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countAbc challenge

Posted on 2016-09-21
9
Medium Priority
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Last Modified: 2016-09-22
Hi,


I am working on below challenge.
http://codingbat.com/prob/p161124


I wrote as below and failing some tests.
Expected      Run            
countAbc("abc") → 1      0      X      
countAbc("abcxxabc") → 2      0      X      
countAbc("abaxxaba") → 2      0      X      
countAbc("ababc") → 2      1      X      
countAbc("abxbc") → 0      0      OK      
countAbc("aaabc") → 1      0      X      
countAbc("hello") → 0      0      OK      
countAbc("") → 0      0      OK      
countAbc("ab") → 0      0      OK      
countAbc("aba") → 1      0      X      
countAbc("aca") → 0      0      OK      
countAbc("aaa") → 0      0      OK      
other tests
X      


public int countAbc(String str) {
  
  
 CharSequence cs1="abc";
 CharSequence cs2="aba";
  if(str.length()==0||str.length()==1||str.length()==2){
    return 0;
  }
  if(!str.contains(cs1)||!str.contains(cs2)){
    return 0;
  }
  else return 1+countAbc(str.substring(3));
}

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how to fix? any alternate approaches or improvements? please advise
0
Comment
Question by:gudii9
  • 5
  • 3
9 Comments
 
LVL 7

Author Comment

by:gudii9
ID: 41809446
public int countAbc(String str) {
  
  
 CharSequence cs1="abc";
 
 CharSequence cs2="aba";
  if(str.length()==0||str.length()==1||str.length()==2){
    return 0;
  }


  else if(!str.contains(cs1)||!str.contains(cs2)){
    return 0;
  }
  else return 1+countAbc(str.substring(1));
}

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above also failing some tests
0
 
LVL 37

Accepted Solution

by:
zzynx earned 1000 total points
ID: 41810187
This works:

public int countAbc(String str) {
  String stringABC = "abc";    
  String stringABA = "aba";

  if (str.length()==0||str.length()==1||str.length()==2){
    return 0;
  }

  if(str.startsWith(stringABC)){
    return 1 + countAbc(str.substring(3));
  } else if(str.startsWith(stringABA)){
    return 1 + countAbc(str.substring(2));
  }
  return countAbc(str.substring(1));
}

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0
 
LVL 36

Assisted Solution

by:ste5an
ste5an earned 1000 total points
ID: 41810195
(str.length()==0||str.length()==1||str.length()==2)

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Really?
0
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LVL 37

Expert Comment

by:zzynx
ID: 41810211
No.
if (str.length()<3)

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:)
0
 
LVL 7

Author Comment

by:gudii9
ID: 41810797
  else if(!str.contains(cs1)||!str.contains(cs2)){

Open in new window


contains charsequence approach wont work here?
https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#contains(java.lang.CharSequence)
0
 
LVL 7

Author Comment

by:gudii9
ID: 41810806
public int countAbc(String str) {
  String stringABC = "abc";    
  String stringABA = "aba";

  if (str.length()<3){
    return 0;
  }

  if(str.startsWith(stringABC)){
    return 1 + countAbc(str.substring(3));
  } else if(str.startsWith(stringABA)){
    return 1 + countAbc(str.substring(2));
  }
  return countAbc(str.substring(1));
}

Open in new window


startsWith how is working here why we are moving to str.substring(2));
instead

str.substring(1));
0
 
LVL 7

Author Comment

by:gudii9
ID: 41810811
return 1 + countAbc(str.substring(3));
why one return has 3 in if loop as above and one has 2 as below in else if?

  return 1 + countAbc(str.substring(2));

abc need not be at start right? it could be at middle also?
0
 
LVL 37

Expert Comment

by:zzynx
ID: 41810949
If we find the substring "aba", that last 'a' could be the start of another "aba" or "abc" and the challenge wants us to count that. That's why
0
 
LVL 7

Author Comment

by:gudii9
ID: 41811225
got it
0

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