Pattern printing program

Pattern Number #81

1
A B
2 3 4
C D E F
5 6 7 8 9

The program for the pattern is written in C# programming language and will accept a number as input. The loops will iterate based on the number of entry.
 
Pattern printing program

Note: If you are new to C# and Console Application. Try to code First C# Program

Note: Read articles on how to use Loops and Conditions.

Let’s find out a simple and easy way to code these pattern.

Practical Implementation:

using System;

namespace patternProblem
{
    class Pattern81
    {
        public static void Main()
        {           
            Pattern81 obj = new Pattern81();
            //Different solutions of creating the pattern.
           
            /*(--Tip use the solution one at a time--)*/

            //Solution1
            obj.Solution1();

            //Solution2
            //obj.Solution2();
        }

        public void Solution1()
        {
            Console.WriteLine("Enter iteration times: ");
            int n = Convert.ToInt32(Console.ReadLine());
            int j;
            int temp1 = 1;
            int temp2 = 1;

            Console.WriteLine("-----Output-----");
            Console.WriteLine();

            for (int i = 1; i <= n; i++)
            {
                for (j = 1; j <= i; j++)
                {
                    if (i % 2 == 0)
                    {
                        Console.Write((char)(64 + temp1) + " ");
                        temp1++;
                    }
                    else
                    {
                        Console.Write(temp2 + " ");
                        temp2++;
                    }
                }
                Console.WriteLine();
            }
            Console.ReadKey();
        }

        public void Solution2()
        {
            Console.WriteLine("Enter iteration times: ");
            int n = Convert.ToInt32(Console.ReadLine());
            int j;
            char ch = 'A';
            int temp = 1;

            Console.WriteLine("-----Output-----");
            Console.WriteLine();

            for (int i = 1; i <= n; i++)
            {
                for (j = 1; j <= i; j++)
                {
                    if (i % 2 == 0)
                    {
                        Console.Write(ch + " ");
                        ch++;
                    }
                    else
                    {
                        Console.Write(temp + " ");
                        temp++;
                    }
                }
                Console.WriteLine();
            }

            Console.ReadKey();
        }
    }
}

Output:
The input number here is 5. So, the program will iterate 5 times. The output is shown below:

Enter iteration times:
5
-----Output-----

1
A B
2 3 4
C D E F
5 6 7 8 9
Press any key to continue . . .


For any query, comment us below.

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