A shrinking repeating number pattern starts with many copies of 1, then raises the digit while the row gets shorter — so you see 11111, 2222, 333, and so on.
Remember
Rule: outer i = 1..rows; inner j = i..rows; print i
11111
2222
333
44
5 ← rows = 5
Unlike Program 10 (digit falls while length grows) or Program 11 (digit and count both equal i counting down), this pattern counts the digit up while the row shrinks.
Approach
How to Solve It
Outer loop from 1 to rows. Inner loop from current i to rows, printing i each time. End the line after the inner loop.
Method
Idea
Best for
Nested loops
Outer up, inner i..rows, print i
Learning, interviews, exams
String.repeat
One println per row
When the bounds already make sense
Pseudocode
Pseudocode
for i from 1 to rows:
for j from i to rows:
print i
print newline
Cheat sheet
Goal
Pattern
Outer (digit rises)
for (int i = 1; i <= rows; i++)
Inner (repeat count)
for (int j = i; j <= rows; j++) System.out.print(i);
End the row
System.out.println();
Repeats on row i
rows - i + 1
Shortcut
String.valueOf(i).repeat(rows - i + 1)
Vs Program 10
Same length rule; digit counts up, not down
Printing Numbers vs Starting a New Line
API
Effect
Use for
System.out.print
Stays on the same line
Each digit i
System.out.println
Ends the current line
After the inner loop
Print digits without a newline, then end the row once.
Try it
Live Preview
Change the row count and the shrinking repeating pattern updates instantly.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live resultrows = 5 · 15 digits
11111
2222
333
44
5
Trace
Worked Walkthrough
Trace three outer values when rows = 5 — watch the digit rise while the repeat count falls.
Outer i
Inner j
Repeats
Prints
1
1..5
5
11111
3
3..5
3
333
5
5..5
1
5
The printed value is always the outer i; the inner loop only controls how many times it appears.
Code
Java Programs
Three complete programs: fixed rows = 5, Scanner input, and a String.repeat shortcut. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded height — outer up from 1, inner from i to 5, print i.
Java
public class ShrinkingRepeatingPattern {
public static void main(String[] args) {
int rows = 5;
for (int i = 1; i <= rows; i++) {
for (int j = i; j <= rows; j++) {
System.out.print(i);
}
System.out.println();
}
}
}
Output
11111
2222
333
44
5
How It Works
1. Outer sets the digit.i runs from 1 to rows — that value is what you print.
2. Inner sets the count.j runs from i to rows, so the digit appears rows - i + 1 times.
3. End the row. Call System.out.println() only after the inner loop finishes.
Example 2 — User Input Rows
Read the row count at runtime. Prefer hasNextInt() before nextInt() in real apps.
Java
import java.util.Scanner;
public class ShrinkingRepeatingPatternInput {
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 = i; j <= rows; j++) {
System.out.print(i);
}
System.out.println();
}
sc.close();
}
}
Output (when user enters 4)
Enter the number of rows: 4
1111
222
33
4
How It Works
1. Same loop core. Only the source of rows changes from a literal to Scanner.
2. Entering 4 stops early. You get four rows ending at a single 4.
3. Validate in real apps. Prefer checking hasNextInt() and requiring a positive height (tip below).
Safer input tip
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 — String.repeat Shortcut
Build each row in one call once the rows - i + 1 count is clear.
Java
public class ShrinkingRepeatingString {
public static void main(String[] args) {
int rows = 5;
for (int i = 1; i <= rows; i++) {
System.out.println(String.valueOf(i).repeat(rows - i + 1));
}
}
}
Output
11111
2222
333
44
5
How It Works
1. Same outer loop.i still counts up from 1 to rows.
2. Count formula.rows - i + 1 is exactly how many times the nested inner loop would run.
3. Learn loops first. Prefer Examples 1–2 in class; use repeat as a later shortcut (Java 11+).
Edge Cases & Pitfalls
Check these before calling the solution done.
Print value
Print i, not j
Printing j reprints a left-shifted ascending run, not a repeated-digit pattern.
Inner bound
Inner must run i..rows
Using 1..i instead reprints Program 9 (1, 22, 333), not this shape.
println inside
Do not put println inside the inner loop
That prints one digit per line and destroys the pattern.
Bad input
Validate with hasNextInt
nextInt() throws on letters — prefer hasNextInt() and require rows >= 1.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(n²)
O(1)
String.repeat (Example 3)
O(n²)
O(n) per row string
Digits printed are n + (n − 1) + … + 1 = n(n + 1) / 2 — quadratic in n. Direct prints need only the loop variables; repeat builds a short-lived row string.
Remember
Key Takeaways
Rule: outer i = 1..rows, inner j = i..rows, print i.
Shrink: each row drops one copy while the digit increases.
Break the row: call println only after the inner loop finishes.
Complexity:O(n²) time; O(1) extra space with direct prints.
One line: print digit i exactly rows - i + 1 times while i counts up.
Frequently Asked Questions
A shrinking repeating number pattern: for rows=5 you get 11111, 2222, 333, 44, 5.
The outer loop increases i from 1 to rows. The inner loop runs from i to rows, so it executes (rows - i + 1) times — fewer repeats as i grows.
System.out.print(i) stays on the same line. System.out.println() ends the current line. Digits use print; the row break uses println after the inner loop.
Program 10 counts the digit down (5, 44, 333). Program 12 counts the digit up while the row still shrinks (11111, 2222, 333).
Program 11 repeats i exactly i times counting down (55555, 4444). Program 12 uses inner i..rows so digit rises while length falls.
Yes. System.out.println(String.valueOf(i).repeat(rows - i + 1)) prints a full row in one call. Nested loops are better for learning; repeat is a handy shortcut later.
O(n²) for n rows because total prints are 1 + 2 + ... + n = n(n+1)/2.
One row prints a single digit 1.
🤔
Did you know?
Row digit i repeats rows − i + 1 times. The outer loop counts up from 1, while the inner loop runs from i to rows, so each row shrinks — still O(n²) total prints.
