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Java Program to Print Left Pascal Triangle Character Pattern

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In this program we print a left Pascal triangle of letters. A downward left triangle shrinks from ABCDE to A, and a left triangle below it grows back to ABCDE, with both lined up on the right side.

Like every pattern program, it is built from nested loops: the outer loop picks the row, and the inner loops print what goes on that row. Java stores a char as a number, so adding to 'A' and casting back with (char) gives the following letters.

1. The Pattern

For 5 rows the program prints:

Pattern
ABCDE
 ABCD
  ABC
   AB
    A
    A
   AB
  ABC
 ABCD
ABCDE

2. How the Program Works

Here is what each printed line is made of when numberOfRows is 5:

LineLeading spacesPrinted after the spaces
10ABCDE
21ABCD
32ABC
43AB
54A
64A
73AB
82ABC
91ABCD
100ABCDE
  • Step 1 prints row - 1 spaces and then numberOfRows - row + 1 letters, so the rows get shorter.
  • Step 2 prints numberOfRows - row spaces and then row letters, so the rows get longer again.
  • Both parts print a row with a single A, so A appears twice in the middle.
  • System.out.println() moves to the next line after each row.

3. Java Program

Pattern20.java 
public class Pattern20 {
    public static void main(String[] args) {
        int numberOfRows = 5;

        // Step 1: downward left triangle, rows get shorter
        for (int row = 1; row <= numberOfRows; row++) {
            for (int column = 1; column < row; column++) {
                System.out.print(" ");
            }
            for (int column = 1; column <= numberOfRows - row + 1; column++) {
                System.out.print((char) ('A' + column - 1));
            }
            System.out.println();
        }

        // Step 2: left triangle, rows get longer again
        for (int row = 1; row <= numberOfRows; row++) {
            for (int column = 1; column <= numberOfRows - row; column++) {
                System.out.print(" ");
            }
            for (int column = 1; column <= row; column++) {
                System.out.print((char) ('A' + column - 1));
            }
            System.out.println();
        }
    }
}
Output
ABCDE
 ABCD
  ABC
   AB
    A
    A
   AB
  ABC
 ABCD
ABCDE

4. Print the Pattern for Any Number of Rows

To let the user choose the size of the pattern, read the number of rows with a Scanner. The loops stay the same.

Pattern20Input.java 
import java.util.Scanner;

public class Pattern20Input {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter the number of rows: ");
        int numberOfRows = scanner.nextInt();
        scanner.close();

        // Step 1: downward left triangle, rows get shorter
        for (int row = 1; row <= numberOfRows; row++) {
            for (int column = 1; column < row; column++) {
                System.out.print(" ");
            }
            for (int column = 1; column <= numberOfRows - row + 1; column++) {
                System.out.print((char) ('A' + column - 1));
            }
            System.out.println();
        }

        // Step 2: left triangle, rows get longer again
        for (int row = 1; row <= numberOfRows; row++) {
            for (int column = 1; column <= numberOfRows - row; column++) {
                System.out.print(" ");
            }
            for (int column = 1; column <= row; column++) {
                System.out.print((char) ('A' + column - 1));
            }
            System.out.println();
        }
    }
}
Output (with sample input)
Enter the number of rows: 4
ABCD
 ABC
  AB
   A
   A
  AB
 ABC
ABCD
Note: there are only 26 letters. Past Z, (char) ('A' + 26) is [, not a letter, so keep the number of rows small enough that the pattern never needs a letter after Z.

5. Summary

  • The outer loop picks the row; the inner loops print the spaces and the letters for that row.
  • (char) ('A' + k) gives the letter k places after A.
  • Keep numberOfRows in a variable, so the same program can print the pattern in any size.

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