asked on

min concatination challenge

Hi,

I am trying below challenge

http://codingbat.com/prob/p105745

My code is like below.

public String minCat(String a, String b) {
int aLen=a.length();
int bLen=b.length();
if(aLen==bLen){
return a+b;
}

if(aLen>bLen){
return a+b;
}


if(aLen<bLen){
return a+b;
}
  return null;
}

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i am failing all test cases. Please advise on how to modify and improve my code.

Thanks in advance.

ASKER CERTIFIED SOLUTION

Qlemo

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gudii9

ASKER

String minCat(String a, String b) {
		int aLen=a.length();
		int bLen=b.length();
		int minSize=0;
		String s="";
		if(aLen==bLen){
			s=a+b;
		return s;
		}
		//s1.compareTo(s)

		/*You are always just concatenating the two strings, so of course that fails.
		You need to take the smaller size, and use only the last n characters of the longer string.

		If I would write the code, I would not care about checking which string is longer. I would do something like (in pseudo code):
		   minsize = minimum of aLen and bLen
		   return minsize chars from end of a and minsize chars from end of b

		That works for all cases.*/

		if(aLen>bLen){
		//return minSize=bLen.length();
			s= a.substring(aLen-bLen)+b.substring(bLen-bLen);
			return s;
		}


		if(aLen<bLen){
		s= a.substring(aLen-aLen)+b.substring(bLen-aLen);
		}
		  return s;
		}

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Like above?

public class Test41 {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
	String s=	minCat("at","hi");
	System.out.println("s is-->"+s);

	}

	public static String minCat(String a, String b) {
		int aLen=a.length();
		int bLen=b.length();
		int minSize=0;
		String s="";
		if(aLen==bLen){
			s=a+b;
		return s;
		}
		//s1.compareTo(s)

		/*You are always just concatenating the two strings, so of course that fails.
		You need to take the smaller size, and use only the last n characters of the longer string.

		If I would write the code, I would not care about checking which string is longer. I would do something like (in pseudo code):
		   minsize = minimum of aLen and bLen
		   return minsize chars from end of a and minsize chars from end of b

		That works for all cases.*/

		if(aLen>bLen){
		//return minSize=bLen.length();
			s= a.substring(aLen-bLen)+b.substring(bLen-bLen);
			return s;
		}


		if(aLen<bLen){
		s= a.substring(aLen-aLen)+b.substring(bLen-aLen);
		}
		  return s;
		}
}

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Now i was able to pass all the tests. How can i improve my code.
Please advise

Qlemo

int minLen = Math.min(aLen, bLen);
return a.substring(Math.max(0, aLen-minLen)) + b.substring(Math.max(0, bLen-minLen));

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is what I was talking about

ozo

Is the Math.max necessary?

Qlemo

Dunno for Java, but in most languages negative string positions result in errors.

SOLUTION

ozo

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Qlemo

No :-\ Got you, we can and should omit the max.

int minLen = Math.min(aLen, bLen);
return a.substring(aLen-minLen) + b.substring(bLen-minLen);

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SOLUTION

zzynx

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gudii9

ASKER

i see instead of two if loops like i wrote now we can do in one return statement with substring as we already got minLen from Math.min function right?
please advise

Qlemo

You didn't have any loop. You used three IFs to catch all cases. That or the mathematical approach like shown by me are the common solutions.

zzynx

right?

Right! Instead of your two IFs.

zzynx

Thanx 4 axxepting