Thread: LOOPS

Need some hints and ideas to code the following problems. I have done question 5. it's just for the reference for question 6.Thanks

4) The Evil E is back and it brought back some friends! Write a method that will acknowledge that the letter E is evil. Each letter E will eliminate the characters around it. But the Evil Es will not eliminate its friends. So it will not eliminate the letters e or E.

evilEs("HELLO") --> " E LO"
evilEs("dEep") --> " Eep"
evilEs("Evil") --> "E il"
evilEs("bean") --> "bean"
evilEs("AbEEmgmEeodE") --> "A EE g Eeo E"
evilEs("ELEPHANT") --> "E E HANT"
evilEs("HomepagE") --> "Homepa E"

5) Write a method that will encode a given string in the following format.

encode("Hello") -> "leHlo"
encode("abcdef") -> "fdbace"
encode("ILoveJava") -> "vJvLIoeaa"
encode("mnopqrstuv") -> "vtrpnmoqsu"


6) Write a method that will undo the encoding in question 5. So this method will decode a String that is returned in the format of question 5.
(This one is much more complicated than question 5)

decode("leHlo") -> "Hello"
decode("fdbace") -> "abcdef"
decode("vJvLIoeaa") -> "ILoveJava"

@Abhilash: please review The Problem with Spoon-Feeding and comply.

Thank you.

I had no idea what spoon-feeding is until I have clicked to your spoon feeding link. But do not worry guys, I didn't just copy and paste; I have understood the whole problem.
Your help and comments are much appreciated @Abhilash, Greg and Blue Eyes

--- Update ---

Here's my next question
3) Take 2 Strings and compare each char. If the chars are the same it will concatenate a y if they are different it will concatenate an n. It will then return the new concatenated String. The length of the returned string will always be the length of the longest string.

sameYN(“gold”, “bold”)  “nyyy”
sameYN(“abcd”, “abcd”)  “yyyy”
sameYN(“copy”, “copies”)  “yyynnn”
sameYN(“abcxyz”, “mnoxyzabc”)  “nnnyyynnn”

The following is the solution I came up with. But one of my friend(he won't help) keeps on telling me there's one more scenario to be taken care of.

public static String sameYN(String str1, String str2)
    {
        String output = "";
        int len1 = str1.length();
        int len2 = str2.length();
        if(len1 > len2)
        {
            for(int i = 0; i < len2; i++)
            {
                if(str1.charAt(i) == str2.charAt(i))
                    output = output+"y";
                else 
                    output = output+"n";
            }
        }
        else if(len2 > len1)
        {
            for(int i = 0; i < len1; i++)
            {
                if(str1.charAt(i) == str2.charAt(i))
                    output = output+"y";
                else 
                    output = output+"n";
            }
        }
        else 
        {
            for(int i = 0; i < len1; i++)
            {
                if(str1.charAt(i) == str2.charAt(i))
                    output = output+"y";
                else 
                    output = output+"n";
            }
        }
        for(int k = 0; k < Math.abs(len1 - len2); k++)
        {
            output = output +  "n";
        }
 
        return output;
    }

Hey! no problem. If you have any more problems, I'll help anytime, but this time I will explain instead of giving the full code. So that you do the program your way and understand it.

The following is the solution I came up with. But one of my friend(he won't help) keeps on telling me there's one more scenario to be taken care of.

Read the assignment and check that you've satisfied every requirement. If you find that you've missed one and can't solve it, then come back and let us know specifically what you need help with.

I don't know if I'm right but I guess the "Scenario" involves cases

Eg:-
sameYN(“BOLD”, “bold”) returns "nnnn" instead of "yyyy"