<div>
<pre><code class="code-10 nolinks" id="code-20-41813369-1">public int strDist(String str, String sub) {
 if (str.indexOf(sub) == -1) return 0;
 if (str.substring(0, sub.length()).equals(sub)
 &amp;&amp; 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);
 }
 

}
</code></pre>
 
above gives below error 
 

<blockquote>
Compile problems: 
 
 
Error: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;else (! str.substring(0, sub.length() ).equals(sub) ){ 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;^ 
Syntax error, insert &quot;AssignmentOperator ArrayInitializer&quot; to complete ArrayInitializerAssignement 
 
 
see Example Code to help with compile problems
</blockquote>
 
please advise
</div>

<div>
<pre><code class="code-10 nolinks" id="code-20-41815256-1">public int strDist(String str, String sub) {
 if (str.indexOf(sub) == -1) return 0;
 if (str.substring(0, sub.length()).equals(sub)
 &amp;&amp; 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;
}
</code></pre>
 
i had condition in else which is wrong 
 
above gives below errors 

<blockquote>
 
 
Expected &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Run &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;catcowcat&quot;, &quot;cat&quot;) → 9 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;9 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;OK &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;catcowcat&quot;, &quot;cow&quot;) → 3 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1000 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;X &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;cccatcowcatxx&quot;, &quot;cat&quot;) → 9 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1000 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;X &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;abccatcowcatcatxyz&quot;, &quot;cat&quot;) → 12 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1000 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;X &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;xyx&quot;, &quot;x&quot;) → 3 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;3 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;OK &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;xyx&quot;, &quot;y&quot;) → 1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1000 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;X &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;xyx&quot;, &quot;z&quot;) → 0 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;OK &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;z&quot;, &quot;z&quot;) → 1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;OK &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;x&quot;, &quot;z&quot;) → 0 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;OK &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;&quot;, &quot;z&quot;) → 0 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;OK &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;hiHellohihihi&quot;, &quot;hi&quot;) → 13 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;13 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;OK &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;hiHellohihihi&quot;, &quot;hih&quot;) → 5 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1000 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;X &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;hiHellohihihi&quot;, &quot;o&quot;) → 1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1000 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;X &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
strDist(&quot;hiHellohihihi&quot;, &quot;ll&quot;) → 2 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;1000 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;X &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
other tests 
X &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
</blockquote>
</div>

<div>
<pre><code class="code-10 nolinks" id="code-20-41815272-1">public int strDist(String str, String sub) {
 if (str.indexOf(sub) == -1) return 0;
 if (str.substring(0, sub.length()).equals(sub)
 &amp;&amp; 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;

}
</code></pre>
 
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
</div>

<div>
<blockquote>
any improvements or alternate approaches.
</blockquote>
to avoid repeating statements you could do: 
 

<pre><code class="code-10 nolinks" id="code-20-41816014-1">int len = str.length();
if (str.startsWith(sub)&amp;&amp;str.endsWith(sub) );
{
 return len;
}
int pos = str.indexOf(sub);
if (pos == -1) 
 return 0;
if (pos &gt; 0)
{
 return StrDist(str.substring(pos), sub);
}

return StrDist(str.substring(0, len-1), sub);
</code></pre>
 
note, you should return 0 if the string doesn't contain the sub string. 
 
Sara
</div>

<div>
i like above approach with pos and also starts and endwith 
 
what would be pos value &nbsp;for last return. when does it goes into below statement 
 

<pre><code class="code-1 nolinks" id="code-20-41816259-1">return StrDist(str.substring(0, len-1), sub);
</code></pre>
 
is it pos&lt;0 but -1 is already covered above?
</div>

<div>
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 &lt;&nbsp;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: 
 

<pre><code class="code-10 nolinks" id="code-20-41816515-1">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
}
</code></pre>
 
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
</div>

<div>
<pre><code class="code-10 nolinks" id="code-20-41817986-1">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
}
</code></pre>
 
above gives below error 
 
Compile problems: 
 
 
Error: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;return StrDist(str.substring(0, len-1), lensub) 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;^ 
Syntax error, insert &quot;;&quot; to complete ReturnStatement 
 
 
see Example Code to help with compile problems 
 
 
please advise
</div>

<div>
<blockquote>
Syntax error, insert &quot;;&quot; to complete ReturnStatement
</blockquote>
 
the compiler told you what to do. 
 
Sara
</div>

<div>
<pre><code class="code-10 nolinks" id="code-20-41818316-1">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
}
</code></pre>
i fixed that and then i got below error 
Compile problems: 
 
 
Error: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;str = str,substring(pos); 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ^ 
Syntax error on token &quot;,&quot;, ; expected 
 
 
see Example Code to help with compile problems 
 
now 

<pre><code class="code-10 nolinks" id="code-20-41818316-2">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
}
</code></pre>
 
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
</div>

<div>
<blockquote>
str = str,substring(pos);
</blockquote>
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
</div>

<div>
yes I corrected comma with . error. 
 
now I got new error like method not defined 

<pre><code class="code-10 nolinks" id="code-20-41818381-1">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
}
</code></pre>
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)
</div>

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

<pre><code class="code-10 nolinks" id="code-20-41818386-1">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
}
</code></pre>
 
saying duplicate method 
 
 
Error: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;int StrDist(String str, int lensub) 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 
Duplicate method StrDist(String, int) in type Shell
</div>

<div>
<pre><code class="code-10 nolinks" id="code-20-41818392-1">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
}
</code></pre>
 
changing variable names did not help due to elbow same error 
 
Compile problems: 
 
 
Error: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;int StrDist(String str, int lensub) 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 
Duplicate method StrDist(String, int) in type Shell 
 
 
 
 
see Example Code to help with compile problems
</div>

<div>
<blockquote>
Error:The method strDist(String, String) is undefined 
</blockquote>
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
</div>

<div>
I think you somehow lost your way on this. &nbsp;Let's go back to what Sara posted above here &nbsp;at 
<a href="https://www.experts-exchange.com/questions/28971911/strDist-challenge.html?anchorAnswerId=41816014#a41816014" rel="ugc">https://www.experts-exchange.com/questions/28971911/strDist-challenge.html?anchorAnswerId=41816014#a41816014</a>&nbsp; &nbsp; 
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. 
<pre><code class="code-10 nolinks" id="code-20-41818969-1">public int strDist(String str, String sub) {
 int len = str.length();
 if (str.startsWith(sub)&amp;&amp;str.endsWith(sub) ){
 return len;
 }
 int pos = str.indexOf(sub);
 if (pos == -1) return 0;
 if (pos &gt; 0){
 return strDist(str.substring(pos), sub);
 }
 return strDist(str.substring(0, len-1), sub);
}
</code></pre>
 &nbsp; My code is &nbsp;at &nbsp; 
<a href="https://www.experts-exchange.com/questions/28971911/strDist-challenge.html?anchorAnswerId=41815304#a41815304" rel="ugc">https://www.experts-exchange.com/questions/28971911/strDist-challenge.html?anchorAnswerId=41815304#a41815304</a>
</div>

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

<pre><code class="code-10 nolinks" id="code-20-41819036-1">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
</code></pre>
 
Is above approach is fixable?
</div>

<div>
<pre><code class="code-10 nolinks" id="code-20-41819037-1">public int strDist(String str, String sub) {
 int len = str.length();
 if (str.startsWith(sub)&amp;&amp;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 &gt; 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??
}
</code></pre>
</div>

<div>
<blockquote>
How to make code approach in ID: 41816515 to make it work as it gives below error
</blockquote>
 I don't know. That is &nbsp;Sara's code.
</div>

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

<div>
<pre><code class="code-10 nolinks" id="code-20-41819048-1">public int strDist(String str, String sub) {
 if (str.indexOf(sub) == -1) return 0;
 if (str.substring(0, sub.length()).equals(sub)
 &amp;&amp; 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;

}
</code></pre>
 
how my above approach is different from your below approach? 

<pre><code class="code-10 nolinks" id="code-20-41819048-2">public int strDist(String str, String sub) {
 int len = str.length();
 if (str.startsWith(sub)&amp;&amp;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 &gt; 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??
}
</code></pre>
</div>

<div>
<blockquote>
how my above approach is different from your below approach?
</blockquote>
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. 
&nbsp; &nbsp;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
</div>

<div>
<blockquote>
returns 0 if sub is not contained in str what is well-defined
</blockquote>
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?
</div>

<div>
no, i was referring to 
 
&nbsp; &nbsp;return 1000; 
 
of your code what is not well defined compared to 
 
&nbsp; &nbsp;if (pos == -1) &nbsp; &nbsp; return 0; 
 
what would return length = 0 if sub was not contained into str. 
 
Sara
</div>

<div>
<pre><code class="code-10 nolinks" id="code-20-41821818-1">public int strDist(String str, String sub) {
 if (str.indexOf(sub) == -1) return 0;
 if (str.substring(0, sub.length()).equals(sub)
 &amp;&amp; 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);

}
</code></pre>
 
above passes all tests as well. 
 
i moved else if code to else
</div>

<div>
<pre><code class="code-10 nolinks" id="code-20-41821878-1">public int strDist(String str, String sub) {
 int len = str.length();
 if (str.startsWith(sub)&amp;&amp;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 &gt; 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??
}
</code></pre>
 
not clear on above code construct 
 
 
1. if start and end is sub then &nbsp;return length 
2. if -1 ie sub not found return 0 
3. if pos &gt;&nbsp;0 then doing recursion call again passing shorter string starting from index of sub ie pos and sub 
4. else we are calling 
&nbsp;return strDist(str.substring(0, len-1), sub); i was not clear in which case it comes to this last else???
</div>

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

<div>
in that case why recusion method called by passing str.length()-1 in substring starting from 0?? 
 
&nbsp;return strDist(str.substring(0,str.length()-1), sub);
</div>

<div>
<blockquote>
why recusion method called by passing str.length()-1 in substring starting from 0??
</blockquote>
 
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
</div>

<div>
<blockquote>
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.
</blockquote>
 
last else is same as below case right? 
 
&nbsp;else if(! str.substring( str.length()-sub.length() ).equals(sub) ){ 
&nbsp; &nbsp; &nbsp; &nbsp; 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
</div>

<div>
???? 
 
1. condition: if (str.startsWith(sub) &amp;&amp;&nbsp;str.endsWith(sub) ) 
 
that is the final positive case 
 
2. condition: int pos = str.indexOf(sub); &nbsp; 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 &gt;&nbsp;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
</div>

<div>
<blockquote>
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.
</blockquote>
 
you mean for below case 
 
strDist(&quot;cccatcowcatxx&quot;, &quot;cat&quot;) → 9 
cccatcowcat 
so it cuts right to left last xx 
then 
so above bold becomes 9 length right?
</div>

<div>
<blockquote>
so it cuts right to left last xx
</blockquote>
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
</div>

<div>
<div class="content wysiwyg-content">
Hi, 
 
I am working on below challenge. 
<a href="http://codingbat.com/prob/p195413" rel="ugc">http://codingbat.com/prob/p195413</a> 
Psedo code: 
1. find length of str starting and ending with sub. 
2. return the length 
 
I wrote my code as below and not passing all tests 

<pre><code class="code-10 nolinks" id="code-10-28971911-1">public int strDist(String str, String sub) {
 if (str.length() &lt;2){ 
	 	return 0;
	 		}
	 if (str.startsWith( sub) {
	 	return 1 + strDist(str.substring(2));
	 	}
	 	else {
		 return strDist(str.substring(1));
	 	 
	 	}
}
</code></pre>
i am getting below error 
 
Compile problems: 
 
 
Error: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;return 1 + strDist(str.substring(2)); 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ^^^^^^^ 
The method strDist(String, String) in the type Shell is not applicable for the arguments (String) 
 
 
see Example Code to help with compile problems 
Any improvements or alternate approaches? &nbsp; &nbsp; &nbsp; 
 
please advise
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Hi,

I am working on below challenge.
http://codingbat.com/prob/p195413
Psedo code:
1. find length of str starting and ending with sub.
2. return the length

I wrote my code as below and not passing all tests
(CODE)
i am getting below error

Compile problems:


Error:	return 1 + strDist(str.substring(2));
	           ^^^^^^^
The method strDist(String, String) in the type Shell is not applicable for the arguments (String)


see Example Code to help with compile problems
Any improvements or alternate approaches?      

please advise

strDist challenge

Java is a platform-independent, object-oriented programming language and run-time environment, designed to have as few implementation dependencies as possible such that developers can write one set of code across all platforms using libraries. Most devices will not run Java natively, and require a run-time component to be installed in order to execute a Java program.

Java

Java Enterprise Edition (Java EE) is a specification defining a collection of Java-based server and client technologies and how they interoperate. Java EE specifies server and client architectures and uses profiles to define technology sets targeted at specific classes of applications. All Java EE profiles share a set of common features, such as naming and resource injection, packaging rules and security requirements.

Java EE

Programming includes both the specifics of the language you’re using, like Visual Basic, .NET, Java and others, but also the best practices in user experience and interfaces and the management of projects, version control and development. Other programming topics are related to web and cloud development and system and hardware programming.