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second 6 is 7 challenge

Posted on 2014-11-09
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Last Modified: 2014-11-16
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
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Comment
Question by:gudii9
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13 Comments
 
LVL 84

Assisted Solution

by:ozo
ozo earned 250 total points
ID: 40431925
Count the number of times that 6, 6 appears, plus the number of times that 6, 7 appears
0
 
LVL 26

Expert Comment

by:dpearson
ID: 40432049
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;
}

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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
 
LVL 7

Author Comment

by:gudii9
ID: 40435959
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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LVL 84

Expert Comment

by:ozo
ID: 40436525
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})
0
 
LVL 26

Expert Comment

by:dpearson
ID: 40436563
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
 
LVL 7

Author Comment

by:gudii9
ID: 40437907
, 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
0
 
LVL 7

Author Comment

by:gudii9
ID: 40437910
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
 
LVL 84

Expert Comment

by:ozo
ID: 40438908
Can you think of a condition that is true when two 6's are next to each other?
0
 
LVL 7

Author Comment

by:gudii9
ID: 40441094
 if(nums[i]==6&&nums[i+1]==7)

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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
0
 
LVL 84

Expert Comment

by:ozo
ID: 40441115
Perhaps after trying http://codingbat.com/prob/p121853
you can then come back to http://codingbat.com/prob/p110019 with a new idea.
0
 
LVL 26

Accepted Solution

by:
dpearson earned 250 total points
ID: 40441656
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
 
LVL 7

Author Comment

by:gudii9
ID: 40442647
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.
0
 
LVL 26

Expert Comment

by:dpearson
ID: 40445992
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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