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I have got a question to do in which i need to find distance between letters.
eg: If the input letters are A and E
On trying to reach E from A we need to cross B C D. Hence the number of letters in between E and A is 3. The distance is number of letters in between E and A + 1, which is equal to 4.
I wrote the code as follows:
I got the answer for some of the inputs. But i got a wrong answer for W A

What I have tried:

Java
import java.lang.Character;
import java.util.Scanner;
public class Main
{
public static void main(String[] args)throws Exception
{
Scanner in=new Scanner(System.in);
char a=in.next().charAt(0);
char d=in.next().charAt(0);
System.out.println(Math.abs(Character.toLowerCase(a)-Character.toLowerCase(d)));
}
}
Posted
Updated 7-Oct-16 10:57am
v3
Comments
[no name] 7-Oct-16 9:59am    
Do you have a question for us that we could help you with concerning your homework assignment?
Member 12781602 7-Oct-16 11:38am    
For the input W A, I need to get 4 as the output,not 22. How should I edit to get it so?
[no name] 7-Oct-16 11:47am    
Use your debugger to figure out why you are getting 4 but CPallini is getting 22.
[no name] 7-Oct-16 14:01pm    
Wait a minute, I just re-read your statement. Why is it that you think that the distance between W and A is only 4 and not 22? Either you are not telling us something or you misread your homework assignment.
Richard MacCutchan 7-Oct-16 10:35am    
As I understand the alphabet W has a higher value than A, so your algorithm will come up with a negative number. Go on from there.

1 solution

If you need to consider wrap araound, then the code becomes a little bit more complex.
I hope you are able to simplify it, though.

Java
import java.util.Scanner;
public class Main
{
  public static void main(String[] args) throws Exception
  {
    Scanner in = new Scanner(System.in);
    char a = in.next().charAt(0);
    char b = in.next().charAt(0);
    a = Character.toLowerCase(a);
    b = Character.toLowerCase(b);
    int d1, d2;

    if ( a <= b )
    {
      d1 = b - a; // direct distance
      d2 = 26 - b + a; // distance considering wrap around
    }
    else
    {
      d1 = a - b; // direct distance
      d2 = 26 - a + b; // distance considering wrap around
    }
    int  d; // now the distance is the smallest between 'direct' and 'wrapped around'
    if ( d1 <= d2)
      d = d1;
    else
      d = d2;

    System.out.println(d);
  }
}
 
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Comments
Richard Deeming 7-Oct-16 16:36pm    
That looks over-complicated to me. Assuming the inputs are valid letters, why not simply do:

int d = Math.abs(a - b);
if (d > 13) { d = 26 - d; }
CPallini 7-Oct-16 16:50pm    
Nice.
Richard MacCutchan 8-Oct-16 12:29pm    
'A' to 'W' comes out as 4 instead of 22.
It should be
if (a > b) { d = 26 - d; }
Richard Deeming 8-Oct-16 12:31pm    
If you read the comments to the question, that's what the OP wanted. :)
Richard MacCutchan 8-Oct-16 12:33pm    
No he wants W to A to be 4, not A to W.

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