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string challenge first and n times last

Hi,

I am working on below coding challenge
http://codingbat.com/prob/p174148
i wrote as below
public String nTwice(String str, int n) {
int i=0;
int len=str.length();
String last=str.substring(len-1);
if(len>n){
for(int j=0;j<n;j++)
{
len=len+len;

}
return str.substring(0,1)+len;
}
  return "";
}

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My test cases are failing as below
Expected	Run		
nTwice("Hello", 2) → "Helo"	"H20"	X	    
nTwice("Chocolate", 3) → "Choate"	"C72"	X	    
nTwice("Chocolate", 1) → "Ce"	"C18"	X	    
nTwice("Chocolate", 0) → ""	"C9"	X	    
nTwice("Hello", 4) → "Hellello"	"H80"	X	    
nTwice("Code", 4) → "CodeCode"	""	X	    
nTwice("Code", 2) → "Code"	"C16"	X	    
other tests

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i would like to know how can improve on my above code.. Not sure if that is what challenge is expecting me to do when it says

Given a string and an int n, return a string made of the first and last n chars from the string. The string length will be at least n.

nTwice("Hello", 2) → "Helo"
nTwice("Chocolate", 3) → "Choate"
nTwice("Chocolate", 1) → "Ce
.Please advise.Thanks in advance
0
gudii9
Asked:
gudii9
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2 Solutions
 
ozoCommented:
I'm not sure why you are doing len=len+len;
or str.substring(0,1)+len
but I think you earlier did a similar challenge for n=1 (or was it n=2)
0
 
ozoCommented:
nTwice("Hello", 2) → "Helo"
first 2 "He", last 2 "lo"
nTwice("Chocolate", 3) → "Choate"
first 3 "Cho", last 3 "ate"
nTwice("Chocolate", 1) → "Ce"
first 1 "C", last 1 "e"
0
 
n2fcCommented:
Correct answer below:

public String nTwice(String str, int n) {
int len=str.length();

String first=str.substring(0,n);
String last=str.substring(len-n,len);
return first+last;
}

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n2fcCommented:
Another solution:


public String nTwice(String str, int n) {
   return str.substring(0,n)+str.substring(str.length()-n,str.length());
}

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0
 
n2fcCommented:
Both of the above solutions do the same thing...
The second combines into a single statement, with no intermediate results...


You concatenate the 1st n characters with the last n characters of the supplied string.
0
 
gudii9Author Commented:
I understand the challenge correctly now. I thought I should display n times from first followed by n times last. If that is challenge how to do it.  Please advise
0
 
n2fcCommented:
Challenge is to display the 1st & last "n" characters in the given string...

See if you can follow the code I supplied above to accomplish this...

Both check out OK on your "codingbat.com" page...
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gudii9Author Commented:
public String nTwice(String str, int n) {
int i=0;
int len=str.length();
String last=str.substring(len-1);
if(len>=n){
return str.substring(0,n)+str.substring(len-n);

}
  return "";
}

i modified as above. Does that looks fine. Please advise
0
 
n2fcCommented:
It works...
BUT...

1) "int i=0"   is unnecessary... Why bother? It is NEVER used!
SAME goes for:
     "String last=str.substring(len-1);"

2) You do NOT have to test for GIVEN conditions!  You test for "if (len>=n)"...
UNNECESSARY... See the problem statement...  
"The string length will be at least n. "   Therefore you do NOT need to test this in your program!


In coding, the rule is always:   KISS= "Keep it SIMPLE, S..."

I would SIMPLIFY as follows:

public String nTwice(String str, int n) {
  int len=str.length();
  return str.substring(0,n)+str.substring(len-n);
}

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0
 
ozoCommented:
Since len is only used once, I might further simplify to
public String nTwice(String str, int n) {
  return str.substring(0,n)+str.substring(str.length()-n);
}

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0
 
n2fcCommented:
ozo...
Thanks for your input...

I already mentioned that in my first few posts...
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