technorsh / pattern

Lab Assignment ( Java Programming )

Lab Assignment

Pattern :

* 1 2 3 4
1 * 1 2 3
1 2 * 1 2
1 2 3 * 1

Code :

import java.util.Scanner;

class Pattern{
    public static void main( String args[]){

        System.out.print("Enter Size of Pattern : ");

        Scanner sc = new Scanner(System.in); // Scanner - For input from the user
        int n = sc.nextInt(); 

        int counter;

        for( int i = 0; i < n; i++ ){
            counter = 1;
            for( int j = 0; j < n; j++, counter++ ){
                
                if( i == j ){     
                    System.out.print("*");
                    counter = 1;
                }

                System.out.print(counter);
                
            }
            System.out.print("\n");
        }
        
    }
}

Coder : Rahul Sharma ( 20234747039 ), Naveen Rohilla ( 20234747029 )

✨ Logic :

We are printing the * whenever we are getting row and column number same, then after * we are printing the number starting from 1, according to the size of the pattern. Also before * we are printing the number starting from 1 till the row and column number are not the same.

✨ Description :

Let n be the size of the pattern, row denoted by i and the column denoted by j.

According to the pattern ( given in the question ) column is always more than 1, so according to our logic we are printing * and 1 together in a single iteration. So programmatically, you will see we are iterating to n in both the loop, except of having the different number of rows & columns.

We are first iterating it by rows i = 0 to n ( 1 row to nth row ) and whenever we are iterating, we are initialising the counter to 1.

Within the first loop we are iterating with another loop i.e., columns j = 0 to n ( 1 column to nth ) column and whenever we are encountering the condition where i == j, so we are just printing * and re-initializing it to 1, so that we can begin with 1 to the remaining size of pattern after the *.

So it will keep going on till we end up with a condition ( i >= n ).

Let take an example for the pattern :

Size of pattern, n = 4.

i = 4
j = 4

Initially, we have i = 0 and j = 0 to n

When we are going to the loop we are initializing the counter = 1

So, when i = 0 & j = 0 , we encounter i == j condition, then we will print * and reinitialize the counter = 1
and keep printing and incrementing the counter till we reach the nth column.

After the 1st iteration,

    * 1 2 3 4

Similarly for i = 1 and j = 0 to n, we are doing the same thing.

After the 2nd iteration,

    * 1 2 3 4
    1 * 1 2 3

Finally after nth iteration, we will have our pattern, like this

    * 1 2 3 4
    1 * 1 2 3
    1 2 * 1 2
    1 2 3 * 1
    

Output :