A growing reverse triangle starts with 1 and grows by one digit on the left each row — digits on row i run from i down to 1.
Remember
Rule: for i = 1 to n
print j from i down to 1
1
21
321
4321
54321 ← n = 5
Unlike Program 5 (1, 12, 123), here the inner loop counts down — same growth, reversed digit order.
Approach
How to Solve It
Grow the row index upward; on each row, print from that index down to 1.
Method
Idea
Best for
Fixed n
Outer i = 1..n; inner j = i..1
Labs and demos
Scanner input
Same loops; read n at runtime
Interactive practice
Spaced digits
print(j + " ")
Easier reading
Pseudocode
Pseudocode
for i from 1 to n:
for j from i down to 1:
print j
new line
Cheat sheet
Goal
Pattern
Set size
int n = 5;
Outer loop
for (int i = 1; i <= n; i++)
Inner loop
for (int j = i; j >= 1; j--)
Print digit
System.out.print(j);
End row
System.out.println();
Digits on row i
Exactly i
Printing Numbers vs Starting a New Line
API
Effect
Use for
System.out.print
Stays on the same line
Each digit in the row
System.out.println
Ends the line
After the inner loop finishes
Build the row with print, then break once.
Try it
Live Preview
Change n and the growing reverse triangle updates instantly.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live resultn = 5 · digits = 15
1
21
321
4321
54321
Trace
Worked Walkthrough — n = 3
Trace each outer value of i and the digits printed on that row.
i
Inner j
Printed row
1
1
1
2
2 1
21
3
3 2 1
321
Each step raises the start of the countdown by 1, so one more digit appears on the left.
Code
Java Programs
Three complete programs: fixed n = 5, Scanner input, and spaced digits. Use View Output to reveal sample results.
Example 1 — Fixed n = 5
Hard-coded height — outer loop grows; each row prints i..1.
Java
public class GrowingReverseTrianglePattern {
public static void main(String[] args) {
int n = 5;
for (int i = 1; i <= n; i++) {
for (int j = i; j >= 1; j--) {
System.out.print(j);
}
System.out.println();
}
}
}
Output
1
21
321
4321
54321
How It Works
1. Outer loop.i runs from 1 to n — each row is one digit longer.
2. Inner loop. For each i, print j from i down to 1 — first row is 1, then 21, and so on.
3. Newline.println() after the inner loop starts the next row.
Example 2 — Scanner Input
Read n at runtime. Same nested-loop core.
Java
import java.util.Scanner;
public class GrowingReverseTriangleInput {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter the number of rows: ");
int n = sc.nextInt();
if (n < 1) return;
for (int i = 1; i <= n; i++) {
for (int j = i; j >= 1; j--) {
System.out.print(j);
}
System.out.println();
}
sc.close();
}
}
Output (when user enters 4)
Enter the number of rows: 4
1
21
321
4321
How It Works
1. Prompt and guard. Read n; exit early if it is less than 1.
2. Same loops. Only the source of n changes — outer up, inner i..1.
3. Safer input tip. Prefer:
Safer input
if (!sc.hasNextInt()) {
System.out.println("Enter a positive integer.");
return;
}
int n = sc.nextInt();
if (n < 1) {
System.out.println("Enter a positive integer.");
return;
}
Example 3 — Spaced Digits
Same loops; print a space after each digit for easier reading.
Java
public class GrowingReverseTriangleSpaced {
public static void main(String[] args) {
int n = 5;
for (int i = 1; i <= n; i++) {
for (int j = i; j >= 1; j--) {
System.out.print(j + " ");
}
System.out.println();
}
}
}
Output
1
2 1
3 2 1
4 3 2 1
5 4 3 2 1
How It Works
1. Same shape. Outer and inner bounds match Examples 1 and 2.
2. Formatting only.j + " " separates digits — useful when numbers grow past single digits.
3. Trailing space. Each row may end with a space; that is normal for this simple style.
Edge Cases & Pitfalls
Check these before calling the solution done.
j ascending
Wrong digit order
Using j = 1..i prints 1, 12, 123 — that is Program 5, not this pattern.
println inside
Vertical digits
Calling println inside the inner loop prints one digit per line. Keep it after the inner loop.
j from n
Wrong shape
Starting at j = n every row builds a different pattern — use j = i so row length equals i.
n = 1
Single digit
Output is just 1 on one line.
n ≤ 0
Empty output
The outer loop never runs. Guard interactive input with n >= 1.
Bad input
Use hasNextInt
nextInt() throws on letters — prefer hasNextInt() and require a positive integer.
Analysis
Time and Space Complexity
Program
Time
Extra space
Examples 1–3
O(n²)
O(1)
Total digits printed: 1 + 2 + … + n = n(n+1)/2 → still O(n²).
Remember
Key Takeaways
Rule: outer i = 1..n; print j = i..1 on each row.
Growth: one new digit appears on the left each row.
Vs Program 5: same outer loop; flip the inner direction to reverse digits.
Complexity:O(n²) for n rows.
One line: grow the row length upward, and count down from the row index to 1 on each line.
Frequently Asked Questions
Each row must print digits from i down to 1. Starting at j = i and using j-- gives 321 on row 3 instead of 123.
The outer loop runs i from 1 to n. For each row i, the inner loop runs j from i down to 1 and prints j, then println ends the row.
Program 5 prints 1, 12, 123 (inner loop j = 1..i ascending). Program 7 prints 1, 21, 321 (same outer loop, inner loop j = i..1 descending).
Program 6 uses a descending outer loop and prints 5, 45, 345 (inner j = i..n). Program 7 uses an ascending outer loop and counts j down on each row (1, 21, 321).
Yes. Use System.out.print(j + " ") inside the inner loop — see Example 3.
O(n²) for n rows. Total printed digits are 1+2+…+n = n(n+1)/2.
Use sc.hasNextInt() before sc.nextInt() so bad input does not throw InputMismatchException.
The outer loop never runs, so nothing is printed. Validate and prompt again if you want a clear user message.
🤔
Did you know?
Outer loop grows row index i; inner loop prints j from i down to 1 — 1, 21, 321, … O(n²) for n rows.
