Java: Display the following character rhombus structure
Java Conditional Statement: Exercise-26 with Solution
Write a Java program to display the following character rhombus structure.
Test Data
Input the number: 7
Pictorial Presentation:

Sample Solution:
Java Code:
import java.util.Scanner;
public class Exercise26 {
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
System.out.println("Input the number: ");
int n = sc.nextInt();
int count = 1;
int count2 = 1;
char c = 'A';
for (int i = 1; i < (n * 2); i++)
{
for (int spc = n - count2; spc > 0; spc--)
//print space
{
System.out.print(" ");
}
if (i < n)
{
count2++;
}
else
{
count2--;
}
for (int j = 0; j < count; j++)
{
System.out.print(c);//print Character
if (j < count / 2)
{
c++;
} else
{
c--;
}
}
if (i < n)
{
count = count + 2;
}
else
{
count = count - 2;
}
c = 'A';
System.out.println();
}
}
}
Sample Output:
Input the number: 7 A ABA ABCBA ABCDCBA ABCDEDCBA ABCDEFEDCBA ABCDEFGFEDCBA ABCDEFEDCBA ABCDEDCBA ABCDCBA ABCBA ABA A
Flowchart:

Java Code Editor:
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Java: Tips of the Day
IsPowerOfTwo
Checks if a value is positive power of two.
To understand how it works let's assume we made a call IsPowerOfTwo(4).
As value is greater than 0, so right side of the && operator will be evaluated.
The result of (~value + 1) is equal to value itself. ~100 + 001 => 011 + 001 => 100. This is equal to value.
The result of (value & value) is value. 100 & 100 => 100.
This will value the expression to true as value is equal to value.
public static boolean isPowerOfTwo(final int value) { return value > 0 && ((value & (~value + 1)) == value); }
Ref: https://bit.ly/3sA5d4I
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