string challenge without first and last

Hi,

I am working on below coding challenge
http://codingbat.com/prob/p174254
i wrote as below


public String withouEnd2(String str) {
if(str.length()>=2){
return str.substring(1,str.length()-1);

}
return "";
 
}

My test cases are passing
i would like to know how can improve on my above code. For string length 1or 2 i returned "". Not sure if that is what challenge is expecting me to do.Please advise.Thanks in advance
LVL 7
gudii9Asked:
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ozoCommented:
return str.length()>2?str.substring(1,str.length()-1):"";
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ariohCommented:
It's a good practice to test values for null
return str != null && str.length() > 2 ? str.substring( 1, str.length() - 1 ) : "";
0

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gudii9Author Commented:
return str.length()>2?str.substring(1,str.length()-1):"";

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does above one supposed to be as below with a = sign

return str.length()>=2?str.substring(1,str.length()-1):"";

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please advise
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ozoCommented:
when str.length() is 2, str.substring(1,str.length()-1) is "", so either one gives the same answer, but "" seems the more efficient way to handle that case.
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zzynxSoftware engineerCommented:
>> Not sure if that is what challenge is expecting me to do
Yes it is. Since it states:
withouEnd2("ab") → ""

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and in the tests are also included:
withouEnd2("a") → ""

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and
withouEnd2("") → ""

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So, why doubting?
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gudii9Author Commented:
when str.length() is 2, str.substring(1,str.length()-1) is "", so either one gives the same answer, but "" seems the more efficient way to handle that case.

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can you please elaborate on this. I was not clear on this.
public String withouEnd2(String str) {
if(str.length()>=2){
return str.substring(1,str.length()-1);

}
return "";
  
}

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my code passed all test cases including "ab" eventhough i used str.length()>=2 condition. I did not return "" for this right instead returning  str.substring(1,str.length()-1);

please advise
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ozoCommented:
when str is "ab", str.substring(1,str.length()-1) is ""
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zzynxSoftware engineerCommented:
1. See the documentation of the substring method
It has this examples:
 "hamburger".substring(4, 8) returns "urge"
 "smiles".substring(1, 5) returns "mile"

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and also this explanation:
Parameters:
    beginIndex - the beginning index, inclusive.
    endIndex - the ending index, exclusive.

When your string is "ab" it has a length of 2. So, in fact
str.substring(1,str.length()-1)

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becomes
str.substring(1, 1);

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That says, "take the substring that starts with the 2nd character, but not including the 2nd character".
Well, in that case you end up with nothing being the empty string "".

So writing

public String withouEnd2(String str) {
    if(str.length()>=2) {
      return str.substring(1,str.length()-1);
    }
    return "";
}

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is functionally the same as writing

public String withouEnd2(String str) {
    if(str.length()>2) {
      return str.substring(1,str.length()-1);
    }
    return "";
}

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gudii9Author Commented:
is functionally the same as writing
but if my string is "ab" for below code it wont enter into the if loop right since 2>2 is false
public String withouEnd2(String str) {
    if(str.length()>2) {
      return str.substring(1,str.length()-1);
    }
    return "";
}

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Please advise
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ozoCommented:
Yes,  2>2 is false so it would do  return "";

which would have the same result as return str.substring(1,str.length()-1);
both have the effect of withouEnd2("ab") → ""
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zzynxSoftware engineerCommented:
>> but if my string is "ab" for below code it wont enter into the if loop right since 2>2 is false
That's right.
So, it skips the if block and returns "".
And if it would have gone into the if block it would also return "".
So: it's functionally the same
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zzynxSoftware engineerCommented:
Thanx 4 axxepting
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