<div>
<pre><code class="code-1 nolinks" id="code-20-40433114-1"><span>  String letter = str.substring(i,i+1) ;</span>
<span>  if(i == 0 || i == str.length()-1 || !letter.equals(&quot;x&quot;))</span>
<span>    sb.append(letter) ;</span>
</code></pre>
<br />
I am not very clear on above two lines. if it first or last or not x then apend to stringbuider sb(is buffer same as string builder..i see you used word buffer in your comment)<br />
<br />
<br />
what happens if other than below case<br />

<blockquote>
if it first or last or not x then apend to stringbuider sb
</blockquote>
I see all tests do pass.<br />
Please advise
</div>


<div>
<blockquote>
You might want sb to be built from the str that is passed in, rather than always using the same &quot;str&quot;<br />
You may also want to make the code work when more than one character should be deleted.
</blockquote>
<br />
can you please advise on this.
</div>


<div>
<blockquote>
(is buffer same as string builder..i see you used word buffer in your comment)
</blockquote>
<br />
Yes StringBuffer and StringBuilder are (essentially) identical. &nbsp;You should generally use StringBuilder. &nbsp;It's a newer implementation and slightly better than StringBuffer, but both would work.<br />
<br />

<blockquote>
what happens if other than below case<br />
if it first or last or not x then apend to stringbuider sb
</blockquote>
<br />
We're looking to build up a new string that contains:<br />
1) The first character<br />
2) The last character<br />
3) Any character that's not an x (between the first and last characters)<br />
<br />
So this test<br />
if(i == 0 || i == str.length()-1 || !letter.equals(&quot;x&quot;))<br />
<br />
does all 3 of those tests in one go.<br />
<br />
If any of them are true, we add the character into the string we'll return.<br />
Otherwise, we don't do anything and we just move on to the next character.<br />
<br />
Does that make sense?<br />
<br />
Doug
</div>


<div>
<blockquote>
We're looking to build up a new string that contains:<br />
1) &nbsp;The first character<br />
2) The last character
</blockquote>
<br />
it should like below right? please advise<br />
<br />

<blockquote>
We're looking to build up a new string that contains:<br />
1) Not The first character<br />
2) Not The last character
</blockquote>
</div>


<div>
<blockquote>
3) Any character that's not an x (between the first and last characters)
</blockquote>
<br />
i was clear on above point. But not below points<br />

<blockquote>
We're looking to build up a new string that contains:<br />
1) &nbsp;The first character<br />
2) The last character
</blockquote>
please advise
</div>


<div>
<blockquote>
an &quot;x&quot; at the very start or end should not be removed. 
</blockquote>
So an &quot;x&quot; at the very start or end should be included.<br />
Also, a non &quot;x&quot; at the very start or end should be included.<br />
So any character at the very start or end should be included.
</div>


<div>
<pre><code class="code-1 nolinks" id="code-20-40444060-1"><span> if(i == 0 || i == str.length()-1 || !letter.equals(&quot;x&quot;))</span>
<span>    sb.append(letter) </span>
</code></pre>
<br />
can we break to few more lines and simplify above code to exactly see where is beolw things happening clearly<br />
<br />

<blockquote>
So an &quot;x&quot; at the very start or end should be included.<br />
Also, a non &quot;x&quot; at the very start or end should be included.<br />
So any character at the very start or end should be included.
</blockquote>
please advise
</div>


<div>
String letter = str.substring(i,i+1) ;<br />
if(i == 0 || i == str.length()-1 || !letter.equals(&quot;x&quot;))<br />
&nbsp; &nbsp; sb.append(letter) <br />
i == 0 → letter at the very start <br />
i == str.length()-1 → letter at the very end<br />
!letter.equals(&quot;x&quot;) → letter a non &quot;x&quot; <br />
These cover all the cases when character should be included
</div>


<div>
<blockquote>
can we break to few more lines and simplify above code to exactly see where is beolw things happening clearly<br />

</blockquote>
Sure - you can always break up an if statement into separate lines if you prefer.<br />
<br />
So this:<br />
<br />

<pre><code class="code-1 nolinks" id="code-20-40445983-1"><span>if(i == 0 || i == str.length()-1 || !letter.equals(&quot;x&quot;))</span>
<span>    sb.append(letter) </span>
</code></pre>
<br />
becomes this:<br />
<br />

<pre><code class="code-1 nolinks" id="code-20-40445983-2"><span>if (i == 0)</span>
<span>  // We're at the start</span>
<span>  sb.append(letter) ;</span>
<span>else if (i == str.length()-1)</span>
<span>  // We're at the end</span>
<span>  sb.append(letter) ;</span>
<span>else if (!letter.equals(&quot;x&quot;))</span>
<span>  // It's not an 'x'</span>
<span>  sb.append(letter) ;</span>
</code></pre>
<br />
It's more code, but it does the same thing.
</div>


<div>
If first and last even though x include.<br />
Where we are giving above instruction
</div>


<div>
<blockquote>
If first and last even though x include.<br />
Where we are giving above instruction<br />

</blockquote>
<br />
The problem statement had this:<br />
&nbsp; &nbsp;&gt;&gt; Except an &quot;x&quot; at the very start or end should not be removed.
</div>


<div>
<blockquote>
If first and last even though x include.<br />
Where we are giving above instruction
</blockquote>
<br />
i mean in the code where we are taking care of it
</div>


<div>
<blockquote>
else if (!letter.equals(&quot;x&quot;))<br />
&nbsp; // It's not an 'x'<br />
&nbsp; sb.append(letter) ;
</blockquote>
<br />
in the above code we are saying if no x(in any positions other than first and last) then add to string builder right?<br />
<br />
<br />

<pre><code class="code-1 nolinks" id="code-20-40447220-1"><span>if(i == 0 || i == str.length()-1 || !letter.equals(&quot;x&quot;))</span>
<span>    sb.append(letter) </span>
</code></pre>
<br />
<br />
do not we should have like below instead of above<br />
if(i == 0 &amp;&amp;&nbsp;i == str.length()-1 &amp;&amp;&nbsp;!letter.equals(&quot;x&quot;))<br />
&nbsp; &nbsp; sb.append(letter)
</div>


<div>
<blockquote>
else if (!letter.equals(&quot;x&quot;))<br />
&nbsp; // It's not an 'x'<br />
&nbsp; sb.append(letter) ;
</blockquote>
<br />
so whatever comes inside above else if is the i when we are not at first and last index right
</div>


<div>
i == 0 &amp;&amp;&nbsp;i == str.length()-1 <br />
can only be true when str.length() is 1
</div>


<div>
<blockquote>
so whatever comes inside above else if is the i when we are not at first and last index right 
</blockquote>
given <br />
if (i == 0)<br />
&nbsp; // We're at the start<br />
&nbsp; sb.append(letter) ;<br />
else if (i == str.length()-1)<br />
&nbsp; // We're at the end<br />
&nbsp; sb.append(letter) ;<br />
else if (!letter.equals(&quot;x&quot;))<br />
&nbsp; // It's not an 'x'<br />
&nbsp; sb.append(letter) ;<br />
<br />
the cases of the i when we are at first and last index are handled in the first two else statements, so if we get to the last else, &nbsp;we are not at first and last index
</div>


<div>
<blockquote>
i == 0 &amp;&amp;&nbsp;i == str.length()-1 <br />
can only be true when str.length() is 1
</blockquote>
our solution we are not using &amp;&amp;&nbsp;but using || as below<br />
I wonder what is relevance of &amp;&amp;&nbsp;for this solution.<br />

<blockquote>
if(i == 0 || i == str.length()-1 || !letter.equals(&quot;x&quot;))
</blockquote>
</div>


<div>
For this problem you want an &quot;or&quot; because we add the current character to the result if <b>any</b> of the conditions are true.<br />
You want &quot;and&quot; when <b>all</b> of the conditions need to be true for the character to be added.<br />
<br />
Doug
</div>


<div>
<pre><code class="code-1 nolinks" id="code-20-40453144-1"><span>if (i == 0)</span>
<span>  // We're at the start</span>
<span>  sb.append(letter) ;</span>
<span>else if (i == str.length()-1)</span>
<span>  // We're at the end</span>
<span>  sb.append(letter) ;</span>
<span>else if (!letter.equals(&quot;x&quot;))</span>
<span>  // It's not an 'x'</span>
<span>  sb.append(letter) ;</span>
</code></pre>
Above if -else if-else if construct resembles to me like 'and' not 'or'. Please advise with some simple example
</div>


<div>
The above if -else if-else if construct does sb.append(letter) &nbsp;when any of the conditions are true.<br />
The || construct also does sb.append(letter) &nbsp;when any of the conditions are true.<br />
<br />
<i>i == 0 || i == str.length()-1</i> is true when either <i>i == 0</i> is true or <i>i == str.length()-1</i> is true.<br />
<i>i == 0 &amp;&amp;&nbsp;i == str.length()-1</i> is only true when both <i>i == 0</i> is true and <i>i == str.length()-1</i> is true.<br />
(which can only happen when str.length()==1)
</div>


<div>
<blockquote>
The above if -else if-else if construct does sb.append(letter) &nbsp;when any of the conditions are true.<br />
The || construct also does sb.append(letter) &nbsp;when any of the conditions are true.
</blockquote>
now i know how to visualize those if -else if-else constructs, basically inside out, as inside common sb.append(letter) &nbsp;appends when any one of conditions are met(not all)
</div>


<div>
similar to how we break <br />
if(i == 0 || i == str.length()-1 || !letter.equals(&quot;x&quot;)<br />
<br />
as simple steps of &nbsp;if -else if-else if construct as below<br />

<pre><code class="code-1 nolinks" id="code-20-40455475-1"><span>if (i == 0)</span>
<span>  // We're at the start</span>
<span>  sb.append(letter) ;</span>
<span>else if (i == str.length()-1)</span>
<span>  // We're at the end</span>
<span>  sb.append(letter) ;</span>
<span>else if (!letter.equals(&quot;x&quot;))</span>
<span>  // It's not an 'x'</span>
<span>  sb.append(letter) ;</span>
</code></pre>
<br />
how to break into simple steps?<br />
if(i == 0 &amp;&amp;&nbsp;i == str.length()-1 &amp;&amp;&nbsp;!letter.equals(&quot;x&quot;)
</div>


<div>
if( i == 0 ){<br />
&nbsp; &nbsp; &nbsp;if( i == str.length()-1 ){<br />
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if( !letter.equals(&quot;x&quot;) ){<br />
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;sb.append(letter) ;<br />
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }<br />
&nbsp; &nbsp; &nbsp;}<br />
}
</div>


<div>
<div class="content wysiwyg-content">
Hi,<br />
<br />
I am trying below example<br />
<a href="http://codingbat.com/prob/p171260" rel="ugc">http://codingbat.com/prob/p171260</a><br />
<br />
I wrote my code as below<br />

<pre><code class="code-10 nolinks" id="code-10-28553993-1"><span>public String stringX(String str) {</span>
<span>StringBuilder sb=new StringBuilder(&quot;str&quot;);</span>
<span>for(int i=0;i&lt;str.length()-1;i++)</span>
<span>{</span>
<span>if(str.substring(i,i+1).equals(&quot;x&quot;))</span>
<span>sb.delete(i,i+1);</span>
<span>return sb.toString();</span>
<span>}</span>
<span>return sb.toString();</span>
<span>}</span>
</code></pre>
<br />
Please advise on how to fix and improve the my code. Thanks in advance
</div>
</div>


Hi,

I am trying below example
http://codingbat.com/prob/p171260

I wrote my code as below
(CODE)

Please advise on how to fix and improve the my code. Thanks in advance

removing x 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

A programming language is a formal constructed language designed to communicate instructions to a machine, particularly a computer. Thousands of different programming languages have been created, mainly in the computer field, and many more still are being created every year. The description of a programming language is usually split into the two components of syntax (form) and semantics (meaning). Some languages are defined by a specification document (for example, the C programming language is specified by an ISO Standard), while other languages (such as Perl) have a dominant implementation that is treated as a reference. Some languages have both, with the basic language defined by a standard and extensions taken from the dominant implementation being common.