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has12 challenge

Posted on 2016-08-03
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Last Modified: 2016-08-05
Hi,

I am working on below challenge
http://codingbat.com/prob/p169260

Psedo code description of approach :
1. loop though given array
2. check where is 1
3. check where is 2
4. if 2place is somewhere further down than by 1 return true
5 else return false
I wrote my code as below
public boolean has12(int[] nums) {
  int onePlace=0;
  int twoPlace=0;
  for(int i=0;i<nums.length;i++){
    if(nums[i]==1){
      onePlace=i;
      
    }
     if(nums[i]==2){
      twoPlace=i;
      
    }
  }
  return (twoPlace>onePlace);
}

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I am passing all tests

How to improve my design, approach, code? please advise
0
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Question by:gudii9
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13 Comments
 
LVL 32

Expert Comment

by:phoffric
ID: 41741758
>> return true if there is a 1 in the array with a 2 somewhere later in the array

By hand, what should the answer be for:
5 1 6 2 7 7 8 1 0 9

Does your program agree with your by hand answer?
0
 
LVL 35

Expert Comment

by:ste5an
ID: 41741874
First of all, there is a test missing.

What does your function return, when you're having the arrays [ 2 ] or [ 0, 2 ] ?

So you need to handle that.

Also the code path is not optimal. Your test function describes an implicit termination criteria for the loop.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41742465
actually my code passes all tests.
public boolean has12(int[] nums) {
  int onePlace=0;
  int twoPlace=0;
  for(int i=0;i<nums.length;i++){
    if(nums[i]==1){
      onePlace=i;
      
    }
     if(nums[i]==2){
      twoPlace=i;
      
    }
  }
  return (twoPlace>onePlace);
}

Open in new window

Expected      Run            
has12([1, 3, 2]) → true      true      OK      
has12([3, 1, 2]) → true      true      OK      
has12([3, 1, 4, 5, 2]) → true      true      OK      
has12([3, 1, 4, 5, 6]) → false      false      OK      
has12([3, 1, 4, 1, 6, 2]) → true      true      OK      
has12([2, 1, 4, 1, 6, 2]) → true      true      OK      
has12([2, 1, 4, 1, 6]) → false      false      OK      
has12([3, 5, 9]) → false      false      OK      
has12([3, 5, 1]) → false      false      OK      
has12([3, 2, 1]) → false      false      OK      
has12([1, 2]) → true      true      OK      
has12([1, 1]) → false      false      OK      
has12([1]) → false      false      OK      
has12([]) → false      false      OK      
other tests
OK      
i think i am good on this unless you want to make some suggestions on improvement, corrections?
By hand, what should the answer be for:
5 1 6 2 7 7 8 1 0 9//yes

Does your program agree with your by hand answer?// let me start eclipse in debug mode and double check it
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LVL 32

Accepted Solution

by:
phoffric earned 1000 total points
ID: 41742602
>> By hand, what should the answer be for:  5 1 6 2 7 7 8 1 0 9//yes

>>  Does your program agree with your by hand answer?// let me start eclipse in debug mode and double check it
OK. Reason I gave you this case is that your program should produce a different result. Also, for this small program, it is useful to walk through the program by hand to see what result you come up with. Then walk through in the debugger. If you do that, you get three results, not all the same. Which one(s) is/are correct?
0
 
LVL 32

Expert Comment

by:phoffric
ID: 41742605
It is not necessarily true that the code tester is running through all the edge (corner) cases.
Only thing that counts is a description of the problem (including their example cases which we have seen sometimes helps to clarify ambiguities).
0
 
LVL 35

Assisted Solution

by:ste5an
ste5an earned 1000 total points
ID: 41742668
i think i am good on this unless you want to make some suggestions on improvement, corrections?

In terms of TDD? Yes. When all tests are green, then its okay. But only for a black box test.

As we have your source we can do a white box test and feed it with specially crafted samples. Like phoffric's [ 1, 2, 1] or my [ 2 ] and [ 0, 2 ].  Have you evaluate [ 2 ] and [ 0, 2] by hand?

You have asked for further enhancements:
Well passing this tests correctly is such an enhancement.
Another one is the given termination criteria.
And last but not least there is a simple restructuring possible to minimize the number of if-tests.

How to improve my design, approach, code? please advise

Think about our points. Correct those cases above. Ask again when the rest is unclear.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41742760
Another one is the given termination criteria.
And last but not least there is a simple restructuring possible to minimize the number of if-tests.
sure. how to achieve above?
0
 
LVL 35

Expert Comment

by:ste5an
ID: 41742799
Well, I've crafted my test interval for a certain purpose.

Solving this is step 1. Cause it will give you the necessary hint for the other two points.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41742837
other two points.

which points? i was not clear?
0
 
LVL 35

Expert Comment

by:ste5an
ID: 41743022
Sure. But do you have already a solution which gives the correct result for [ 2 ] and [ 0, 2 ]?
0
 
LVL 7

Author Comment

by:gudii9
ID: 41743267
>> By hand, what should the answer be for:  5 1 6 2 7 7 8 1 0 9//yes

>>  Does your program agree with your by hand answer?// let me start eclipse in debug mode and double check it
OK. Reason I gave you this case is that your program should produce a different result. Also, for this small program, it is useful to walk through the program by hand to see what result you come up with. Then walk through in the debugger. If you do that, you get three results, not all the same. Which one(s) is/are correct?



Given an array of ints, return true if there is (not like all 1's to say 5 1 6 2 7 7 8 1 0 9 false due to bolded 1 )a 1 in the array with a 2 somewhere later in the array.

has12([1, 3, 2]) → true
has12([3, 1, 2]) → true
has12([3, 1, 4, 5, 2]) → true
0
 
LVL 32

Expert Comment

by:phoffric
ID: 41743643
>> Given an array of ints, return true if there is (not like all 1's to say 5 1 6 2 7 7 8 1 0 9 false due to bolded 1 )a 1 in the array with a 2 somewhere later in the array.

I didn't understand your comment.

You answered true for the manual, by hand, check. I agree since there is a 1 in the array followed by a 2. (But I suppose that one could try to interpret the question as:
  • if there is more than one 1 in the array than every 1 must be followed by a 2; or
  • if there is more than one 1 in the array than at least the last 1 must be followed by a 2; or
  • other interpretations

But I agree with your "by hand" answer.
In that case, your posted program does not give True because the last 1 is not followed by a 2.
Can you post your own solution so that your new program agrees with your "by hand" result?
0
 
LVL 35

Expert Comment

by:ste5an
ID: 41743799
Have changed your code?

public boolean has12(int[] nums) {
    for (int i = 0; i < nums.length - 1; i++) {
        if (nums[i] == 1) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[j] == 2) {
                    return true;
                }
            }
        }
    }
    return false;
}

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