Challenge #47 explanation

CHALLENGE SOLVA

Coding made simple

Challenge No. 47

ROHIT KANOJIYA🇮🇳
TODAY'S Challenge : 47 COMMENT Your Code 👇 Date : 30/04/2020
link to the challenge

Hello everyone, here is an explanation to my code made to answer the challenge above. The code is written in Java language for now I hope it will help someone.

Explanation

To begin with, let's analyse the pattern given in the challenge. The input is ABCDE and the pattern is:

a) Pattern width: Pattern width is the length of the string input.That is, the number of individual characters in the string inclusive of white spaces and symbols. In this case it is:
"A" + "B" + "C" + "D" + "E" = 5
b) Pattern height: You realise pattern heigth is twice the length of the string but less than 1.
In our case the length of the string is 5 hence the height will be:
(5*2) - 1 = 9
c) Pattern: If you look keenly to the pattern you will find out that every line begin at a space more or less and each character is spaced out.
In the first half, every line begin at a space more and with the next subsequent character in the string while the other half its vice versa

First half

    ^s0A  B  C  D  E
    ^s0s1B  C  D  E
    ^s0s1s2C  D  E
    ^s0s1s2s3D  E
    ^s0s1s2s3s4E

Second half

    ^s0s1s2s3s4E
    ^s0s1s2s3D  E
    ^s0s1s2C  D  E
    ^s0s1B  C  D  E
    ^s0A  B  C  D  E

s represents the space characters

That being said, let's jump directly into the fun part of it. The code is quite simple but it requires you have a background knowledge in Java language.

Firstly, you have to declare your class and the main method. and import the Scanner class to get the user input

    
    import java.util.Scanner;
    
    public class Program {
    
        public static void main (String[] args) {
        
            
        }
    }

Secondly, lets declare some variables and take user input using the Scanner and System.in objects.
Variable input will hold the user input while s will help make the pattern
After taking the input will remove any white space so as to have a flowless pattern using the replace method.

                
    import java.util.Scanner;
    
    public class Program {
    
        public static void main (String[] args) {
        
            String input;
            int s = 0;
            
            input = new Scanner(System.in).nextLine().
                    replace(" ", "");
        }
    }

Thirdly, notify the user the input he or she entered. The System.out object will help us accomplish this.

                
    import java.util.Scanner;
    
    public class Program {
    
        public static void main (String[] args) {
        
            String input;
            int s = 0;
            
            input = new Scanner(System.in).nextLine().
                    replace(" ", "");
                    
            System.out.println(Input word:  + input);
        }
    }

We are now done with the first part, let's go right away to the pattern itself.
Now to do this, will use the for-loop statement. Remember what we analysed earlier; the height, width and the pattern, review analysis we gonna use just that.

                
    import java.util.Scanner;
    
    public class Program {
    
        public static void main (String[] args) {
        
            String input;
            int s = 0;
            
            input = new Scanner(System.in).nextLine().
                    replace(" ", "");
                    
            System.out.println(Input word:  + input);
            
            // S T A R T  O F  P A T T E R N
            for(int y=1; y<=input.length()*2-1; y++) { // this will itterate through the rows(lines)
            
                for(int x=0; x<input.length(); x++) { // this will itterate through the columns(characters)
                
                    if(x>=s)
                        System.out.print(input.charAt(x) + " "); // print characters
                    else
                        System.out.print(" ");
                }
                s = (y < input.length())? ++s: --s;  // updating the space pattern
                System.out.println();  // finishing off the row(line)
            }
            // E N D  O F  P A T T E R N
        }
    }

This is what is happening in the for-loop code above.

The first for-loop will loop through the lines of the pattern see pattern. The code
y<=input.length()*2 -1 will limit the loop to the height of the pattern.

The second for-loop will loop through the characters of the string input. The loop is limited to the length of the string.

variable s is initialized to zero since the first character of the string is at position zero. s is not static but is changing with each line so that we can be able to identify the character that we are supposed to start with.

The if statement will check if x has reached the starting point and if true, it will start printing the characters otherwise it will print white space.

Challenge #47 explained