A left-shifted descending number triangle prints digits from the row start i through rows. Each next row begins one higher, so the line shortens from the left.
Remember
Rule: for i from 1 to rows,
print j from i to rows (no spaces)
12345
2345
345
45
5 ← rows = 5
The only new idea vs Program 1 is the inner start: j = i instead of always starting at 1. Same nested-loop skeleton, different shape.
Approach
How to Solve It
Outer loop picks the start digit; inner loop prints through rows.
Method
Idea
Best for
Nested loops
Outer i, inner j = i..rows
Learning, interviews, exams
StringBuilder
Append digits, then println the row
When you want one print per row
Pseudocode
Pseudocode
for i from 1 to rows:
for j from i to rows:
print j
print newline
Cheat sheet
Goal
Pattern
Pick row start
for (int i = 1; i <= rows; i++)
Print i..rows
for (int j = i; j <= rows; j++) System.out.print(j);
End the row
System.out.println();
Spaced digits
System.out.print(j + " ");
One print per row
Append to StringBuilder, then println(row)
Vs Program 1
Here: j = i; Program 1: print 1..i
Printing Numbers vs Starting a New Line
API
Effect
Use for
System.out.print
Stays on the same line
Each digit j
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 left-shifted triangle updates instantly.
Whole numbers from 1 to 12. Tap a chip or type a value — the preview redraws as you go.
Live resultrows = 5 · 15 digits
12345
2345
345
45
5
Trace
Worked Walkthrough — rows = 4
Trace each row’s start, range, and printed line.
i
Inner range
Digits
Printed row
1
1..4
4
1234
2
2..4
3
234
3
3..4
2
34
4
4..4
1
4
Total digits: 4 + 3 + 2 + 1 = 10 = n(n + 1) / 2 — that is why time is O(n²).
Code
Java Programs
Three complete programs: fixed rows = 5, Scanner input, and a StringBuilder row builder. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded height — outer loop picks start i, inner prints i..rows.
Java
public class LeftShiftedNumberTriangle {
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(j);
}
System.out.println();
}
}
}
Output
12345
2345
345
45
5
How It Works
1. Outer loop picks the start.i runs from 1 to rows — that value is where each row begins.
2. Inner loop prints the run.j goes from i to rows with System.out.print(j) (no spaces).
3. Break the line.System.out.println() after the inner loop starts the next row.
When i = 3, the inner loop prints 345. When i = 5, it prints just 5.
Example 2 — Rows Input
Read rows at runtime. Prefer hasNextInt() before nextInt() in real apps.
Java
import java.util.Scanner;
public class LeftShiftedNumberTriangleInput {
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(j);
}
System.out.println();
}
sc.close();
}
}
Output (when user enters 4)
Enter the number of rows: 4
1234
234
34
4
How It Works
1. Prompt and read. Ask for a height, then store it with sc.nextInt().
2. Same nested-loop core. Only the source of rows changes — the i..rows rule is identical to Example 1.
3. Safer input tip. Prefer:
Safer input
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 — StringBuilder Row Builder
Append each digit, then print the full row once with println.
Java
public class LeftShiftedNumberString {
public static void main(String[] args) {
int rows = 5;
for (int i = 1; i <= rows; i++) {
StringBuilder row = new StringBuilder();
for (int j = i; j <= rows; j++) {
row.append(j);
}
System.out.println(row);
}
}
}
Output
12345
2345
345
45
5
How It Works
1. Same bounds. Outer i and inner j = i..rows match Example 1.
2. Build, then print.row.append(j) collects digits; println(row) prints the whole line.
3. Same shape. Useful when exams want one print per row — loop bounds stay visible either way.
Edge Cases & Pitfalls
Check these before calling the solution done.
j = 1
Wrong start
Starting the inner loop at 1 prints a growing triangle (Program 1 style), not the left-shifted shape. Use j = i.
println inside
Column of digits
If println is inside the inner loop, each digit lands on its own line. Use print for digits; println only after the inner loop.
Missing println
One long line
Omitting println glues every digit onto a single endless line.
rows = 1
Single 1
Output is just 1 — a good sanity check.
rows ≤ 0
Empty output
The outer loop never runs. Validate and require rows ≥ 1.
Bad input
Use hasNextInt
nextInt() throws on letters — prefer hasNextInt() and require a positive whole number.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(n²)
O(1)
StringBuilder (Example 3)
O(n²)
O(n) per row buffer
Total digit prints: n + (n − 1) + … + 1 = n(n + 1) / 2 — still quadratic in n.
Remember
Key Takeaways
Rule: for each i from 1 to rows, print digits i..rows.
Inner start:j = i is what creates the left shift — not spaces.
Break the row: call println only after the inner loop.
Complexity:O(n²) time; O(1) extra space with direct prints.
One line: for each start i, print i..rows with print, then println().
Frequently Asked Questions
The inner loop starts at j = i, not j = 1. When i increases (1, 2, 3, …), each row begins at that value and prints until rows.
Because each next row starts at a higher i, numbers below i are not printed anymore — the triangle shortens from the left.
System.out.print(j) 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 1 grows by printing 1..i each row. Program 2 starts each row at i and prints i..rows — same total prints, different shape.
Change the rows variable or read it with Scanner. The outer loop runs from 1 to rows, and the inner loop prints j from i to rows.
Yes. Append each digit to a StringBuilder, then println the row — same shape with a different output style. See Example 3.
O(n²) for n rows because total prints are n+(n−1)+…+1 = n(n+1)/2.
Use sc.hasNextInt() before sc.nextInt(), require rows ≥ 1, and reject non-numeric input — see Example 2 notes.
🤔
Did you know?
Each row starts at i and prints through rows, so the left edge shifts right each line — still O(n²) total prints for n rows.
