sameFirstLast

hi,
I am trying below challenge
http://codingbat.com/prob/p118976

i am failing in one test case
public boolean sameFirstLast(int[] nums) {
int numsLen=nums.length;
int first=nums[0];
int last=nums[numsLen-1];
if(numsLen>=1&&first==last)
return true;
return false;
  
}

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Expected	Run		
sameFirstLast({1, 2, 3}) → false	false	OK	    
sameFirstLast({1, 2, 3, 1}) → true	true	OK	    
sameFirstLast({1, 2, 1}) → true	true	OK	    
sameFirstLast({7}) → true	true	OK	    
sameFirstLast({}) → false	Exception:java.lang.ArrayIndexOutOfBoundsException: 0 (line number:3)	X	    
sameFirstLast({1, 2, 3, 4, 5, 1}) → true	true	OK	    
sameFirstLast({1, 2, 3, 4, 5, 13}) → false	false	OK	    
sameFirstLast({13, 2, 3, 4, 5, 13}) → true	true	OK	    
sameFirstLast({7, 7}) → true	true	OK	    
other tests

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please advise
LVL 7
gudii9Asked:
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zzynxConnect With a Mentor Software engineerCommented:
If you write:
 if (first == last) {
       return true ;
 }
 return false ;

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When runing this piece of code, your computer will evaluate the expression first == last.
The value of an expression is a boolean. That's a variable that can hold the value 'true' or 'false'.
When the result/value of that expression is true, it will return true.
When the result/value of that expression is false, it will return false.
So, in fact, it can just return the result/value of that expression.
You write that as:
return first == last;

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0
 
gudii9Author Commented:
public boolean sameFirstLast(int[] nums) {

int numsLen=nums.length;

if(numsLen==0){

return false;
}
int first=nums[0];
int last=nums[numsLen-1];
if(numsLen>=1&&first==last)
return true;
return false;
  
}

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i think i oredere edge case and passed all tests. I wonder how to improve my code. please advise
0
 
dpearsonConnect With a Mentor Commented:
I would say your code actually looks fine.

The only part that could be written more clearly is to replace these 3 lines:

...
if(numsLen>=1&&first==last)
return true;
return false;
...

With this one line:

return (first == last) ;

There's no need to check that "numsLen >= 1" because you already tested for length 0 at the top of the method - so the length here must be >= 1.

And rather than:
   if (first == last)
       return true ;
  return false ;

you can just return whether "first == last"
   return (first == last) ;

Make sense?

Whole program becomes:

public boolean sameFirstLast(int[] nums) {

int numsLen=nums.length;

if(numsLen==0){

return false;
}
int first=nums[0];
int last=nums[numsLen-1];
return (first == last) ;
}

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0
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ozoCommented:
numsLen>=1&& is unneccessary
and at that point, you can just
return first==last;

or, using numsLen>=1&&  you can make the if(numsLen==0){ unnecessary
0
 
gudii9Author Commented:
And rather than:
   if (first == last)
       return true ;
  return false ;

you can just return whether "first == last"
   return (first == last) ;

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instead of doing in 2 steps like checking and then returning we are doing in one step in return itself comparing so that i will automatically give true false in return statement right?
please advise
0
 
ozoConnect With a Mentor Commented:
first == last is true when the operands are equal, and false when they are not equal
0
 
dpearsonCommented:
instead of doing in 2 steps like checking and then returning we are doing in one step in return itself comparing so that i will automatically give true false in return statement right?

Yes that's right.  We're replacing 3 lines of code with 1 - which is good because it's simpler.  And we're also removing a conditional (the "if" test) so the code always executes the same lines of code - which is also good.

Rather than:
   if (first == last) {
      // Sometimes we run code in here
      return true ;
   }
   // Sometimes we run code here
   return false ;

We have now:
   // Always run this code
   return (first == last) ;

It's best to avoid branches and conditional statements in your code where possible.  It makes bugs less likely.

Doug
0
 
zzynxSoftware engineerCommented:
Thanx 4 axxepting
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