strDist challenge

please advise

public int strDist(String str, String sub) {
   if (str.indexOf(sub) == -1) return 0;
    if (str.substring(0, sub.length()).equals(sub)
            && str.substring(str.length() - sub.length())
            .equals(sub)){
        return str.length();
            }
    else (! str.substring(0, sub.length() ).equals(sub) ){
        return strDist(str.substring(1), sub);
    }
    

}

Select allOpen in new window

above gives below error

Compile problems:


Error:      else (! str.substring(0, sub.length() ).equals(sub) ){
                                                          ^
Syntax error, insert "AssignmentOperator ArrayInitializer" to complete ArrayInitializerAssignement


see Example Code to help with compile problems

please advise

public int strDist(String str, String sub) {
   if (str.indexOf(sub) == -1) return 0;
    if (str.substring(0, sub.length()).equals(sub)
            && str.substring(str.length() - sub.length())
            .equals(sub)){
        return str.length();
            }
    else if(! str.substring(0, sub.length() ).equals(sub) ){
        return strDist(str.substring(1), sub);
    }
  
  else return 1000;
}

Select allOpen in new window

i had condition in else which is wrong

above gives below errors

Expected      Run            
strDist("catcowcat", "cat") → 9      9      OK      
strDist("catcowcat", "cow") → 3      1000      X      
strDist("cccatcowcatxx", "cat") → 9      1000      X      
strDist("abccatcowcatcatxyz", "cat") → 12      1000      X      
strDist("xyx", "x") → 3      3      OK      
strDist("xyx", "y") → 1      1000      X      
strDist("xyx", "z") → 0      0      OK      
strDist("z", "z") → 1      1      OK      
strDist("x", "z") → 0      0      OK      
strDist("", "z") → 0      0      OK      
strDist("hiHellohihihi", "hi") → 13      13      OK      
strDist("hiHellohihihi", "hih") → 5      1000      X      
strDist("hiHellohihihi", "o") → 1      1000      X      
strDist("hiHellohihihi", "ll") → 2      1000      X      
other tests
X      

public int strDist(String str, String sub) {
   if (str.indexOf(sub) == -1) return 0;
    if (str.substring(0, sub.length()).equals(sub)
            && str.substring(str.length() - sub.length())
            .equals(sub)){
        return str.length();
            }
    else if(! str.substring(0, sub.length() ).equals(sub) ){
        return strDist(str.substring(1), sub);
    }
  
  else if(! str.substring( str.length()-sub.length() ).equals(sub) ){
        return strDist(str.substring(0,str.length()-1), sub);
}
else return 1000;

}

Select allOpen in new window

above passes all tests.

any improvements or alternate approaches.

Basically pesedo code:

1. check first few letter same as sub and also last few letters same as sub if true return big string length
2. if first few letters are same as sub then do recursion of same method by passing one less character from front of big string along with sub
3. if last few letters are same as sub then do recursion of same method by passing one less character from back along with sub

any improvements or alternate approaches.

to avoid repeating statements you could do:

int len = str.length();
if (str.startsWith(sub)&&str.endsWith(sub) );
{
     return len;
}
int pos = str.indexOf(sub);
if (pos == -1)    
      return 0;
if (pos > 0)
{
     return StrDist(str.substring(pos), sub);
}

return StrDist(str.substring(0, len-1), sub);

Select allOpen in new window

note, you should return 0 if the string doesn't contain the sub string.

Sara

i like above approach with pos and also starts and endwith

what would be pos value  for last return. when does it goes into below statement

return StrDist(str.substring(0, len-1), sub);

Select allOpen in new window

is it pos<0 but -1 is already covered above?

the return value is the length of the maximum string that begins with sub and ends with sub.

the final return is when both the begin and the end are matching or when the sub is not in str at all. for the last case you should return -1 or 0. both these values would indicate failure since otherwise the smallest return value is length of sub.

pos < 0 must be checked to find out whether str contains at least one occurrance of sub. for all further calls this is redundant. same the indexOf is onl used in the first call but not later.

because of that you could optimize like that:

int StrDist(String str, int lensub)
{
     int len = str.length();
     if (str.endsWith(str.substring(0, lensub)))
     {
           return len;
     }
     return StrDist(str.substring(0, len-1), lensub)
}

int StrDist(String str, String sub)
{
    int pos = str.indexOf(sub);
    if (pos == -1)    
        return 0;
    str = str,substring(pos);
    return StrDist(str, sub.length());  // calls the first StrDist function
}

Select allOpen in new window

the first StrDist function expects a string and a length, and only checks whether the left tail of str is equal to the right tail. this function is the one that was called recursively.

the second StrDist is the initial call where we check whether the sub is in the string at all and where the first position is. for all further calls the string only would be changed at the right.

Sara

int StrDist(String str, int lensub)
{
     int len = str.length();
     if (str.endsWith(str.substring(0, lensub)))
     {
           return len;
     }
     return StrDist(str.substring(0, len-1), lensub)
}

int StrDist(String str, String sub)
{
    int pos = str.indexOf(sub);
    if (pos == -1)    
        return 0;
    str = str,substring(pos);
    return StrDist(str, sub.length());  // calls the first StrDist function
}

Select allOpen in new window

above gives below error

Compile problems:


Error:      return StrDist(str.substring(0, len-1), lensub)
                                                    ^
Syntax error, insert ";" to complete ReturnStatement


see Example Code to help with compile problems


please advise

Syntax error, insert ";" to complete ReturnStatement

the compiler told you what to do.

Sara

int StrDist(String str, int lensub)
{
     int len = str.length();
     if (str.endsWith(str.substring(0, lensub)))
     {
           return len;
     }
     return StrDist(str.substring(0, len-1), lensub);
}

int StrDist(String str, String sub)
{
    int pos = str.indexOf(sub);
    if (pos == -1)    
        return 0;
    str = str,substring(pos);
    return StrDist(str, sub.length());  // calls the first StrDist function
}

Select allOpen in new window

i fixed that and then i got below error
Compile problems:


Error:      str = str,substring(pos);
               ^
Syntax error on token ",", ; expected


see Example Code to help with compile problems

now

int StrDist(String str, int lensub)
{
     int len = str.length();
     if (str.endsWith(str.substring(0, lensub)))
     {
           return len;
     }
     return StrDist(str.substring(0, len-1), lensub);
}

int StrDist(String str, String sub)
{
    int pos = str.indexOf(sub);
    if (pos == -1)    
        return 0;
    str = str.substring(pos);
    return StrDist(str, sub.length());  // calls the first StrDist function
}

Select allOpen in new window

above gives below error
Compile problems:


Error:The method strDist(String, String) is undefined

see Example Code to help with compile problems


please advise.
i saw above method though? not sure why compiler complaining

str = str,substring(pos);

didn't you see that it is a comma left of substring. did you ever encounter a comma left of substring? again, the Compiler told you what is wrong.

you really could read the compiler errors. and when taking them literally, you have a good chance to correct the errors by your own...

Sara

yes I corrected comma with . error.

now I got new error like method not defined

int StrDist(String str, int lensub)
{
     int len = str.length();
     if (str.endsWith(str.substring(0, lensub)))
     {
           return len;
     }
     return StrDist(str.substring(0, len-1), lensub);
}

int StrDist(String str, String sub)
{
    int pos = str.indexOf(sub);
    if (pos == -1)    
        return 0;
    str = str.substring(pos);
    return StrDist(str, sub.length());  // calls the first StrDist function
}

Select allOpen in new window

Compile problems:


Error:The method strDist(String, String) is undefined



see Example Code to help with compile problems


I think recursion method needs string and int as argument
int StrDist(String str, int lensub)


where as the method defined has String and String as arguents?

int StrDist(String str, String sub)

when I changed second argument from string to int getting different error

int StrDist(String str, int lensub)
{
     int len = str.length();
     if (str.endsWith(str.substring(0, lensub)))
     {
           return len;
     }
     return StrDist(str.substring(0, len-1), lensub);
}

int StrDist(String str, int lensub)
{
    int pos = str.indexOf(sub);
    if (pos == -1)    
        return 0;
    str = str.substring(pos);
    return StrDist(str, sub.length());  // calls the first StrDist function
}

Select allOpen in new window

saying duplicate method


Error:      int StrDist(String str, int lensub)
          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Duplicate method StrDist(String, int) in type Shell

int StrDist(String str, int lensub)
{
     int len = str.length();
     if (str.endsWith(str.substring(0, lensub)))
     {
           return len;
     }
     return StrDist(str.substring(0, len-1), lensub);
}

int StrDist(String str2, int lensub2)
{
    int pos = str.indexOf(sub);
    if (pos == -1)    
        return 0;
    str = str.substring(pos);
    return StrDist(str, sub.length());  // calls the first StrDist function
}

Select allOpen in new window

changing variable names did not help due to elbow same error

Compile problems:


Error:      int StrDist(String str, int lensub)
          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Duplicate method StrDist(String, int) in type Shell




see Example Code to help with compile problems

Error:The method strDist(String, String) is undefined

don't know enough of java whether it allows same function name and different arguments.

you better go back to the last solution which passes all tests with one StrDist function taking two String arguments.

Sara

I think you somehow lost your way on this.  Let's go back to what Sara posted above here  at
https://www.experts-exchange.com/questions/28971911/strDist-challenge.html?anchorAnswerId=41816014#a41816014   
The only mistakes were wrongly capitalize the method name and added a semi-colon where it didn't belong. If we correct those mistakes, then it works ok.

public int strDist(String str, String sub) {
  int len = str.length();
  if (str.startsWith(sub)&&str.endsWith(sub) ){
     return len;
  }
  int pos = str.indexOf(sub);
  if (pos == -1)     return 0;
  if (pos > 0){
     return strDist(str.substring(pos), sub);
  }
  return strDist(str.substring(0, len-1), sub);
}

Select allOpen in new window

  My code is  at  
https://www.experts-exchange.com/questions/28971911/strDist-challenge.html?anchorAnswerId=41815304#a41815304

How to make code approach in ID: 41816515 to make it work as it gives below error

int StrDist(String str, int lensub)
{
     int len = str.length();
     if (str.endsWith(str.substring(0, lensub)))
     {
           return len;
     }
     return StrDist(str.substring(0, len-1), lensub);
}

int StrDist(String str2, int lensub2)
{
    int pos = str.indexOf(sub);
    if (pos == -1)    
        return 0;
    str = str.substring(pos);
    return StrDist(str, sub.length());  // calls the first StrDist function
}

Select all
 
Open in new window

changing variable names did not help due to elbow same error

Compile problems:


Error:      int StrDist(String str, int lensub)
          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Duplicate method StrDist(String, int) in type Shell

Select allOpen in new window

Is above approach is fixable?

public int strDist(String str, String sub) {
  int len = str.length();
  if (str.startsWith(sub)&&str.endsWith(sub) ){//if start and end with sub simply returning str lenght //which makes sense
     return len;
  }
  int pos = str.indexOf(sub);
  if (pos == -1)     return 0;//if no sub returning 0 which is ok
  if (pos > 0){//if sub i.e pos is positive number exist then returning recursive function call till reaches //which base case??
     return strDist(str.substring(pos), sub);
  }
  return strDist(str.substring(0, len-1), sub);//when (at what condition  like pos==0??)does control comes here??
}

Select allOpen in new window

How to make code approach in ID: 41816515 to make it work as it gives below error

I don't know. That is  Sara's code.

looks like that is not possible. I tried many ways but could not make it to work

public int strDist(String str, String sub) {
   if (str.indexOf(sub) == -1) return 0;
    if (str.substring(0, sub.length()).equals(sub)
            && str.substring(str.length() - sub.length())
            .equals(sub)){
        return str.length();
            }
    else if(! str.substring(0, sub.length() ).equals(sub) ){
        return strDist(str.substring(1), sub);
    }
  
  else if(! str.substring( str.length()-sub.length() ).equals(sub) ){
        return strDist(str.substring(0,str.length()-1), sub);
}
else return 1000;

}

Select allOpen in new window

how my above approach is different from your below approach?

public int strDist(String str, String sub) {
  int len = str.length();
  if (str.startsWith(sub)&&str.endsWith(sub) ){//if start and end with sub simply returning str lenght //which makes sense
     return len;
  }
  int pos = str.indexOf(sub);
  if (pos == -1)     return 0;//if no sub returning 0 which is ok
  if (pos > 0){//if sub i.e pos is positive number exist then returning recursive function call till reaches //which base case??
     return strDist(str.substring(pos), sub);
  }
  return strDist(str.substring(0, len-1), sub);//when (at what condition  like pos==0??)does control comes here??
}

Select allOpen in new window

how my above approach is different from your below approach?

the second code
- avoids to have the same function call, for example strlen, twice.
- uses functions startsWith and endsWith
- uses the start position of the first occurrence of sub for the next substring to pass.
   this spares some recursive calls if sub is not at position 0 or position 1.
- returns 0 if sub is not contained in str what is well-defined while returning 1000 is not.

Sara

returns 0 if sub is not contained in str what is well-defined

i was not clear on above. can you please elaborate?
Are you referring below line
return strDist(str.substring(0, len-1), sub);

and the zero above?

no, i was referring to

   return 1000;

of your code what is not well defined compared to

   if (pos == -1)     return 0;

what would return length = 0 if sub was not contained into str.

Sara

public int strDist(String str, String sub) {
   if (str.indexOf(sub) == -1) return 0;
    if (str.substring(0, sub.length()).equals(sub)
            && str.substring(str.length() - sub.length())
            .equals(sub)){
        return str.length();
            }
    else if(! str.substring(0, sub.length() ).equals(sub) ){
        return strDist(str.substring(1), sub);
    }
  
 /* else if(! str.substring( str.length()-sub.length() ).equals(sub) ){
        return strDist(str.substring(0,str.length()-1), sub);
}*/
else return strDist(str.substring(0,str.length()-1), sub);

}

Select allOpen in new window

above passes all tests as well.

i moved else if code to else

public int strDist(String str, String sub) {
  int len = str.length();
  if (str.startsWith(sub)&&str.endsWith(sub) ){//if start and end with sub simply returning str lenght //which makes sense
     return len;
  }
  int pos = str.indexOf(sub);
  if (pos == -1)     return 0;//if no sub returning 0 which is ok
  if (pos > 0){//if sub i.e pos is positive number exist then returning recursive  function call till reaches //which base case??
     return strDist(str.substring(pos), sub);
  }
  return strDist(str.substring(0, len-1), sub);//when (at what condition  like pos==0??)does control comes here??
}

Select allOpen in new window

not clear on above code construct


1. if start and end is sub then  return length
2. if -1 ie sub not found return 0
3. if pos > 0 then doing recursion call again passing shorter string starting from index of sub ie pos and sub
4. else we are calling
 return strDist(str.substring(0, len-1), sub); i was not clear in which case it comes to this last else???

last else is my biggest confusion in terms of when the code flow reaches there in which scenario?

in that case why recusion method called by passing str.length()-1 in substring starting from 0??

 return strDist(str.substring(0,str.length()-1), sub);

why recusion method called by passing str.length()-1 in substring starting from 0??

since the left side of str alreaydy starts with sub, we only have to shorten the right end until the remaining string ends with sub. note, that would at least be the case when str == sub. then both begin and end matches to sub and returned length is length of sub.

Sara

since the left side of str alreaydy starts with sub, we only have to shorten the right end until the remaining string ends with sub. note, that would at least be the case when str == sub. then both begin and end matches to sub and returned length is length of sub.

last else is same as below case right?

 else if(! str.substring( str.length()-sub.length() ).equals(sub) ){
        return strDist(str.substring(0,str.length()-1), sub);


you mean like this right


since the left side of str alreaydy not starts with sub, we only have to shorten the right end until the remaining string ends with sub. note, that would at least be the case when str == sub. then both begin and end matches to sub and returned length is length of sub

????

1. condition: if (str.startsWith(sub) && str.endsWith(sub) )

that is the final positive case

2. condition: int pos = str.indexOf(sub);   if (pos == -1 )

this is the case when sub is not contained in str (at all). we return 0 length if so.


3. condition: int pos = str.indexOf(sub); .... if (pos > 0)

sub was contained in str but not at the begin. therefore we call recursively a substring of str where we cut the first 'pos' characters.

4. condition: none of the first 3 conditions has applied yet.

therefore str starts with sub and does NOT end with sub (so far).

Sara

since the left side of str alreaydy starts with sub, we only have to shorten the right end until the remaining string ends with sub. note, that would at least be the case when str == sub. then both begin and end matches to sub and returned length is length of sub.

you mean for below case

strDist("cccatcowcatxx", "cat") → 9
cccatcowcat
so it cuts right to left last xx
then
so above bold becomes 9 length right?

so it cuts right to left last xx

the first cut is on the left side. it cuts 'cc' with the first call since pos == 2.

the right cut happens with two recursive calls, each of them cutting one 'x'.

if there was a 'right-find' function we could spare one of these calls.

Sara