, as [vowels, consonants].

In plain English

Lower-case each letter and check membership in the vowel set; everything else that's a letter is a consonant.

Examples

Example 1

Input: s = "hello"
Output: [2, 3]
Note: e,o vowels; h,l,l consonants.

Example 2

Input: s = "xyz"
Output: [0, 3]
Note: No vowels.

Example 3

Input: s = "aeiou"
Output: [5, 0]
Note: All vowels.

Constraints

0 ≤ s.length ≤ 10^4
Letters only

Clarifying questions2First question: Is 'y' a vowel?Explore questions

Q1.Is 'y' a vowel?
Usually no for this problem; easy to include if specified.
Q2.Are there non-letters?
If so I skip them rather than counting as consonants.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

For each letter, normalise case and test if it's one of a,e,i,o,u; increment vowels or consonants accordingly.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you use set membership and exclude non-letters.

Where candidates slip

Treating any non-vowel (including spaces/digits) as a consonant.

How to stand out

Note the design lets you add 'y' or other alphabets trivially.

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

Vowel-set membership

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

How do you test if a character is one of a,e,i,o,u cleanly?

2Hint 2Explore hint

Lower-case first so 'A' and 'a' share one check.

3Hint 3Explore hint

Only count consonants for characters that are actually letters.

A · Core idea

Classify each letter by set membership in the fixed vowel set {a,e,i,o,u}: a vowel if it's in the set, otherwise (if it's a letter) a consonant.

Why normalise case first. Lower-casing each character means one membership test handles both 'A' and 'a'. Without it you'd need to check ten characters instead of five.

Why a set, not a chain of ORs. Membership in a small fixed alphabet is clearest as a set lookup; it also generalises if the alphabet grows (e.g. treating 'y' as a vowel).

B · Visual dry run

Interactive dry run

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

s = "Hi!"

letter, not vowel

Statecons=1

Final result[1, 1]

C · Code with teaching comments

solution.javaJava

Vowel-set membership — optimal

1
public int[] countVowelsConsonants(String s) {
2
    int vowels = 0, consonants = 0;
3
    String set = "aeiou";
4
    for (char ch : s.toLowerCase().toCharArray()) {
5
        if (Character.isLetter(ch)) {
6
            if (set.indexOf(ch) >= 0) vowels++;
7
            else consonants++;
8
        }
9
    }
10
    return new int[]{vowels, consonants};
11
}

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

ComplexityO(n) time, O(1) space — one case-fold and one membership test per character.

E · Edge cases and pitfalls5s = "xyz" — No vowels → [0, 3].Explore edge cases

Edge cases handled

s = "xyz"No vowels → [0, 3].
s = "aeiou"All vowels → [5, 0].
s = "a1 b"Digit and space skipped → vowels 1, consonants 1.

Pitfalls

Counting uppercase vowels as consonants (normalise case).
Counting spaces/digits as consonants.

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.

Counting uppercase vowels as consonants (normalise case first).
Counting spaces or digits as consonants.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Set membership classify

Code pattern to memorize

if ch in 'aeiou' vow++ elif isLetter(ch) cons++

Rules to remember

1Lower-case first.
2Vowel ⟺ in {a,e,i,o,u}.
3Other letters → consonant.

Use this pattern when

Classifying over a small fixed alphabet

Interview takeaway

Fixed small alphabets are best tested with set membership.

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: Set membership classify. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Count Upper and Lower
Same classify-and-count with ASCII ranges.