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sameFirstLast
hi,
I am trying below challenge
http://codingbat.com/prob/p118976
i am failing in one test case
please advise
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;
}
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
please advise
Asked:
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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;
}
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:
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) ;
}
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
and at that point, you can just
return first==last;
or, using numsLen>=1&& you can make the if(numsLen==0){ unnecessary
And rather than:
if (first == last)
return true ;
return false ;
you can just return whether "first == last"
return (first == last) ;
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
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
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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:
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