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easyBasicsSometimes asked

Count Upper and Lower

Basics · 6 min

Given a string s, return how many uppercase and lowercase letters it has, as [upper, lower].

In plain English

Classify each character by its ASCII range and bump the matching counter.

Examples

Example 1

Input: s = "Hello World"
Output: [2, 8]
Note: H,W upper; rest letters lower.

Example 2

Input: s = "ABC"
Output: [3, 0]
Note: All upper.

Example 3

Input: s = "abc123"
Output: [0, 3]
Note: Digits ignored.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions1First question: Count only ASCII letters?Explore questions

Q1.Count only ASCII letters?
Yes; Unicode letters would need isUpperCase/isLowerCase.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Two counters; for each char test the 'A'..'Z' and 'a'..'z' ranges and increment accordingly. Non-letters are skipped.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you classify by range and exclude non-letters.

Where candidates slip

Including non-letters in a count.

How to stand out

Note the shared classify-and-count pattern across these problems.

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

Range classify

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Two ASCII ranges separate the cases — which are they?

2Hint 2Explore hint

One counter per case; skip non-letters.

3Hint 3Explore hint

Return the counts in the agreed order.

A · Core idea

Classify each character by ASCII range — 'A'..'Z' for uppercase, 'a'..'z' for lowercase — and tally two counters. Non-letters fall in neither range and are skipped.

Why ranges, not isUpper/isLower only. The explicit range checks are O(1) and language-agnostic; Character.isUpperCase works too. Either way, the key is to exclude digits, spaces, and punctuation from both counts.

Why two counters. Uppercase and lowercase are disjoint categories; one counter each, one pass. Return them in the agreed order ([upper, lower]).

This is the same classify-and-count shape as vowels/consonants and positive/negative.

B · Visual dry run

Interactive dry run

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

s = "Hi!"

A-Z

Stateupper=1

Final result[1, 1]

C · Code with teaching comments

solution.javaJava

Range classify — optimal

1
public int[] countCase(String s) {
2
    int upper = 0, lower = 0;
3
    for (char ch : s.toCharArray()) {
4
        if (ch >= 'A' && ch <= 'Z') upper++;
5
        else if (ch >= 'a' && ch <= 'z') lower++;
6
    }
7
    return new int[]{upper, lower};
8
}

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

ComplexityO(n) time, O(1) space — one or two range checks and an increment per character.

E · Edge cases and pitfalls5s = "ABC" — → [3, 0].Explore edge cases

Edge cases handled

s = "ABC"→ [3, 0].
s = "abc123"Digits ignored → [0, 3].
s = ""→ [0, 0].

Pitfalls

Counting digits or spaces as letters.
Returning the counts in the wrong order.

Practice

Try the solution yourself

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

Editable code previewpublic int[] countCase(String s) { int upper = 0, lower = 0; for (char ch : s.toCharArray()) { if (ch >= 'A' && ch <= 'Z') upper++; else if (ch >= 'a' && ch <= 'z') lower++;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 digits or spaces as letters.
Returning counts in the wrong order.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

ASCII-range classify

Code pattern to memorize

if A<=ch<=Z upper++ elif a<=ch<=z lower++

Rules to remember

1Upper: A-Z.
2Lower: a-z.
3Skip non-letters.

Use this pattern when

Bucketing characters by case/category

Interview takeaway

Two ASCII ranges separate uppercase from lowercase.

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: ASCII-range classify. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

easyBasicsSometimes asked

Count Upper and Lower

Basics · 6 min

Given a string s, return how many uppercase and lowercase letters it has, as [upper, lower].

In plain English

Classify each character by its ASCII range and bump the matching counter.

Examples

Example 1

Input: s = "Hello World"
Output: [2, 8]
Note: H,W upper; rest letters lower.

Example 2

Input: s = "ABC"
Output: [3, 0]
Note: All upper.

Example 3

Input: s = "abc123"
Output: [0, 3]
Note: Digits ignored.

Constraints

0 ≤ s.length ≤ 10^4

Clarifying questions1First question: Count only ASCII letters?Explore questions

Q1.Count only ASCII letters?
Yes; Unicode letters would need isUpperCase/isLowerCase.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Two counters; for each char test the 'A'..'Z' and 'a'..'z' ranges and increment accordingly. Non-letters are skipped.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you classify by range and exclude non-letters.

Where candidates slip

Including non-letters in a count.

How to stand out

Note the shared classify-and-count pattern across these problems.

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

Range classify

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Two ASCII ranges separate the cases — which are they?

2Hint 2Explore hint

One counter per case; skip non-letters.

3Hint 3Explore hint

Return the counts in the agreed order.

A · Core idea

Classify each character by ASCII range — 'A'..'Z' for uppercase, 'a'..'z' for lowercase — and tally two counters. Non-letters fall in neither range and are skipped.

Why two counters. Uppercase and lowercase are disjoint categories; one counter each, one pass. Return them in the agreed order ([upper, lower]).

This is the same classify-and-count shape as vowels/consonants and positive/negative.

B · Visual dry run

Interactive dry run

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

s = "Hi!"

A-Z

Stateupper=1

Final result[1, 1]

C · Code with teaching comments

solution.javaJava

Range classify — optimal

1
public int[] countCase(String s) {
2
    int upper = 0, lower = 0;
3
    for (char ch : s.toCharArray()) {
4
        if (ch >= 'A' && ch <= 'Z') upper++;
5
        else if (ch >= 'a' && ch <= 'z') lower++;
6
    }
7
    return new int[]{upper, lower};
8
}

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

ComplexityO(n) time, O(1) space — one or two range checks and an increment per character.

E · Edge cases and pitfalls5s = "ABC" — → [3, 0].Explore edge cases

Edge cases handled

s = "ABC"→ [3, 0].
s = "abc123"Digits ignored → [0, 3].
s = ""→ [0, 0].

Pitfalls

Counting digits or spaces as letters.
Returning the counts in the wrong order.

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 digits or spaces as letters.
Returning counts in the wrong order.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

ASCII-range classify

Code pattern to memorize

if A<=ch<=Z upper++ elif a<=ch<=z lower++

Rules to remember

1Upper: A-Z.
2Lower: a-z.
3Skip non-letters.

Use this pattern when

Bucketing characters by case/category

Interview takeaway

Two ASCII ranges separate uppercase from lowercase.

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: ASCII-range classify. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups