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second 6 is 7 challenge

Hi,

i am  trying below challenge

http://codingbat.com/prob/p110019

I have not understood on problem statement when they say like
count instances where the second "6" is actually a 7.

i wrote as below

public int array667(int[] nums) {
int count=0;
  for(int i=0;i<nums.length-1;i++)
  {
  if(nums[i]==nums[i+1])
  count++;
  }
  return count;
}

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some test cases are failing as below
Expected      Run            
array667({6, 6, 2}) → 1      1      OK         
array667({6, 6, 2, 6}) → 1      1      OK         
array667({6, 7, 2, 6}) → 1      0      X         
array667({6, 6, 2, 6, 7}) → 2      1      X         
array667({1, 6, 3}) → 0      0      OK         
array667({6, 1}) → 0      0      OK         
array667({}) → 0      0      OK         
array667({3, 6, 7, 6}) → 1      0      X         
array667({3, 6, 6, 7}) → 2      1      X         
array667({6, 3, 6, 6}) → 1      1      OK         
array667({6, 7, 6, 6}) → 2      1      X         
array667({1, 2, 3, 5, 6}) → 0      0      OK         
array667({1, 2, 3, 6, 6}) → 1      1      OK         

please advise
0
gudii9
Asked:
gudii9
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2 Solutions
 
ozoCommented:
Count the number of times that 6, 6 appears, plus the number of times that 6, 7 appears
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dpearsonCommented:
Your code detects whenever there are 2 identical numbers appearing one after another in the array.

So your code would detect
6,6   (which is correct)
but also
3,3  (which is not)

Also the instructions say you need to find pairs like this:
6,6
or
6,7

So you want code something like this:
public int array667(int[] nums) {
int count=0;
  for(int i=0;i<nums.length-1;i++)
  {
  if(nums[i]==6 && nums[i+1] == 7)
  count++;
  }
  return count;
}

Open in new window

which would detect one of those pairs - the 6,7 combination.

So you need to add code to make it also recognize the 6,6 combination.
0
 
gudii9Author Commented:
public int array667(int[] nums) {
int count=0;
  for(int i=0;i<nums.length-1;i++)
  {
  if(nums[i]==nums[i+1]|| nums[i]==6&&nums[i+1]==7)
  count++;
  }
  return count;
}

Open in new window

i wrote as above . It pass all the tests.
does it looks ok?
0
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ozoCommented:
Although it passes all the tests at the challenge link, the count would be different from the solution given at the challenge link if it was tested with
array667({1,1})
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dpearsonCommented:
Yeah - like ozo pointed out, you don't really want this

if (nums[i]==nums[i+1] ... )

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in the test as that's just checking if two numbers in the sequence are identical.

The problem is asking for specifically checking for 6s and 7s, not any two numbers.

Doug
0
 
gudii9Author Commented:
, you don't really want this

By removing that as below

public int array667(int[] nums) {
int count=0;
  for(int i=0;i<nums.length-1;i++)
  {
  if(nums[i]==6&&nums[i+1]==7)
  count++;
  }
  return count;
}

Open in new window


i am failing test cases as below
Expected	Run		
array667({6, 6, 2}) → 1	0	X	    
array667({6, 6, 2, 6}) → 1	0	X	    
array667({6, 7, 2, 6}) → 1	1	OK	    
array667({6, 6, 2, 6, 7}) → 2	1	X	    
array667({1, 6, 3}) → 0	0	OK	    
array667({6, 1}) → 0	0	OK	    
array667({}) → 0	0	OK	    
array667({3, 6, 7, 6}) → 1	1	OK	    
array667({3, 6, 6, 7}) → 2	1	X	    
array667({6, 3, 6, 6}) → 1	0	X	    
array667({6, 7, 6, 6}) → 2	1	X	    
array667({1, 2, 3, 5, 6}) → 0	0	OK	    
array667({1, 2, 3, 6, 6}) → 1	0	X	    

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Please advise
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gudii9Author Commented:
return the number of times that two 6's are next to each other

how to incorporate above in to if condition . please advise
0
 
ozoCommented:
Can you think of a condition that is true when two 6's are next to each other?
0
 
gudii9Author Commented:
 if(nums[i]==6&&nums[i+1]==7)

Open in new window



as of now i am able to check 6 and 7 are adjacent. Since i am using 6 in this condition not sure how to check if another 6 also adjacent. Not able to think. Also i wonder why thinking is hard sometimes. please advise
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ozoCommented:
Perhaps after trying http://codingbat.com/prob/p121853
you can then come back to http://codingbat.com/prob/p110019 with a new idea.
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dpearsonCommented:
Can you see what this tests for?

  if(nums[i]==6 && (nums[i+1] == 6 || nums[i+1] == 7))

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And how we can use it in the whole method like this?

public int array667(int[] nums) {
int count=0;
  for(int i=0;i<nums.length-1;i++)
  {
  if(nums[i]==6 && (nums[i+1] == 6 || nums[i+1] == 7))
  count++;
  }
  return count;
}

Open in new window

0
 
gudii9Author Commented:
public int array667(int[] nums) {
int count=0;
  for(int i=0;i<nums.length-1;i++)
  {
  if(nums[i]==6 && (nums[i+1] == 6 || nums[i+1] == 7))
  count++;
  }
  return count;
}

Open in new window


i like above solution where we can check same index for 2 different values 6 as well as 7. This is new perspective opened for me.
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dpearsonCommented:
This is new perspective opened for me.
Excellent.  That's the whole point of expert's exchange :)

Glad you've got this now.

Doug
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