Print this Alphabetical Pattern using 2 loops

user_15174609 0
Viewed 104

I need to print this pattern in java using only 2 loops one for the row and the second for the column.

A B C D E 
 B C D E 
  C D E 
   D E 
    E  
   D E 
  C D E 
 B C D E 
A B C D E

What I've tried

public class Main{
    public static void main(String[] args) {
        int n = 5;
        char c = 'A';
        for(int i=1; i<=n; i++){
            int k=1;
            for(int j=1; j<=(n*2)-1; j++){
                if(j>=i && j<=((n*2)-i) && k!=0){
                    System.out.print(c);
                    k=0;
                } else {
                    System.out.print(" ");
                    k=1;
                    c++;
                }
            }
            c = 'A';
            System.out.println();
        }   
    }
}

Output: This logic is printing half of the pattern correctly as I'm unable to understand how to built for the second half.

loops java
2 Answers

Try using the same logic as you used to print the top half, with the outer for-loop reversed, i.e., change:

for (int i = 1; i <= n; i++) {

to:

for (int i = n - 1; i >= 1; i--) {

Full code:

class Main {
  public static void main(String[] args) {
    printHourGlass(5);
  }

  public static void printHourGlass(int n) {
    char c = 'A';
    for (int i = 1; i <= n; i++) {
      int k = 1;
      for (int j = 1; j <= (n * 2) - 1; j++) {
        if (j >= i && j <= ((n * 2) - i) && k != 0) {
          System.out.print(c);
          k = 0;
        } else {
          System.out.print(" ");
          k = 1;
          c++;
        }
      }
      c = 'A';
      System.out.println();
    }
    for (int i = n - 1; i >= 1; i--) {
      int k = 1;
      for (int j = 1; j <= (n * 2) - 1; j++) {
        if (j >= i && j <= ((n * 2) - i) && k != 0) {
          System.out.print(c);
          k = 0;
        } else {
          System.out.print(" ");
          k = 1;
          c++;
        }
      }
      c = 'A';
      System.out.println();
    }
  }
}

Output:

A B C D E
 B C D E 
  C D E  
   D E   
    E    
   D E   
  C D E  
 B C D E 
A B C D E
user_63282560

To make this work for any terminating letter from 'B' to 'Z', define some values.

String indent = "                                     ";
char endLetter = 'E';
int end = endLetter - 'A';

The first part is easy and fairly straight forward.

  • use substring with indent to control indenting.
  • then print the characters incrementing the inner and outer loop until done.
for (int k = 0; k <= end; k++) {
    System.out.print(indent.substring(0,k));
    for (char c = (char)('A' + k); c <= endLetter; c++) {
        System.out.print(c + " ");
    }
    System.out.println();
}

prints

A B C D E 
 B C D E 
  C D E 
   D E 
    E

The trick is to reverse the process, continuing the outer loop.

  • the outer loop now goes from 0 to end*2+1
  • and k is replaced by s
    • using int s = k <= end ? k : end*2 - k;
    • s becomes 0 1 2 3 4 ... n ... n-1 n-2 ... 4 3 2 1 0
for (int k = 0; k < end*2+1; k++) {
    int s = k <= end ? k : end*2 - k;
    System.out.print(indent.substring(0,s));
    for (char c = (char) ('A' + s); c <= endLetter; c++) {
        System.out.print(c + " ");
    }
    System.out.println();
}

prints

A B C D E 
 B C D E 
  C D E 
   D E 
    E 
   D E 
  C D E 
 B C D E 
A B C D E

Note that in Java 11 String.repeat was added so that could also be used to indent the appropriate lines. So the printing of the indent could be System.out.print(" ".repeat(s));

Note: ASCII letters are adjacent from a-z with upper and lower case in different positions of the ASCII chart. Since they are treated as ints in the char type they can be mathematically altered to be other characters. So 'A'+1 is a 'B' and so forth. That is why the inner loops above work as they do.

user_15525340
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