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easyBasicsSometimes asked

Pyramid Star Pattern

Basics · 6 min

Given n, return a centred pyramid of *: row i has n-i leading spaces and 2i-1 stars, rows separated by newlines.

In plain English

Each row mixes spaces then stars; the counts depend on the row index and n.

Examples

Example 1

Input: n = 3
Output: " *\n ***\n*****"
Note: 1,3,5 stars centred.

Example 2

Input: n = 1
Output: "*"
Note: Single.

Example 3

Input: n = 2
Output: " *\n***"
Note: 1 then 3.

Constraints

1 ≤ n ≤ 1000

Clarifying questions1First question: Centred pyramid or right triangle?Explore questions

Q1.Centred pyramid or right triangle?
Centred — so each row has leading spaces.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

For row i (1..n): append `n-i` spaces, then `2i-1` stars, then a newline between rows.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you can derive per-row counts from the row index.

Where candidates slip

Wrong space/star arithmetic producing an off-centre shape.

How to stand out

State the formulas (n−i spaces, 2i−1 stars) and why stars are odd.

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

Spaces + stars per row

TimeO(n²)SpaceO(n²)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Each row has spaces THEN stars — how many of each?

2Hint 2Explore hint

Spaces shrink as stars grow: derive both from the row index.

3Hint 3Explore hint

Stars per row are odd (2i−1) to keep the pyramid centred.

A · Core idea

A centred pyramid needs two inner loops per row: one for the leading spaces that push the row toward the centre, and one for the stars. The arithmetic linking the row index to the counts is the lesson.

Deriving the counts. For row i (1..n): leading spaces = n − i (fewer as you go down), and stars = 2i − 1 (the odd sequence 1, 3, 5, …) so the pyramid is symmetric about the centre. Plugging i into these formulas is the whole exercise.

Why 2i − 1 stars. An odd count guarantees a single centre column, giving the clean triangular silhouette; even counts wouldn't centre.

Newline discipline. As with all patterns, newlines go between rows.

B · Visual dry run

Interactive dry run

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

n = 3

i=1

2 spaces, 1 star

State" *"

Final result" *\n ***\n*****"

C · Code with teaching comments

solution.javaJava

Spaces + stars per row — optimal

2 teaching notes

1
public String pyramid(int n) {
2
    StringBuilder sb = new StringBuilder();
3
    for (int i = 1; i <= n; i++) {
4
        for (int s = 0; s < n - i; s++) sb.append(' ');
5
        for (int j = 0; j < 2 * i - 1; j++) sb.append('*');
6
        if (i < n) sb.append('\n');
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(n²) timeO(n²) spaceOpen the invariant and proofExplore the proof

The arithmetic linking row index to space/star counts is what this pattern teaches.

ComplexityO(n²) time, O(n²) space — row i emits (n−i) spaces and (2i−1) stars; summed over rows it's O(n²).

E · Edge cases and pitfalls4n = 1 — "*".Explore edge cases

Edge cases handled

n = 1"*".
n = 2" *\n***".

Pitfalls

Wrong space/star formula (off-centre or wrong widths).
Forgetting the spaces entirely.

Practice

Try the solution yourself

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

Editable code previewpublic String pyramid(int n) { StringBuilder sb = new StringBuilder(); for (int i = 1; i <= n; i++) { for (int s = 0; s < n - i; s++) sb.append(' '); for (int j = 0; j < 2 * i - 1; j++) sb.append('*');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.

Wrong space/star formula (off-centre or wrong widths).
Forgetting the spaces entirely.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Indexed spaces + stars

Code pattern to memorize

row i: (n-i) spaces + (2i-1) stars

Rules to remember

1Spaces = n − i.
2Stars = 2i − 1.
3Two inner loops.

Use this pattern when

Centred / index-derived patterns

Interview takeaway

Centred patterns derive counts from the row index.

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: Indexed spaces + stars. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Right Triangle Star
Drop the spaces for a left-aligned triangle.

easyBasicsSometimes asked

Pyramid Star Pattern

Basics · 6 min

Given n, return a centred pyramid of *: row i has n-i leading spaces and 2i-1 stars, rows separated by newlines.

In plain English

Each row mixes spaces then stars; the counts depend on the row index and n.

Examples

Example 1

Input: n = 3
Output: " *\n ***\n*****"
Note: 1,3,5 stars centred.

Example 2

Input: n = 1
Output: "*"
Note: Single.

Example 3

Input: n = 2
Output: " *\n***"
Note: 1 then 3.

Constraints

1 ≤ n ≤ 1000

Clarifying questions1First question: Centred pyramid or right triangle?Explore questions

Q1.Centred pyramid or right triangle?
Centred — so each row has leading spaces.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

For row i (1..n): append `n-i` spaces, then `2i-1` stars, then a newline between rows.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you can derive per-row counts from the row index.

Where candidates slip

Wrong space/star arithmetic producing an off-centre shape.

How to stand out

State the formulas (n−i spaces, 2i−1 stars) and why stars are odd.

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

Spaces + stars per row

TimeO(n²)SpaceO(n²)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Each row has spaces THEN stars — how many of each?

2Hint 2Explore hint

Spaces shrink as stars grow: derive both from the row index.

3Hint 3Explore hint

Stars per row are odd (2i−1) to keep the pyramid centred.

A · Core idea

Why 2i − 1 stars. An odd count guarantees a single centre column, giving the clean triangular silhouette; even counts wouldn't centre.

Newline discipline. As with all patterns, newlines go between rows.

B · Visual dry run

Interactive dry run

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

n = 3

i=1

2 spaces, 1 star

State" *"

Final result" *\n ***\n*****"

C · Code with teaching comments

solution.javaJava

Spaces + stars per row — optimal

2 teaching notes

1
public String pyramid(int n) {
2
    StringBuilder sb = new StringBuilder();
3
    for (int i = 1; i <= n; i++) {
4
        for (int s = 0; s < n - i; s++) sb.append(' ');
5
        for (int j = 0; j < 2 * i - 1; j++) sb.append('*');
6
        if (i < n) sb.append('\n');
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(n²) timeO(n²) spaceOpen the invariant and proofExplore the proof

The arithmetic linking row index to space/star counts is what this pattern teaches.

ComplexityO(n²) time, O(n²) space — row i emits (n−i) spaces and (2i−1) stars; summed over rows it's O(n²).

E · Edge cases and pitfalls4n = 1 — "*".Explore edge cases

Edge cases handled

n = 1"*".
n = 2" *\n***".

Pitfalls

Wrong space/star formula (off-centre or wrong widths).
Forgetting the spaces entirely.

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.

Wrong space/star formula (off-centre or wrong widths).
Forgetting the spaces entirely.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Indexed spaces + stars

Code pattern to memorize

row i: (n-i) spaces + (2i-1) stars

Rules to remember

1Spaces = n − i.
2Stars = 2i − 1.
3Two inner loops.

Use this pattern when

Centred / index-derived patterns

Interview takeaway

Centred patterns derive counts from the row index.

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: Indexed spaces + stars. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Right Triangle Star
Drop the spaces for a left-aligned triangle.