trjhtr

I recently did a little task for fun that required me to create a class to determine if a word is a palindrome or not. I know this is quite simple, but I'd be interested to know if a real developer (I'm not a programmer) would approach the task in the same way, or if I make any amateur mistakes/do anything stupid.

My code passed the automated tests on the website, but I understand those aren't perfect.

class Palindrome

 public static function isPalindrome($word)
 
 $word = strtolower($word);
 $wordLength = strlen($word);
 $wordSplitPoint = ceil($wordLength / 2);

 if ($wordLength % 2 == 0) 
 $firstHalf = substr($word, 0, $wordSplitPoint);
 $secondHalf = substr($word, $wordSplitPoint, $wordSplitPoint);
 else 
 $firstHalf = substr($word, 0, $wordSplitPoint-1);
 $secondHalf = substr($word, $wordSplitPoint, $wordSplitPoint);
 

 $secondHalfReversed = strrev($secondHalf);

 if ($firstHalf == $secondHalfReversed) 
 return TRUE;
 else 
 return FALSE;
 
 


echo Palindrome::isPalindrome('Deleveled');

Others have pointed out a fix you could make to your implementation. But there is a way to write this which is much simpler. That takes advantage of a simplified definition of palindrome: "a word which reads the same forwards and backwards".

Now I haven't done PHP in years, and don't have a test environment, so this might have issues, but the basic idea should be clear.

public static function isPalindrome($word)

 return $word == strrev($word);

Update

Based on suggestions in the comments, this would be nicer:

function isPalindrome($word) 
 $word = preg_replace('/[^a-zA-Z]/', '', $word);
 $word = strtolower($word);
 return $word == strrev($word);

That will now correctly identify "Level" (ignoring capitalization) and "Madam, I'm Adam." (ignoring spaces and punctuation.)

To start with id remove the code that you written twice by re-arranging the if statement into this

if ($wordLength % 2 == 0) 
 $firstHalf = substr($word, 0, $wordSplitPoint);
 else 
 $firstHalf = substr($word, 0, $wordSplitPoint-1);


$secondHalf = substr($word, $wordSplitPoint, $wordSplitPoint);

Then for the return value I would use;

return $firstHalf == $secondHalfReversed;

This is about all i would change

This is a very good program. There is only one thing I'd change:

if ($firstHalf == $secondHalfReversed) 
 return TRUE;
 else 
 return FALSE;

You don't really need that if because the top part only returns TRUE when the condition also is TRUE. Otherwise, both the condition and the return is FALSE. Change it to:

return $firstHalf == $secondHalfReversed

I'd write this using a traditional for loop as follows:

class Palindrome

 public static function isPalindrome($word)
 
 $wordLength = strlen($word)-1;

 for ($i = 0; $i < $wordLength/2; $i++) 
 if (strtolower($word[$i]) != strtolower($word[$wordLength-$i])) 
 return FALSE;
 
 

 return TRUE;
 


echo Palindrome::isPalindrome('Deleveled');

echo Palindrome::isPalindrome('Deleveled');

You need to test corner cases and try to break your code. For example, you might try with:

• a string of odd size


• a string of even size


• the empty string


• a string with one character

How is size defined anyway? What about encodings? For example, your code works for '101', but not 'é_é'?