Solved

string challenge

Posted on 2014-10-10
14
235 Views
Last Modified: 2014-10-13
Hi,

I was trying below challenge

http://codingbat.com/prob/p196441

I did not understand the requirement to proceed.

Given a non-empty string and an int N, return the string made starting with char 0, and then every Nth char of the string. So if N is 3, use char 0, 3, 6, ... and so on. N is 1 or more.

everyNth("Miracle", 2) → "Mrce"
everyNth("abcdefg", 2) → "aceg"
everyNth("abcdefg", 3) → "adg"


 Please advise
0
Comment
Question by:gudii9
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
14 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 40373338
return str.replaceAll("(.).{0,"+(n-1)+"}","$1");
0
 
LVL 84

Expert Comment

by:ozo
ID: 40373373
everyNth("Miracle", 2) → "Mrce"
everyNth("abcdefg", 2) → "aceg"
everyNth("abcdefg", 3) → "adg"
0
 
LVL 14

Assisted Solution

by:Geisrud
Geisrud earned 100 total points
ID: 40373476
Here's my algorithm/solution (does the same thing as Ozo's solution, but my Java-Fu is weak, unoptimized, and overly verbose):

*Create a new String to hold the result.
*Use a for loop to parse through the string str they give you.
*At each n grab that char and append it to your new string
*Return your string
0
SharePoint Admin?

Enable Your Employees To Focus On The Core With Intuitive Onscreen Guidance That is With You At The Moment of Need.

 
LVL 28

Assisted Solution

by:dpearson
dpearson earned 100 total points
ID: 40374477
You may find a loop like this gets you started...

  for (int pos = 0 ; pos < str.length() ; pos += n) {
      // Add the character at 'pos' to the output
  }

Doug
0
 
LVL 7

Author Comment

by:gudii9
ID: 40375267
public String everyNth(String str, int n) {
String s="";
    for (int pos = 0 ; pos < str.length() ; pos += n) {
      // Add the character at 'pos' to the output
      
      return (s+str.substring(pos-1, pos));
  }
  return null;
}

Open in new window


I tried as above
Test cases failing as below

Expected      Run            
everyNth("Miracle", 2) → "Mrce"      "Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -1 (line number:6)"      X         
everyNth("abcdefg", 2) → "aceg"      "Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -1 (line number:6)"      X         
everyNth("abcdefg", 3) → "adg"      "Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -1 (line number:6)"      X         
everyNth("Chocolate", 3) → "Cca"      "Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -1 (line number:6)"      X         
everyNth("Chocolates", 3) → "Ccas"      "Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -1 (line number:6)"      X         
everyNth("Chocolates", 4) → "Coe"      "Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -1 (line number:6)"      X         
everyNth("Chocolates", 100) → "C"      "Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -1 (line number:6)"


Please advise
0
 
LVL 7

Author Comment

by:gudii9
ID: 40375273
public class TEst13 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		everyNth("hello", 2);
	}
	
	public static String everyNth(String str, int n) {
		String s="";
		    for (int pos = 0 ; pos < str.length() ; pos += n) {
		      // Add the character at 'pos' to the output
		      
		      return (s+str.substring(pos-1, pos));
		  }
		  return null;
		}

}

Open in new window


i get

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: -1
      at java.lang.String.substring(Unknown Source)
      at TEst13.everyNth(TEst13.java:14)
      at TEst13.main(TEst13.java:6)
0
 
LVL 84

Expert Comment

by:ozo
ID: 40375354
If you write it that way, you'd want to start at pos=0

You also don't want to do the return inside the loop
0
 
LVL 7

Author Comment

by:gudii9
ID: 40375662
public class Test10 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		everyNth("hello", 2);
	}
	
	public static String everyNth(String str, int n) {
		String s="";
		String str2 = null;
		    for (int pos = 0 ; pos < str.length() ; pos += n) {
		      // Add the character at 'pos' to the output
		      
		      str2=(s+str.substring(pos-1, pos));
		  }
		  return str2;
		}

}

Open in new window


i moved return to outside. I thought i started at pos is 0 since my for loop checks pos=0.

Please advise
0
 
LVL 84

Expert Comment

by:ozo
ID: 40375666
pos=0
Sorry, I meant to say pos = 1
but rather than adjust that part, it may be preferable to change
(pos-1, pos) to (pos, pos+1)
that way the end condition wouldn't need to be adjusted too.


You are getting closer, but when you see the results, consider also why you are using both s and str2.
0
 
LVL 7

Author Comment

by:gudii9
ID: 40375684
public class Test10 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		String str3=everyNth("hello", 2);
		System.out.println("str3 value is"+str3);
	}
	
	public static String everyNth(String str, int n) {
		String s="";
		String str2 = null;
		    for (int pos = 0 ; pos < str.length() ; pos += n) {//i see length as 5 and pos started at 0
		      // Add the character at 'pos' to the output
		      
		     str2=(s+str.substring(pos, pos+1));
		  }
		  return str2;
		}

}

Open in new window


i modified further.

when i debug on eclipse i see h  then l then o

but my output only gives o
Console output is

str3 value iso
0
 
LVL 84

Expert Comment

by:ozo
ID: 40375688
Do you see why you only get o?  Do you understand what happened to the h then l?
Can you think of a way to get them in the output?
0
 
LVL 7

Author Comment

by:gudii9
ID: 40376946
somehow it is putting only last character not the before ones(it is kind of overriding before ones not sure why yet)


something needs to be tweaked below

str2=(s+str.substring(pos, pos+1));


not sure yet what
0
 
LVL 84

Accepted Solution

by:
ozo earned 300 total points
ID: 40376969
Consider why do you need both str2 and s?

Hint: you don't
0
 
LVL 7

Author Comment

by:gudii9
ID: 40377023
public class Test11 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		String str3=everyNth("Miracle", 2);
		System.out.println("str3 value is-->"+str3);
	}
	
	public static String everyNth(String str, int n) {
		String s="";
		//String str2 = null;
		    for (int pos = 0 ; pos < str.length() ; pos += n) {//i see length as 5 and pos started at 0
		      // Add the character at 'pos' to the output
		      
		     s=(s+str.substring(pos, pos+1));
		  }
		  return s;
		}

}

Open in new window


you are right. i removed the extra one and concatenated to same string 's'

now all test cases are happy

Expected      Run            
everyNth("Miracle", 2) → "Mrce"      "Mrce"      OK         
everyNth("abcdefg", 2) → "aceg"      "aceg"      OK         
everyNth("abcdefg", 3) → "adg"      "adg"      OK         
everyNth("Chocolate", 3) → "Cca"      "Cca"      OK         
everyNth("Chocolates", 3) → "Ccas"      "Ccas"      OK         
everyNth("Chocolates", 4) → "Coe"      "Coe"      OK         
everyNth("Chocolates", 100) → "C"      "C"      OK
0

Featured Post

Online Training Solution

Drastically shorten your training time with WalkMe's advanced online training solution that Guides your trainees to action. Forget about retraining and skyrocket knowledge retention rates.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

If you haven’t already, I encourage you to read the first article (http://www.experts-exchange.com/articles/18680/An-Introduction-to-R-Programming-and-R-Studio.html) in my series to gain a basic foundation of R and R Studio.  You will also find the …
In this post we will learn different types of Android Layout and some basics of an Android App.
The viewer will learn how to implement Singleton Design Pattern in Java.
The viewer will learn how to clear a vector as well as how to detect empty vectors in C++.

690 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question