Inverted V-Shaped Alphabet Pattern in Java
What You’ll Learn
This pattern draws an inverted V: one A at the top, then pairs like B B, C C, and so on, widening as you go down.
We scan a fixed-width row and print letters only where the row letter matches the current column index.
⭐ Pattern Output
Output for 5 rows:
A
B B
C C
D D
E EJava Program (Reference Logic)
Left scan from E to A, then right scan from B to E.
public class Main {
public static void main(String[] args) {
int i, j, k;
for (i = 65; i <= 69; i++) {
for (j = 69; j >= 65; j--) {
if (i == j) System.out.print((char) j);
else System.out.print(" ");
}
for (k = 66; k <= 69; k++) {
if (i == k) System.out.print((char) k);
else System.out.print(" ");
}
System.out.println();
}
}
}🧠 How It Works
Row letter i
for (i = 65; i <= 69; i++) walks each row (ASCII A…E). The current letter is the one that may appear on both legs (except the apex row, which only prints once).
Left block j (E..A)
for (j = 69; j >= 65; j--) prints (char) j only when i == j; otherwise a space. That draws the descending diagonal.
Right block k (B..E)
for (k = 66; k <= 69; k++) mirrors the rule on the ascending side. Starting at 66 (B) skips a duplicate A on the first row while still forming the widening shape.
Symmetry & newline
Both inner loops use the same predicate (i == column) so legs stay aligned. System.out.println() after both blocks ends the row.
Opening downward
Each row adds width between the legs. Five rows with two O(n) passes per row → O(n²) time and O(1) extra space beyond the stream.
Variation — User Input
For rows, set endChar = 'A' + rows - 1 and run the same two scans; the right scan starts at startChar + 1.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter the number of rows: ");
int rows = sc.nextInt();
rows = Math.max(1, Math.min(rows, 26));
char startChar = 'A';
char endChar = (char) ('A' + rows - 1);
for (char i = startChar; i <= endChar; i++) {
for (char j = endChar; j >= startChar; j--) {
if (i == j) System.out.print(j);
else System.out.print(' ');
}
for (char k = (char) (startChar + 1); k <= endChar; k++) {
if (i == k) System.out.print(k);
else System.out.print(' ');
}
System.out.println();
}
sc.close();
}
}💡 Tips for Enhancement
Try These
- Stretch the shape by printing two spaces instead of one
- Fill the inside region for a solid triangle
- Keep
rows <= 26for A..Z only
Avoid
- Starting the right scan at
Aif you want a single apex - Proportional fonts when evaluating alignment
Key Takeaways
Two diagonal checks pick out the two legs.
Right block starts at B so the first row prints only one A.
Width is 2n - 1 for n letters.
O(n²) time for n rows.
❓ Frequently Asked Questions
A doesn’t print twice. From row B onward, both legs can match the same letter.More Java alphabet patterns
Once you can pick diagonal positions with equality checks, you can draw many shapes in a console grid.
If the right block started at A, the top row would print two As (one on each side) unless you changed the ranges.
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