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count8 challlenge

Posted on 2016-09-09
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Last Modified: 2016-10-15
Hi,
I am working on below challenge

http://codingbat.com/prob/p192383

Given a non-negative int n, compute recursively (no loops) the count of the occurrences of 8 as a digit, except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).

count8(8) → 1
count8(818) → 2
count8(8818) → 4

i am not clear on how to find eight count using recursion. please advise
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Question by:gudii9
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13 Comments
 
LVL 28

Assisted Solution

by:Pawan Kumar
Pawan Kumar earned 250 total points
ID: 41792255
--
----

DECLARE @intValue AS VARCHAR(10) = '8818'
 
SELECT SUM( CASE WHEN SUBSTRING(@intValue,number,1) LIKE '[8]' THEN 1 ELSE 0 END ) Count8DIGITS FROM
( 
  SELECT DISTINCT number FROM
  MASTER..SPT_VALUES WHERE number > 0 AND number <= DATALENGTH(@intValue) 
) x

--
--

Open in new window

0
 
LVL 27

Expert Comment

by:rrz
ID: 41793103
This challenge is almost the same as  
https://www.experts-exchange.com/questions/28968896/count7-challenge.html 
There is just an extra conditional here.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41793537
I will try
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LVL 7

Author Comment

by:gudii9
ID: 41793657
public int count8(int n) {
  if (n < 1) {
    return 0;
  }
  if ((n % 10) == 8){ return 1 + count8(n/10);
  }
  else {
  return count8(n/10);
  }
}

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above failing some tests
Expected      Run            
count8(8) → 1      1      OK      
count8(818) → 2      2      OK      
count8(8818) → 4      3      X      
count8(8088) → 4      3      X      
count8(123) → 0      0      OK      
count8(81238) → 2      2      OK      
count8(88788) → 6      4      X      
count8(8234) → 1      1      OK      
count8(2348) → 1      1      OK      
count8(23884) → 3      2      X      
count8(0) → 0      0      OK      
count8(1818188) → 5      4      X      
count8(8818181) → 5      4      X      
count8(1080) → 1      1      OK      
count8(188) → 3      2      X      
count8(88888) → 9      5      X      
count8(9898) → 2      2      OK      
count8(78) → 1      1      OK      
other tests
X      

This challenge is almost the same as  
https://www.experts-exchange.com/questions/28968896/count7-challenge.html 
There is just an extra conditional here.

what is extra conditional compared to count7 challenge here for count8 challenge?
0
 
LVL 27

Expert Comment

by:rrz
ID: 41793738
what is extra conditional
The challenge stated
except that an 8 with another 8 immediately to its left counts double
So, we need something like
    if(n % 10 == 8){
		if(next digit is 8)return 2 + count8(n / 10);
		else return 1 + count8(n / 10);
    }

Open in new window

0
 
LVL 28

Expert Comment

by:Pawan Kumar
ID: 41793766
Enjoy ! Without Recursion and loop;

--

DECLARE @intValue AS VARCHAR(10) = '1818E8'
 
SELECT SUM( CASE WHEN SUBSTRING(@intValue,number,1) LIKE '[8]' THEN 1 ELSE 0 END ) Count8DIGITS FROM
(
  SELECT DISTINCT number FROM
  MASTER..SPT_VALUES WHERE number > 0 AND number <= DATALENGTH(@intValue)
) x

--
0
 
LVL 7

Author Comment

by:gudii9
ID: 41796638
public int count8(int n) {
  int result=0;
  if (n < 1) {
     result=0;
  }
  if ((n % 10) == 8){
    if ((n+1 % 10) == 8){
      result= 2 + count8(n/10);
     // return result;
    }
   else{ 
     result= 1 + count8(n/10);
    // return result;
  }
  }
  return result;
}
 /* else {
  return count8(n/10);
  }*/
   /* if(n % 10 == 8){
		if(next digit is 8)return 2 + count8(n / 10);
		else return 1 + count8(n / 10);*/
    

Open in new window



above fails below tests

Expected      Run            
count8(8) → 1      1      OK      
count8(818) → 2      1      X      
count8(8818) → 4      1      X      
count8(8088) → 4      2      X      
count8(123) → 0      0      OK      
count8(81238) → 2      1      X      
count8(88788) → 6      2      X      
count8(8234) → 1      0      X      
count8(2348) → 1      1      OK      
count8(23884) → 3      0      X      
count8(0) → 0      0      OK      
count8(1818188) → 5      2      X      
count8(8818181) → 5      0      X      
count8(1080) → 1      0      X      
count8(188) → 3      2      X      
count8(88888) → 9      5      X      
count8(9898) → 2      1      X      
count8(78) → 1      1      OK      
other tests
X      

please advise
0
 
LVL 27

Accepted Solution

by:
rrz earned 250 total points
ID: 41796836
 if ((n+1 % 10) == 8)

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That is not what you want to do. That tests whether the present input + 1 has a 8 as its last digit. You want to use n/10 to peek at the input that will follow the present one.
0
 
LVL 7

Author Comment

by:gudii9
ID: 41796901
public int count8(int n) {
  int result=0;
  if (n < 1) {
     result=0;
  }
  if ((n % 10) == 8){
    if ((n/10) == 8){
      result= 2 + count8(n/10);
     // return result;
    }
   else{ 
     result= 1 + count8(n/10);
    // return result;
  }
  }
  return result;
}
 /* else {
  return count8(n/10);
  }*/
   /* if(n % 10 == 8){
		if(next digit is 8)return 2 + count8(n / 10);
		else return 1 + count8(n / 10);*/
    /*That is not what you want to do. That tests whether the present input + 1
    has a 8 as its last digit.
    You want to use n/10 to peek at the input that will follow the present one.*/

Open in new window

like above?it still fails same tests?
0
 
LVL 7

Author Comment

by:gudii9
ID: 41796935
I think i got approach.
public int count8(int n) {
  //int result=0;
  if (n < 1) {
     return 0;
  }
  if (n >=88 && n%100==88){
   
      return 2 + count8(n/10);
     // return result;
    }
    else if( n%10==8){
    // result= 1 + count8(n/10);
    // return result;
    return 1 + count8(n/10);
  }
  return count8(n/10);
  
 // return result;
}
 /* else {
  return count8(n/10);
  }*/
   /* if(n % 10 == 8){
		if(next digit is 8)return 2 + count8(n / 10);
		else return 1 + count8(n / 10);*/
    /*That is not what you want to do. That tests whether the present input + 1
    has a 8 as its last digit.
    You want to use n/10 to peek at the input that will follow the present one.*/

Open in new window


above passed all tests. any improvements or alternate approaches?
0
 
LVL 7

Author Comment

by:gudii9
ID: 41796946
if then if then else
in above fashion code is there else belong to top if or second(last if in this case)?
0
 
LVL 27

Expert Comment

by:rrz
ID: 41796955
  if (n >=88 && n%100==88){

Open in new window

That doesn't look right.
Here is my solution.  
public int count8(int n) {
      if(n == 0)return 0;
    if(n % 10 == 8){
		if((n / 10) % 10 == 8)return 2 + count8(n / 10);
		else return 1 + count8(n / 10);
    }
    else return count8(n / 10);
}

Open in new window

0
 
LVL 7

Author Comment

by:gudii9
ID: 41796961
public int count8(int n) {
      if(n == 0)return 0;
    if(n % 10 == 8){
		if((n / 10) % 10 == 8)return 2 + count8(n / 10);
		else return 1 + count8(n / 10);
    }
    else return count8(n / 10);
}

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i like this approach
0

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