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noX challenge

Posted on 2016-09-15
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Last Modified: 2016-09-20
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
0
Comment
Question by:gudii9
  • 7
  • 6
  • 3
17 Comments
 
LVL 32

Expert Comment

by:phoffric
Comment Utility
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?
0
 
LVL 27

Expert Comment

by:rrz
Comment Utility
0
 
LVL 32

Expert Comment

by:phoffric
Comment Utility
>> 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.
0
 
LVL 27

Expert Comment

by:rrz
Comment Utility
@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.
0
 
LVL 27

Expert Comment

by:rrz
Comment Utility
I'm sorry. I was wrong.   Your choice of the replace method is fine.
0
 
LVL 32

Expert Comment

by:phoffric
Comment Utility
>> 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.
0
 
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Expert Comment

by:rrz
Comment Utility
@phoffric, I just meant that my previous suggestion was not helpful.
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
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?
0
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LVL 27

Assisted Solution

by:rrz
rrz earned 250 total points
Comment Utility
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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LVL 32

Expert Comment

by:phoffric
Comment Utility
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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LVL 27

Expert Comment

by:rrz
Comment Utility
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
not clear on above comments. are we saying not to use recursion if more characters to check?
0
 
LVL 32

Expert Comment

by:phoffric
Comment Utility
@rrz,
I am unclear why you are identifying posts in other questions. Are you asking me to read them and verify them? Or what?
0
 
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Accepted Solution

by:
phoffric earned 250 total points
Comment Utility
@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
0
 
LVL 27

Expert Comment

by:rrz
Comment Utility
@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.
0
 
LVL 7

Author Comment

by:gudii9
Comment Utility
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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