Encode/compress sequence of repeating integers

Question

I have very long integer sequences that look like this (arbitrary length!):

0000000001110002220033333

Now I need some algorithm to convert

Answer 1

Here is a naive implementation of what you want.

$toEncode = '0000000001110002220033333';
$currentChar = '-1';
$length = strlen($toEncode);
$encoded = '';
$currentNbrChar = 0;
for($i = 0; $i < $length; $i++){
  if($toEncode[$i] != $currentChar){
    if($currentChar != '-1'){
      $encoded .= chr(97 + $currentChar).$currentNbrChar;
    }
    $currentNbrChar = 0;
    $currentChar = $toEncode[$i];
  }
  $currentNbrChar ++;
}
if($currentChar != '-1'){
  $encoded .= chr(97 + $currentChar).$currentNbrChar;
}
echo $encoded;

Answer 2

Here's a shorter version:

function smush(str) {
  return str.replace(/((.)\2*)/g, function(_, w, x) {
    return x + w.length;
  });
}

edit oh I see you want to encode with php; sorry I don't know that. Here's a decoder in a similar spirit:

function unsmush(str) {
  return str.replace(/(.)(\d+)/g, function(_, x, n) {
    return new Array(parseInt(n, 10) + 1).join(x);
  });
}

Answer 3

Just FYI, you could probably gzip your data and the browse will automatically unzip it. For most implementations this is going to work better than RLE. But less fun obviously.

Answer 4

It is called Run Length Encoding

Basic encoder in PHP:

function numStringToRle($s){
    $rle = '';
    $count = 1;
    $len = strlen($s);
    for ( $i = 0; $i < $len; $i++ ){
        if ( $i != $len && $s[$i] == $s[$i+1] ){
            $count++;                
        }else{
          $rle .= chr($s[$i] + 97).$count;    
          $count = 1;
        }
    }
    return $rle;
}

Be warned it will preform badly issues with a string like

 123456789123456789

If you were going to be handling a string that may have a lot of individual single characters you would be better to add some complexity and not write the length of the run if the length of the run is 1.

//change
$rle .= chr($s[$i] + 97).$count;    

//to
$rle .= chr($s[$i] + 97).( $count == 1 ? '' : $count );   

//or
$rle .= chr($s[$i] + 97)
if ( $count != 1 ){
    $rle .= $count;
}

Answer 5

$str="0000000001110002220033333";

//$c will count the number of occurances.

$c=1;

$lastInt=substr($str,0,1);

$str=substr($str,1);

$resultStr='';

$loopEnd=strlen($str);


for($i=1; $i<=$loopEnd+1;$i++)

{

    $nowInt=substr($str,0,1);   
    if($lastInt==$nowInt)
    {
        $c++;
        $str=substr($str,1);
    }
    else
    {
        $char=chr((int)$lastInt + 97);
        $resultStr=$resultStr.$char.$c;
        $str=substr($str,1);
        $c=1;
        $lastInt=$nowInt;
    }
}

// we use if condition since for loop will not take the last integer if it repeats.

if($c>1)
{

$char=chr((int)$lastInt + 97);

$resultStr=$resultStr.$char.$c;

}

echo $resultStr;

Answer 6

function compress( $str) {
$strArr = str_split($str.'0');
$count = 0;
$resStr = '';
$strCheck = $strArr[0];
foreach($strArr as $key => $value)
{
    if($strCheck == $value)
    {
       $count++;
    } 
    else
    {
        if($count == 1)
        {
            $strCheck = $value;
            $resStr .= $strArr[$key-1];
            $count=1;
        }
        elseif($count == 2)
        {
            $strCheck = $value;
            $resStr .= $strArr[$key-1].$strArr[$key-1];
            $count=1;
        }
        else
        {
            $strCheck = $value;
            $resStr .= $strArr[$key-1].$count;
            $count=1;
        }
    } 

} 
return $resStr;

}