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array sum2 challenge
Hi,
i am trying below challenge
http://codingbat.com/prob/p190968
i tried as below
i failed few tests. How to fix and improve my code. please advise
i am trying below challenge
http://codingbat.com/prob/p190968
i tried as below
public int sum2(int[] nums) {
int numsSum[]=new int[4];
if(nums.length>=2)
return nums[0]+nums[1];
// else (nums.length==1)
// return nums[0];
return 0;
}
Expected Run
sum2({1, 2, 3}) → 3 3 OK
sum2({1, 1}) → 2 2 OK
sum2({1, 1, 1, 1}) → 2 2 OK
sum2({1, 2}) → 3 3 OK
sum2({1}) → 1 0 X
sum2({}) → 0 0 OK
sum2({4, 5, 6}) → 9 9 OK
sum2({4}) → 4 0 X
other tests
OK
i failed few tests. How to fix and improve my code. please advise
Asked:
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public int sum2(int[] nums) {
int numsSum[]=new int[4];
if(nums.length>=2)
return nums[0]+nums[1];
if(nums.length==1)
return nums[0];
return 0;
}
if i put both if and then return 0 at end i passed all test though. not sure if that is correct way of doing
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public int sum2(int[] nums) {
if (nums.length ==0){return 0;}
if(nums.length==1){return nums[0];}
return nums[0]+nums[1];
}
This is how I would do the ternary option :
public int sum2(int[] nums) {
return nums.length>1?nums[0]+nums[1]:nums.length>0?nums[0]:0;
}
return (nums.length>0?nums[0]:0)+(nums.leng th>1?nums[ 1]:0);
return nums.length>1?nums[0]+nums[1]:nums.l ength>0?nu ms[0]:0;
return nums.length==0?0:nums.length==1?nums [0]:nums[0 ]+nums[1;]
return nums.length<1?0:nums[0]+(nums.length <2?0:nums[ 1]);
i have not clearly understood meaning of above 4 lines.
what it means by
nums[1]:0);
also what it mean by
nums[1;]
also as below
return nums.length<1?0:nums[0]+(n
please advise
(nums.length>1?nums[1]:0);
is nums[1] when nums.length>1, otherwise, it is 0
nums[1;]
should have been
nums[1];
is nums[1] when nums.length>1, otherwise, it is 0
nums[1;]
should have been
nums[1];
>>return (nums.length>0?nums[0]:0)+
return nums.length>1?nums[0]+nums
return nums.length==0?0:nums.leng
return nums.length<1?0:nums[0]+(n
i have not clearly understood meaning of above 4 lines.<<
They are all nested ternary (if-then-else) statements satisfying the requirements;
The first is saying, if the length is greater than 0 return the first number, if not return 0,
then, if the length is greater than 1, plus the second number, else plus 0.
The second is saying, if the length is greater than 1, return the first number plus the second number, then, if the length is greater than 0, return the first number, else return 0.
The third is saying, if the length is less than 1, return 0, else, if the length equals 1, return the first number, else return the first number plus the second number.
The fourth is saying, if the length is less than 1, return 0, else return the first number added to, if the length is less than 2, plus 0, else plus the second number.
return nums.length>1?nums[0]+nums
return nums.length==0?0:nums.leng
return nums.length<1?0:nums[0]+(n
i have not clearly understood meaning of above 4 lines.<<
They are all nested ternary (if-then-else) statements satisfying the requirements;
The first is saying, if the length is greater than 0 return the first number, if not return 0,
then, if the length is greater than 1, plus the second number, else plus 0.
The second is saying, if the length is greater than 1, return the first number plus the second number, then, if the length is greater than 0, return the first number, else return 0.
The third is saying, if the length is less than 1, return 0, else, if the length equals 1, return the first number, else return the first number plus the second number.
The fourth is saying, if the length is less than 1, return 0, else return the first number added to, if the length is less than 2, plus 0, else plus the second number.
return (nums.length>0?nums[0]:0)+// >1 should cover >0 case already in the front right???(nums.leng th>1?nums[ 1]:0);//// >0 should cover >1 case already in the front right???
return nums.length>1?nums[0]+nums[1]:nums.l ength>0?nu ms[0]:0;
The first is saying, if the length is greater than 0 return the first number, if not return 0,
then, if the length is greater than 1, plus the second number, else plus 0.
The second is saying, if the length is greater than 1, return the first number plus the second number, then, if the length is greater than 0, return the first number, else return 0.
return (nums.length>0?nums[0]:0)+>0 covers the nums[0] part of the >1 case result(nums.leng th>1?nums[ 1]:0);//// >0 should cover >1 case already in the front right???
return nums.length>1?nums[0]+numsreturn nums.length>1?([1]:nums.l ength>0?nu ms[0]:0;
nums[0]+nums[1] // this is the >1 case
):(
nums.length>0?nums[0]:0 // this is the not >1 case
);
nums.length>0?nums[0]:0 // this is the not >1 case
above can be writen as below??
nums.length==0?nums[0]:0 // this is the not >1 case
above can be writen as below??
nums.length==0?nums[0]:0 // this is the not >1 case
No, but it can be written as
nums.length==0?0:nums[0]
(because we know that nums.length is never < 0)
nums.length==0?0:nums[0]
(because we know that nums.length is never < 0)
public int sum2(int[] nums) {
return (nums.length==0?0:nums[0])+(nums.length>1?nums[1]:0);
}
above worked fine
public int sum2(int[] nums) {
return (nums.length==0?0:nums[0])+(nums.length>=1?nums[1]:0);
}
why above code failing when i change from
(nums.length>1
to
(nums.length>=1
please advise
return (nums.length==0?0:nums[0])
in above working solution where we are taking care when nums.length is 1 case?
return nums.length==0?0:nums.length==1?nums[0]:nums[0]+nums[1;]
return nums.length<1?0:nums[0]+(nums.length<2?0:nums[1]);
i am bit confused about above 2 lines.How to break down put some brackets around to clearly understand the flow of solution. please advise.
where can i see more nested ternary related examples to understand above lines of codes
>>why above code failing when i change from
(nums.length>1
to
(nums.length>=1<<
Take a look at what this says
return (nums.length==0?0:nums[0])
If the length is 0, return 0, else return the first number plus, if the length is 2 or more, the second number, else plus nothing.
and this says
return (nums.length==0?0:nums[0])
If the length is 0, return 0, else return the first number plus, if the length is 1 or more, the second number, else plus nothing. The problem is that if the length is 1, there is no second number and you will get an out of bounds index error.
BTW, you need to stop piling questions on top of each other too quickly so, when the experts respond, everyone knows which question they're referring to.
(nums.length>1
to
(nums.length>=1<<
Take a look at what this says
return (nums.length==0?0:nums[0])
If the length is 0, return 0, else return the first number plus, if the length is 2 or more, the second number, else plus nothing.
and this says
return (nums.length==0?0:nums[0])
If the length is 0, return 0, else return the first number plus, if the length is 1 or more, the second number, else plus nothing. The problem is that if the length is 1, there is no second number and you will get an out of bounds index error.
BTW, you need to stop piling questions on top of each other too quickly so, when the experts respond, everyone knows which question they're referring to.
where can i see more nested ternary related examples to understand above lines of codes
In just about all the Codingbat questions you've asked.
return (nums.length==0?0:nums[0])+(nums.length>1?nums[1]:0);When nums.length == 1, nums.length==0 is false, so the first term becomes nums[0]
in above working solution where we are taking care when nums.length is 1 case?
When nums.length == 1, nums.length>0 is false, so the second term becomes 0
nums[0]+0 is the required result
//return nums.length<1?0:nums[0]+(n
return
nums.length<1
?
0
:
(
nums[0]
+
(nums.length<2?0:nums[1])
)
;
//return nums.length==0?0:nums.leng
return
nums.length==0
?
0
:
(
nums.length==1
?
nums[0]
:
(nums[0]+nums[1])
)
;
return
nums.length<1
?
0
:
(
nums[0]
+
(nums.length<2?0:nums[1])
)
;
//return nums.length==0?0:nums.leng
return
nums.length==0
?
0
:
(
nums.length==1
?
nums[0]
:
(nums[0]+nums[1])
)
;
public int sum2(int[] nums) {how this solution works when nums.length is 1. In this case it should return
if (nums.length ==0){return 0;}
if(nums.length==1){return nums[0];}
return nums[0]+nums[1];
}
return nums[0]+nums[1];
but we do not have nums[1] when nums.length is 1?
please advise
>>how this solution works when nums.length is 1. In this case it should return
return nums[0]+nums[1];<<
No, it shouldn't, it should return just nums[0]
return nums[0]+nums[1];<<
No, it shouldn't, it should return just nums[0]
//return nums.length==0?0:nums.length==1?nums[0]:nums[0]+nums[1];
return
nums.length==0
?
0
:
[b] ([/b]
nums.length==1
?
nums[0]
:
(nums[0]+nums[1])
[b] )[/b]
;
above highlighted brackets are crucial for my understanding. Now get picture with those brackets without them i lost originally.
How to put those brackets in below single line
return nums.length==0?0:(nums.length==1?nums[0]:num
can i write as above? But i am getting below error
public int sum2(int[] nums) {
return nums.length==0?0:(nums.len
Compile problems:
Error: return nums.length==0?0:(nums.len
^^^^^^^^^^^^
The operator + is undefined for the argument type(s) int, int[]
see Example Code to help with compile problems
please advise
gudii9, Just in case you didn't notice what ozo pointed out, the parenthesis needs to be outside the nums[1]) and not in between nums and the opening bracket.
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public int sum2(int[] nums) {
int numsSum[]=new int[4];
if(nums.length>=2)
return nums[0]+nums[1];
else (nums.length==1)
return nums[0];
}
why i got below error
Compile problems:
Error: else (nums.length==1)
^
Syntax error, insert "AssignmentOperator ArrayInitializer" to complete ArrayInitializerAssignemen
what it means?
please advise