asked on

frontAgain challenge

Hi,

I am working on below challenge
http://codingbat.com/prob/p196652

I wrote my code as below
public boolean frontAgain(String str) {
int len=str.length();
if(str.substring(0,1)==str.substring(len-1) &&str.substring(1,2)==str.substring(len-2,len-1))
{
return true;
}
else
{
return false;
}
}

I am failing with below message
Expected      Run
frontAgain("edited") → true      false      X
frontAgain("edit") → false      false      OK
frontAgain("ed") → true      false      X
frontAgain("jj") → true      false      X
frontAgain("jjj") → true      false      X
frontAgain("jjjj") → true      false      X
frontAgain("jjjk") → false      false      OK
frontAgain("x") → false      Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: 2 (line number:3)      X
frontAgain("") → false      Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: 1 (line number:3)      X
frontAgain("java") → false      false      OK
frontAgain("javaja") → true      false      X
other tests
X
Your progress graph for this problem

How to fix and improve my code. Thanks in advance

ozo

Your code attempts to access non-existent parts of strings of length 0 or 1
Your code attempts to compare strings with ==

SOLUTION

Sathish David Kumar N

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gudii9

ASKER

public boolean frontAgain(String str) {
int len=str.length();

if(len>=4){
if(str.substring(0,1).equals(str.substring(len-1)) &&str.substring(1,2).equals(str.substring(len-2,len-1)))
{
return true;
}
else
{
return false;
}
}
return false;
}

i fixed == part. still failing some test cases. please advise

ozo

Your code would return true for "deited" not "edited"
Your code returns false for any string shorter than length 4

gudii9

ASKER

i wonder what i should do with 2 character strings.

Given a string, return true if the first 2 chars in the string also appear at the end of the string, such as with "edited".

frontAgain("edited") → true
frontAgain("edit") → false
frontAgain("ed") → true

frontAgain("ed") → true
I wonder how above statement should yield to true based on the challenge description. please advise

gudii9

ASKER

public boolean frontAgain(String str) {
int len=str.length();

if(len>=4){
if(str.substring(0,1).equals(str.substring(len-1)) &&str.substring(1,2).equals(str.substring(len-2,len-1)))
{
return true;
}
  else(len<2)
  {
  return false;
  }
  }
  return false;
}

Open in new window

I wonder why i cannot give len<2 condition inside the else block. please advise

gudii9

ASKER

public boolean frontAgain(String str) {
int len=str.length();

if(len>=4){
if(str.substring(0,1).equals(str.substring(len-1)) &&str.substring(1,2).equals(str.substring(len-2,len-1)))
{
return true;
}
}

if(len<=2)
  {
  return true;
  }
  
  
  return false;
}

Open in new window

i passed few other tests still failing some. please advise.
challenge did not mention about str being "" or single character.
please advise

ozo

frontAgain("ed") → true
What are the first two characters of "ed"?
What are the last two character of "ed"?
Are they the same?

ozo

This is how if-then-else statements are constructed http://docs.oracle.com/javase/tutorial/java/nutsandbolts/if.html

gudii9

ASKER

else do not take condition where as if and else if takes.

if (testscore >= 90) {
grade = 'A';
} else if (testscore >= 80) {
grade = 'B';
} else if (testscore >= 70) {
grade = 'C';
} else if (testscore >= 60) {
grade = 'D';
} else {

gudii9

ASKER

frontAgain("ed") → true
What are the first two characters of "ed"?
What are the last two character of "ed"?
Are they the same?

oh i see from both sides ed so both same.

what happens if only one character is there

gudii9

ASKER

Your code would return true for "deited" not "edited"

if(str.substring(0,1).equals(str.substring(len-1)) &&str.substring(1,2).equals(str.substring(len-2,len-1))

i am comparing first with last character and second with last but one character. I wonder what i s the mistake i am doing to make java think as deited
please advise

ozo

Can a string with one character satisfy "the first 2 chars in the string also appear at the end of the string"?

gudii9

ASKER

public boolean frontAgain(String str) {
int len=str.length();

if(len>=4){
if(str.substring(0,1).equals(str.substring(len-2,len-1)) &&str.substring(1,2).equals(str.substring(len-1)))
{
return true;
}
}

if(len<=2)
  {
  return true;
  }
  
  
  return false;
}

Open in new window

i modified as above still failing 3 test cases as below

Expected	Run		
frontAgain("edited") → true	true	OK	    
frontAgain("edit") → false	false	OK	    
frontAgain("ed") → true	true	OK	    
frontAgain("jj") → true	true	OK	    
frontAgain("jjj") → true	false	X	    
frontAgain("jjjj") → true	true	OK	    
frontAgain("jjjk") → false	false	OK	    
frontAgain("x") → false	true	X	    
frontAgain("") → false	true	X	    
frontAgain("java") → false	false	OK	    
frontAgain("javaja") → true	true	OK	    
other tests
OK

Open in new window

please advise

gudii9

ASKER

Can a string with one character satisfy "the first 2 chars in the string also appear at the end of the string"?

i vote more for No since one character string do not have two characters in it

ozo

Granted there is some room for doubt, but the examples
frontAgain("x") → false
frontAgain("") → false
seem to clarify it.

gudii9

ASKER

public boolean frontAgain(String str) {
int len=str.length();

if(len>2){
if(str.substring(0,1).equals(str.substring(len-2,len-1)) &&str.substring(1,2).equals(str.substring(len-1)))
{
return true;
}
}

if(len==2)
  {
  return true;
  }
  
  
  return false;
}

Open in new window

i think i got what you mentioned.
I passed all tests.
How can i improve my code. please advise

ozo

You're doing pretty good now, but there is a simpler way to do the test with one .equals instead of two

You might also combine some other things

awking00

Why compare first character with next to last and second character with last? Just compare the two-character substrings -
public boolean frontAgain(String str) {
int len = str.length();
if (len > 1) {
return str.substring(0,2).equals(str.substring(len - 2));
}
return false;
}

awking00

As ozo suggests, the above code could be combined to make it shorter -
public boolean frontAgain(String str) {
if (str.length() > 1) {
return str.substring(0,2).equals(str.substring(str.length() - 2));
}
return false;
}

ozo

Two .substring operations instead of four is an improvement, but you can also reduce it to one:
return str.length()>=2 && str.endsWith(str.substring(0,2));

If you expect length==2 to be common, it may improve efficiency to add
str.length()==2 ||

gudii9

ASKER

Two .substring operations instead of four is an improvement, but you can also reduce it to one:
   return str.length()>=2 && str.endsWith(str.substring(0,2));

If you expect length==2 to be common, it may improve efficiency to add
str.length()==2 ||

Open in new window

I wonder how to do this

ozo

If you have no reason to expect any particular kind of input to be more common than any other kind of input then it makes little difference.
But if you expect your function to be called more often with one kind of input more often than it is called with other kinds of input, and you have a choice of which kinds of input would flow through the more efficient execution path, then you'd probably prefer to give priority to the kinds of input you expect to see most often.

Here there is no reason for expectation, but there is a choice.

I make the observation that in the case of strings of length 2, you know that str.endsWith(str.substring(0,2)) will automatically be true.
In one sense, it can me more efficient to let the special case be handled as just another instance of the general case,
which may reduce complicated proliferations of branching decisions.
But in another sense, testing the length of a string is a simpler operation than creating a new substring and then checking whether it matches the start of a string.

gudii9

ASKER

I make the observation that in the case of strings of length 2, you know that str.endsWith(str.substring(0,2)) will automatically be true.
In one sense, it can me more efficient to let the special case be handled as just another instance of the general case,
which may reduce complicated proliferations of branching decisions.
But in another sense, testing the length of a string is a simpler operation than creating a new substring and then checking whether it matches the start of a string.

both options seems logicaly valid not sure which one to prefer?
please advise

gudii9

ASKER

public boolean frontAgain(String str) {
    return str.length()>=2 && str.endsWith(str.substring(0,2));
}

Open in new window

i think i did not understood above code. I see all test passed.

what is meaning of below code.

return str.length()>=2 && str.endsWith(str.substring(0,2));

please advise

ASKER CERTIFIED SOLUTION

ozo

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SOLUTION

awking00

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