Print Number Star Half Diamond Pattern Type 4

June 23, 2019

PRINT PATTERN  2 3*3 4*4*4 3*3 2

For any input number N Print the following code – For below code N=4

2
3*3
4*4*4
3*3
2

PREREQUISITE:

Basic knowledge in Java programming, usage of loops.

ALGORITHM:

  1. Take input from user i.e number of lines required (N value) and starting ‘a’ value.
  2. Take two loops one for each line (say ‘i’) and other for each digit in a particular line (say ‘j’). i starts from 1 and j starts from 1.
  3. The program is divided into two sections for increment part (i.e first 4 lines) and decrement part (last 3 lines). Two separate i and j loops are written for each part. In first part i is from a to n and in the second part i is from (n-i) to a.
  4. Here ‘i’ loop is used to access each line from 1 to n and ‘j’ loop is used to print values in each line. j loop is executed until it reaches i value.
  5. Print ‘i’ value until the j loop reaches i value.
  6. Print the final value ‘i’ of each line and go to next line.
  7. Repeat the loop until the condition is failed.

CODE IN JAVA:

[code language=”java”]
import java.lang.*;
import java.io.*;
class Pattern
{
public static void main(String[] args)throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n,i,j,a;
System.out.print("Enter starting value:");
a=Integer.parseInt(br.readLine());
System.out.print("Enter N value:");
n=Integer.parseInt(br.readLine());
for(i=a;i<=n;i++)
{
for(j=a;j<i;j++)
{
System.out.print(i+"*");
}
System.out.println(i);
}
for(i=(n-1);i>=a;i–)
{
for(j=1;j<(i-1);j++)
{
System.out.print(i+"*");
}
System.out.println(i);
}
}
}
[/code]

 

TAKING INPUT:

DISPLAYING OUTPUT: