An alternating odd/even number triangle grows by one number each row, but switches between odd-only and even-only sequences based on the row’s parity.
Remember
Rule: odd rows start at 1; even rows start at 2;
then k += 2 for the rest of the row
1
2 4
1 3 5
2 4 6 8
1 3 5 7 9 ← 5 rows
Row width still grows like a classic triangle (1, 2, 3, … numbers per row). The twist is the start value from i % 2 and the step of +2 that keeps parity fixed on each line.
Approach
How to Solve It
Pick the start with row parity, then print a growing odd or even stream with k += 2.
Method
Idea
Best for
if / else start
Odd → k = 1; even → k = 2
Learning, interviews, exams
Ternary start
k = (i % 2 == 0) ? 2 : 1
Shorter demos once parity clicks
Pseudocode
Pseudocode
for i from 1 to rows:
if i is even:
k = 2
else:
k = 1
for j from 1 to i:
print k
k = k + 2
print newline
Cheat sheet
Goal
Pattern
Walk each row
for (int i = 1; i <= rows; i++)
Pick start
int k = (i % 2 == 0) ? 2 : 1;
Grow width
for (int j = 1; j <= i; j++)
Print + step
System.out.print(k + " "); k += 2;
End the row
System.out.println();
Flip parity
Even → k = 1; odd → k = 2
Printing Numbers vs Starting a New Line
API
Effect
Use for
System.out.print
Stays on the same line
Each number (and its trailing space)
System.out.println
Ends the current line
After the inner loop
Print numbers without a newline, then end the row once.
Try it
Live Preview
Change the height and the odd/even triangle updates instantly — including the triangular number total.
Whole numbers from 1 to 12. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 numbers
1
2 4
1 3 5
2 4 6 8
1 3 5 7 9
Trace
Worked Walkthrough — rows = 4
Trace each row’s parity, start value, and printed numbers.
i
Parity
Start k
Printed row
1
odd
1
1
2
even
2
2 4
3
odd
1
1 3 5
4
even
2
2 4 6 8
Total numbers: 1 + 2 + 3 + 4 = 10 = 4×5/2. That triangular sum is why time is O(n²).
Code
Java Programs
Three complete programs: fixed 5 rows, Scanner input with a ternary start, and flipped parity. Use View Output to reveal sample results.
Example 1 — Fixed 5 Rows
Hard-coded height — if/else picks the start; the inner loop prints and steps by 2.
Java
public class AlternatingOddEvenTriangle {
public static void main(String[] args) {
int rows = 5;
for (int i = 1; i <= rows; i++) {
int k;
if (i % 2 == 0)
k = 2;
else
k = 1;
for (int j = 1; j <= i; j++) {
System.out.print(k + " ");
k += 2;
}
System.out.println();
}
}
}
Output
1
2 4
1 3 5
2 4 6 8
1 3 5 7 9
How It Works
1. Outer loop grows width. Row i prints exactly i numbers.
2. Parity picks the start. Even i → k = 2; odd i → k = 1.
3. Inner loop prints and steps. Print k, then k += 2 so the row stays odd-only or even-only.
When i = 3, k runs 1, 3, 5 → 1 3 5.
Example 2 — Row Count Input
Read the height at runtime. Prefer hasNextInt() before nextInt() in real apps.
Java
import java.util.Scanner;
public class AlternatingOddEvenInput {
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++) {
int k = (i % 2 == 0) ? 2 : 1;
for (int j = 1; j <= i; j++) {
System.out.print(k + " ");
k += 2;
}
System.out.println();
}
sc.close();
}
}
Output (when user enters 4)
Enter the number of rows: 4
1
2 4
1 3 5
2 4 6 8
How It Works
1. Prompt and read. Ask for a row count, then store it with sc.nextInt().
2. Same core, shorter start. The ternary (i % 2 == 0) ? 2 : 1 replaces the if/else block.
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 — Flipped Row Parity
Even rows print odds; odd rows print evens — only the start assignment changes.
Java
public class AlternatingOddEvenFlipped {
public static void main(String[] args) {
int rows = 5;
for (int i = 1; i <= rows; i++) {
int k = (i % 2 == 0) ? 1 : 2;
for (int j = 1; j <= i; j++) {
System.out.print(k + " ");
k += 2;
}
System.out.println();
}
}
}
Output
2
1 3
2 4 6
1 3 5 7
2 4 6 8 10
How It Works
1. Swap the start map. Even i → k = 1; odd i → k = 2.
2. Same print + step. Inner loop and k += 2 are unchanged.
3. Same widths. Only which sequence lands on odd vs even rows flips.
Edge Cases & Pitfalls
Check these before calling the solution done.
k += 1
Mixed parity
If you use k++ instead of k += 2, rows mix odds and evens. Keep the step at 2.
Wrong start
All odds or all evens
Forgetting the parity check and always starting at 1 (or 2) loses the alternating pattern.
println inside
Column of numbers
If println is inside the inner loop, each number lands on its own line. Use print for numbers; println only after the inner loop.
rows = 1
Single 1
Output is just 1 — a good sanity check for the odd-start path.
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 1–3)
O(n²)
O(1)
Total numbers printed = 1 + 2 + … + n = n(n + 1)/2 — still quadratic in n.
Remember
Key Takeaways
Rule: odd rows start at 1; even rows start at 2; then step by 2.
Width grows: row i prints exactly i numbers.
Break the row: call println only after the inner loop.
Complexity:O(n²) time; O(1) extra space.
One line: for each row, pick start from parity, print i numbers stepping by 2, then println().
Frequently Asked Questions
We check i % 2. If i is odd, set k = 1 for odd numbers; if i is even, set k = 2 for even numbers.
Row 2 is even, so k starts at 2. The inner loop prints k then adds 2 twice: 2, then 4.
System.out.print(k + " ") stays on the same line with a trailing space. System.out.println() ends the current line. Numbers use print; the row break uses println after the inner loop.
Row 3 is odd, so k starts at 1 and increments by 2 three times: 1, 3, 5.
After printing k, update k += 2 so the sequence stays odd or even while increasing.
Yes. Swap the condition so even rows start at 1 and odd rows start at 2 — Example 3 on this page shows that flip.
O(n²) for n rows. Total numbers printed equal 1+2+…+n = n(n+1)/2.
Use sc.hasNextInt() before sc.nextInt(), require n ≥ 1, and reject non-numeric input — see Example 2 notes.
🤔
Did you know?
Row parity picks the start value: odd rows begin at 1, even rows at 2. Then k += 2 keeps each row odd-only or even-only — still O(n²) total prints for n rows.
