noX challenge

Hi,

I am working on below challenge
http://codingbat.com/prob/p118230

Psedo code:
1. if array length is zero return ""
2. if array length is 1 and it is x return ""
3. if array length is 1 and it is not x return same str
4. else find location of x and replace with "" if greater than length
5. return replaced string

public String noX(String str) {
  if(str.length()==0){
    return "";
  }
  if(str.length()==1&&"x".equals(str)){
    return "";
  }
   if(str.length()==1&&!"x".equals(str)){
    return str;
  }
   if(str.length()>1){
    return str.replace("x","");
  }
  
  
  return null;
}

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I am passing all tests

How to improve/modify my design, code and any other alternate approaches. please advise
LVL 7
gudii9Asked:
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phoffricCommented:
The instructions say you have to do this challenge recursively.
What would happen if you modified your code to just call replace without all those if statements?
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rrzCommented:
0
phoffricCommented:
>> No recursion is necessary for this challenge.
You are already stating what the author has observed.
But the challenge is to learn how to use recursion.
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rrzCommented:
@phoffric,  gudii9 has already gone through 11 recursion challenges last week.  He will not learn anything by using recursion when it is not useful.  He really needs to learn the API better.
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rrzCommented:
I'm sorry. I was wrong.   Your choice of the replace method is fine.
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phoffricCommented:
>> I'm sorry. I was wrong.   Your choice of the replace method is fine.
Could you elaborate on the meaning of this?

Are you saying that the author's solution is fine by using the replace method?
I have the same question where you wrote this in two other questions from this author.
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rrzCommented:
@phoffric, I just meant that my previous suggestion was not helpful.
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gudii9Author Commented:
public String noX(String str) {
 // public String changeXY(String str) {
    if (str.length() == 0){ 
      return str;
      
    }
   else if (str.charAt(0) == 'x') {
     return "" + noX(str.substring(1));
     
   }
   else{ 
     return str.charAt(0) + noX(str.substring(1));
     
   }
}

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above passes all tests,

any improvement or alternate approaches?
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rrzCommented:
Your code is good. But,  is the following more readable?
public String noX(String str) {
  if ("".equals(str)){ 
    return "";
  }
  if (str.charAt(0) == 'x') {
    return noX(str.substring(1));
  }
  return str.charAt(0) + noX(str.substring(1));
}

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phoffricCommented:
In this question and in others, it looks like you all are recursively examining one char at a time. If the string is 10 thousand chars long, then the stack will have 10000 frames.

How about doing a find operation in noX?
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf%28int%29
Then, if there are only 1% of the undesired char, the stack will only have 100 frames.
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gudii9Author Commented:
not clear on above comments. are we saying not to use recursion if more characters to check?
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phoffricCommented:
@rrz,
I am unclear why you are identifying posts in other questions. Are you asking me to read them and verify them? Or what?
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phoffricCommented:
@gudii9,
>> are we saying not to use recursion
This challenge forces us to use recursion. But you should be aware that each time you recurse, you are creating a function stack frame, which can eat up your allotted stack space if not careful.

If the string is 10 thousand chars long, then your current approach will make 10000 function calls - one for every character, and the stack will have 10000 frames. Now, if you knew, for example, that the undesired char was say 1/10th of 1%, then by using indexOf function to find the undesired char, you would only have to recurse  0.001 * 10000 = 10 calls (i.e., only 10 stack frames).

In fact, I once had to get rid of a spurious undesired char in a Word document, so if using recursion, then we don't want to recurse on every char.

http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf%28int%29
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rrzCommented:
@phoffric and gudii9, I posted those three links to show examples of where I used indexOf in recent challenges. The same approach I used in those examples, can be used in this challenge.
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gudii9Author Commented:
public String noX(String str) {
 // public String changeXY(String str) {
   if("".equals(str)){
     return "";
   }
   else if (str.charAt(0) == 'x') {
     return "" + noX(str.substring(1));
     
   }
   else{ 
     return str.charAt(0) + noX(str.substring(1));
     
   }
}
 /* public String noX(String str) {
  if ("".equals(str)){ 
    return "";
  }
  if (str.charAt(0) == 'x') {
    return noX(str.substring(1));
  }
  return str.charAt(0) + noX(str.substring(1));
}
 
 A*/
/* public String changePi(String str) {
  if (str.equals("") || str.length() < 2) return str;
  if (str.charAt(0) == 'p' && str.charAt(1) == 'i') 
    return "3.14" + changePi(str.substring(2));
  return str.charAt(0) + changePi(str.substring(1));
}*/

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above passes all
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