An inverted right-aligned triangle shrinks star counts while staying flush on the right: row i has i - 1 leading spaces and rows - i + 1 stars.
Remember
Rule: spaces = i - 1, stars = rows - i + 1
*****
****
***
**
* ← 5 rows (spaces shown as blanks)
It combines Program 2’s shrinking stars with Program 3’s right alignment. Every row still has width rows before the newline.
Approach
How to Solve It
Two inner loops per row — spaces then stars — or the same formulas with String.repeat.
Method
Idea
Best for
Nested loops
j < i spaces, then k = i..rows stars
Learning, interviews, exams
String.repeat
Build spaces and stars as whole strings
Shorter demos once loops click (Java 11+)
Pseudocode
Pseudocode
for i from 1 to rows:
for j from 1 to i - 1: // i - 1 spaces
print " " (no newline)
for k from i to rows: // rows - i + 1 stars
print "*" (no newline)
print newline
Cheat sheet
Goal
Pattern
Walk each row
for (i = 1; i <= rows; i++)
Leading spaces
for (j = 1; j < i; j++) System.out.print(" ");
Shrinking stars
for (k = i; k <= rows; k++) System.out.print("*");
Star count form
for (k = 1; k <= rows - i + 1; k++)
Fixed width check
(i - 1) + (rows - i + 1) == rows
Row shortcut
System.out.print(" ".repeat(i - 1)); System.out.println("*".repeat(rows - i + 1));
print vs println
API
Effect
Use for
System.out.print
Stays on the same line
Each space and each *
System.out.println
Ends the current line
After spaces and stars for that row
Try it
Live Preview
Change the row count and the inverted right-aligned triangle updates instantly.
Whole numbers from 1 to 20. Each row has width rows (spaces + stars).
Live result5 rows · 15 stars
*****
****
***
**
*
Trace
Worked Walkthrough — rows = 4
Trace spaces, stars, and total width for each outer-loop value of i.
i
Spaces i - 1
Stars rows - i + 1
Width
Printed row
1
0
4
4
****
2
1
3
4
***
3
2
2
4
**
4
3
1
4
*
Total stars: 4 + 3 + 2 + 1 = 10 = 4×5/2. Width stays 4 on every row.
Code
Java Programs
Three complete programs: fixed rows, Scanner input, and a String.repeat shortcut. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded height — space loop first (j < i), then stars k = i..rows.
Java
public class InvertedRightAligned {
public static void main(String[] args) {
int rows = 5;
for (int i = 1; i <= rows; i++) {
for (int j = 1; j < i; j++) {
System.out.print(" ");
}
for (int k = i; k <= rows; k++) {
System.out.print("*");
}
System.out.println();
}
}
}
Output
*****
****
***
**
*
How It Works
1. Set height.rows = 5 means five lines, each of width 5.
2. Outer loop picks the row.i runs from 1 to rows.
3. Spaces then stars. Print i - 1 spaces (j < i), then stars for k = i..rows (that is rows - i + 1 stars).
4. Break the line.System.out.println() after both inner loops starts the next row.
When i = 1: 0 spaces + 5 stars. When i = 5: 4 spaces + 1 star.
Example 2 — User Input Version
Read the height at runtime with Scanner. Prefer hasNextInt() in real apps (shown in the tip below).
Java
import java.util.Scanner;
public class InvertedRightAlignedInput {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter the number of rows: ");
int rows = sc.nextInt();
for (int i = 1; i <= rows; i++) {
for (int j = 1; j < i; j++) {
System.out.print(" ");
}
for (int k = i; k <= rows; k++) {
System.out.print("*");
}
System.out.println();
}
}
}
Output (when user enters 4)
Enter the number of rows: 4
****
***
**
*
How It Works
1. Prompt and read. Ask for a row count, then store it with sc.nextInt().
2. Same nested-loop core. Only the source of rows changes — the print logic matches Example 1.
if (!sc.hasNextInt()) {
System.out.println("Enter a positive whole number.");
return;
}
int rows = sc.nextInt();
if (rows < 1) {
System.out.println("Enter a positive whole number.");
return;
}
Example 3 — "*".repeat() + Explicit Count
Name the space and star counts, then build each row in two calls (Java 11+).
Java
public class InvertedRightAlignedRepeat {
public static void main(String[] args) {
int rows = 5;
for (int i = 1; i <= rows; i++) {
int spaces = i - 1;
int stars = rows - i + 1;
System.out.print(" ".repeat(spaces));
System.out.println("*".repeat(stars));
}
}
}
Output
*****
****
***
**
*
How It Works
1. Compute both counts.spaces = i - 1 and stars = rows - i + 1 make the invert-and-align rule obvious.
2. Build and print.print the space string, then println the star string (newline included).
3. Learn loops first. Use Examples 1–2 when you need to show nested bounds; treat this as a polish shortcut afterward.
Edge Cases & Pitfalls
Check these before calling the solution done.
j <= i
One extra space
Space loop must be j < i (exactly i - 1 spaces). j <= i breaks the right edge.
Program 3 formulas
Grows instead
rows - i spaces and 1..i stars is Program 3. Here use i - 1 and rows - i + 1.
No spaces
Left-aligned invert
Skipping the space loop gives Program 2. Right alignment needs leading spaces.
rows = 1
Single star
0 spaces + 1 star — same tip case as the other triangle pages.
rows ≤ 0
Empty output
Outer loop never runs. Validate and re-prompt for interactive programs.
Bad input
Use hasNextInt()
nextInt() throws on letters — prefer sc.hasNextInt() and require rows >= 1.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(rows²)
O(1)
String.repeat shortcut (Example 3)
O(rows²)
O(rows) temporary per row string
Each of n rows prints Θ(n) characters (spaces + stars). Star count alone is still n(n+1)/2.
Remember
Key Takeaways
Formulas:i - 1 spaces and rows - i + 1 stars.
Fixed width: spaces + stars = rows on every line.
Break the row:print for spaces/stars; println after both loops.
Complexity:O(n²) time; O(1) extra space for nested loops.
One line: print i - 1 spaces, then rows - i + 1 stars — inverted and flush right.
Frequently Asked Questions
For each row i from 1 to rows, print i minus 1 spaces, then print stars with k running from i to rows inclusive. That prints rows minus i plus 1 stars. Row 1 has no spaces and rows stars; each later row adds one space and removes one star while keeping the same right edge.
The range i through rows has length rows minus i plus 1, which matches the star count. An equivalent loop is k from 1 to rows minus i plus 1.
Program 3 uses (rows - i) spaces and stars 1 through i. Program 4 uses (i - 1) spaces and stars i through rows. Same right alignment; star counts grow in Program 3 and shrink in Program 4.
Program 2 is left-aligned with shrinking stars. Program 4 adds growing leading spaces so the same shrinking star counts stay flush on the right.
System.out.print stays on the same line. System.out.println ends the current line. Spaces and stars use print; the row break uses println after both inner loops.
O(n²) for n rows. Each row prints on the order of n characters; there are n rows.
Yes. System.out.print(" ".repeat(i - 1)); System.out.println("*".repeat(rows - i + 1)); builds each row without explicit inner character loops (Java 11+).
j from 1 to i-1 (written as j < i) prints exactly i - 1 spaces. Using j <= i would add one extra space and break the right edge.
🤔
Did you know?
This pattern merges Program 2’s shrinking star count with Program 3’s right alignment. Every row still has width rows: (i - 1) + (rows - i + 1) = rows.
