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# alternate strings challenge

Posted on 2014-10-18
149 Views
Last Modified: 2014-10-20
Hi,

I was working on below challenge

http://codingbat.com/prob/p165666

I wrote as below

``````public String stringBits(String str) {
for(int i=0;i<str.length()-1;i=i++){
String s="";
return (s+str.substring(i,i+2));
}
return null;
}
``````
I got output as below

Expected      Run
stringBits("Hello") → "Hlo"      "He"      X
stringBits("Hi") → "H"      "Hi"      X
stringBits("Heeololeo") → "Hello"      "He"      X
stringBits("HiHiHi") → "HHH"      "Hi"      X
stringBits("") → ""      "null"      X
stringBits("Greetings") → "Getns"      "Gr"      X
stringBits("Chocoate") → "Coot"      "Ch"      X
stringBits("pi") → "p"      "pi"      X
stringBits("Hello Kitten") → "HloKte"      "He"      X
stringBits("hxaxpxpxy") → "happy"      "hx"      X

How to improve it and fix it. Please advise
0
Question by:gudii9
• 8
• 8
• 3
• +1
22 Comments

LVL 16

Accepted Solution

krakatoa earned 167 total points
ID: 40389339
Try this:

``````public String stringBits(String str) {
StringBuilder sb = new StringBuilder();
for (int y=0;y<(str.length());y=y+2){
sb.append(str.charAt(y));
}
return sb.toString();
}
``````
0

LVL 19

Assisted Solution

n2fc earned 166 total points
ID: 40389345
``````public String stringBits(String str) {
String s="";
for(int i=0;i<str.length();i+=2)
s=s+str.substring(i,i+1);

return s;
}
``````
0

LVL 19

Expert Comment

ID: 40389362
As an alternative:

``````public String stringBits(String str) {
String s="";
for(int i=0; i<str.length(); ++i)
s+=str.substring(i, ++i);

return s;
}
``````

See if you can tell why they BOTH work!!
0

LVL 7

Author Comment

ID: 40389522
``````public String stringBits(String str) {
for(int i=0;i<str.length()-1;i=i++){
String s="";
return (s+str.substring(i,i+2));
}
return null;
}

public String stringBits(String str) {
String s="";
for(int i=0;i<str.length();i+=2)
s=s+str.substring(i,i+1);

return s;
}
``````

comparing my solution with your solution as above both looks almost similar except in the increment i wrote i=i+1 and inside for loop i=i+2 (where as yours i=i+2 and i=i+1) . I wonder what is the difference.

I also made my usual mistake
i<str.length()-1

I though i represent like index so it is 1 less than original length.

Please advise
0

LVL 7

Author Comment

ID: 40389526
for (int y=0;y<(str.length());y=y+2){
sb.append(str.charAt(y));

Is y is not like index (index always starts with 0 so we take str.length()-1) but we are taking y<(str.length())
Please advise
0

LVL 84

Assisted Solution

ozo earned 167 total points
ID: 40389754
Another mistake, is that i=i++ does not do what you seem to think it does
0

LVL 16

Expert Comment

ID: 40389849
Is y is not like index (index  . . .

Right well consider this - the word "Hello". The index starts at zero as you say, so this is character 'H'. Then we add 2 to that, so that points at character 'l', then, because 'y' is now 2, and 2 is still less than 5 (the length of the string 'Hello'), we again add 2 to 'y', making it 4, and take the 4th index character, which is 'o', and so we get 'Hlo'. Now, if 2 is added to 'y', it will exceed the string length, and so this brings the loop to a halt.
0

LVL 16

Expert Comment

ID: 40389872
gudii - do me this favour will you - answer these 5 questions, based on the string "Hello". :

1. What is the index number of the beginning of the string?

2. What is the index number of the end of the string ?

3. What is the length of the string?

4. What is the MAXIMUM value that this expression can return for 'y' ? : y<"Hello".length()

5. What is the difference between the values of your answers to points 4 and point 2 above?
0

LVL 7

Author Comment

ID: 40390034
1. What is the index number of the beginning of the string?
>>0

2. What is the index number of the end of the string ?
>>4

3. What is the length of the string?
>>5
4. What is the MAXIMUM value that this expression can return for 'y' ? : y<"Hello".length()
>>5(since "Hello".length()  is 5)

5. What is the difference between the values of your answers to points 4 and point 2 above?
>>sorry i did not clearly understand this. point 4 gives answer as 5 and point 2 gives answer as 4 so both lenght and index are different and index is always one less than length
0

LVL 7

Author Comment

ID: 40390037
Right well consider this - the word "Hello". The index starts at zero as you say, so this is character 'H'. Then we add 2 to that, so that points at character 'l', then, because 'y' is now 2, and 2 is still less than 5 (the length of the string 'Hello'), we again add 2 to 'y', making it 4, and take the 4th index character, which is 'o', and so we get 'Hlo'. Now, if 2 is added to 'y', it will exceed the string length, and so this brings the loop to a halt.

this explanation is very clear. I was jumping by mistake to next character rather than alternate character whose difference is 2 not1
0

LVL 19

Expert Comment

ID: 40390054
comparing my solution with your solution as above both looks almost similar except in the increment i wrote i=i+1 and inside for loop i=i+2 (where as yours i=i+2 and i=i+1) . I wonder what is the difference.

I also made my usual mistake
i<str.length()-1

I though i represent like index so it is 1 less than original length.

I think your best bet to figure this out would be to "hand test" by making  a table of every assignment change to see where your values are awry...

Think about what you are doing in the loop!

Putting your "return" INSIDE the loop makes NO SENSE!  It would STOP looping the first time it hits it!  Compare to the OTHER solutions offered that build up the proper answer and only then returns that answer AFTER the loop completes...

You also need to understand the way "++i" "i++" and "i=i++" work... go over these operators in a good C or Java language reference...
0

LVL 16

Expert Comment

ID: 40390073
You got my point 4 wrong. Look at it again more carefully.
0

LVL 16

Expert Comment

ID: 40390574
Because not understanding my question 4 is, I believe, the reason you are having trouble, so if you can deal with this point I think you will make progress.
0

LVL 7

Author Comment

ID: 40390831
'y' ? : y<"Hello".length()

Is it is ternary operator(looks like that to me with ? and :

It also does not look like that since there is no two separate entities corresponding to condition true and false.

Please advise
0

LVL 16

Expert Comment

ID: 40390852
y<"Hello".length()

IS THE EXPRESSION!!!!!!!!!!!!!

Im am asking you the max value of y !!!!
0

LVL 84

Expert Comment

ID: 40391394
I think krakatoa means max value of y for which the expression is true
0

LVL 7

Author Comment

ID: 40392598
oh that is  6
0

LVL 84

Expert Comment

ID: 40392611
Is this true?
6<"Hello".length()
0

LVL 7

Author Comment

ID: 40392621
oh i made a mistake. I understand the question now. So the y can go maximum upto 4. It if goes beyond 4 say 5  or 6 etc the condition returns false
0

LVL 16

Expert Comment

ID: 40392815
oh i made a mistake. I understand the question now. So the y can go maximum upto 4. It if goes beyond 4 say 5  or 6 etc the condition returns false

That's it ! y's maximum value can be 4 - the highest number that is less than 5, . And 5, as you know, is the length of the string.
0

LVL 7

Author Comment

ID: 40392825
``````public String stringBits(String str) {
String s="";
for(int i=0;i<str.length();i+=2)
s=s+str.substring(i,i+1);

return s;
}

public String stringBits(String str) {
String s="";
for(int i=0; i<str.length(); ++i)
s+=str.substring(i, ++i);

return s;
}
``````

Both of above scenarios working since we are incrementing first then assigning to i in the second scenario which makes it equivalent to fist scenario
0

LVL 16

Expert Comment

ID: 40393334
Yes, they both work. But your two examples are making the code more difficult to read than it needs to be, because you can make it easier to understand I would say, if you use StringBuilder and follow the example I gave originally.

(It's good that you made the two pieces of code work, but as someone great once said, 'you should make things as simple as possible, and no simpler'.
0

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