Function To Create Regex Matching a Number Range

Question

I am working with the Amazon Mechanical Turk API and it will only allow me to use regular expressions to filter a field of data.

I would like to input an integer range to a function, such as 256-311 or 45-1233, and return a regex that would match only that range.

A regex matching 256-321 would be:

\b((25[6-9])|(2[6-9][0-9])|(3[0-1][0-9])|(32[0-1]))\b

That part is fairly easy, but I am having trouble with the loop to create this regex.

I am trying to build a function defined like this:

function getRangeRegex( int fromInt, int toInt)
{

      return regexString;
}

I looked all over the web and I am surprised that it doesn't look like anyone has solved this in the past. It is a difficult problem...

Thanks for your time.

Answer 1

Here's a quick hack:

<?php

function regex_range($from, $to) {

  if($from < 0 || $to < 0) {
    throw new Exception("Negative values not supported"); 
  }

  if($from > $to) {
    throw new Exception("Invalid range $from..$to, from > to"); 
  }

  $ranges = array($from);
  $increment = 1;
  $next = $from;
  $higher = true;

  while(true) {

    $next += $increment;

    if($next + $increment > $to) {
      if($next <= $to) {
        $ranges[] = $next;
      }
      $increment /= 10;
      $higher = false;
    }
    else if($next % ($increment*10) === 0) {
      $ranges[] = $next;
      $increment = $higher ? $increment*10 : $increment/10;
    }

    if(!$higher && $increment < 10) {
      break;
    }
  }

  $ranges[] = $to + 1;

  $regex = '/^(?:';

  for($i = 0; $i < sizeof($ranges) - 1; $i++) {
    $str_from = (string)($ranges[$i]);
    $str_to = (string)($ranges[$i + 1] - 1);

    for($j = 0; $j < strlen($str_from); $j++) {
      if($str_from[$j] == $str_to[$j]) {
        $regex .= $str_from[$j];
      }
      else {
        $regex .= "[" . $str_from[$j] . "-" . $str_to[$j] . "]";
      }
    }
    $regex .= "|";
  }

  return substr($regex, 0, strlen($regex)-1) . ')$/';
}

function test($from, $to) {
  try {
    printf("%-10s %s\n", $from . '-' . $to, regex_range($from, $to));
  } catch (Exception $e) {
    echo $e->getMessage() . "\n";
  }
}

test(2, 8);
test(5, 35);
test(5, 100);
test(12, 1234);
test(123, 123);
test(256, 321);
test(256, 257);
test(180, 195);
test(2,1);
test(-2,4);

?>

which produces:

2-8        /^(?:[2-7]|8)$/
5-35       /^(?:[5-9]|[1-2][0-9]|3[0-5])$/
5-100      /^(?:[5-9]|[1-9][0-9]|100)$/
12-1234    /^(?:1[2-9]|[2-9][0-9]|[1-9][0-9][0-9]|1[0-2][0-3][0-4])$/
123-123    /^(?:123)$/
256-321    /^(?:25[6-9]|2[6-9][0-9]|3[0-2][0-1])$/
256-257    /^(?:256|257)$/
180-195    /^(?:18[0-9]|19[0-5])$/
Invalid range 2..1, from > to
Negative values not supported

Not properly tested, use at your own risk!

And yes, the generated regex could be written more compact in many cases, but I leave that as an exercise for the reader :)

Answer 2

For anyone else who, like me, was looking for the javascript version of the great @Bart Kiers's production above

//Credit: Bart Kiers 2011
function regex_range(from, to){
        if(from < 0 || to < 0) {
            //throw new Exception("Negative values not supported"); 
            return null;
        }
        if(from > to) {
            //throw new Exception("Invalid range from..to, from > to"); 
            return null;
        }

        var ranges = [];
        ranges.push(from);
        var increment = 1;
        var next = from;
        var higher = true;

        while(true){
            next += increment;
            if(next + increment > to) {
                if(next <= to) {
                    ranges.push(next);
                }
                increment /= 10;
                higher = false;
            }else{ 
                if(next % (increment*10) == 0) {
                    ranges.push(next);
                    increment = higher ? increment*10 : increment/10;
                }
            }

            if(!higher && increment < 10) {
                break;
            }
        }

        ranges.push(to + 1);
        var regex = '/^(?:';

        for(var i = 0; i < ranges.length - 1; i++) {
            var str_from = ranges[i];
            str_from = str_from.toString();
            var str_to = ranges[i + 1] - 1;
            str_to = str_to.toString();
            for(var j = 0; j < str_from.length; j++) {
                if(str_from[j] == str_to[j]) {
                    regex += str_from[j];
                }
                else {
                    regex += "[" + str_from[j] + "-" + str_to[j] + "]";
                }
            }
            regex += "|";
        }

        return regex.substr(0, regex.length - 1 ) + ')$/';
    }

Answer 3

PHP Port of RegexNumericRangeGenerator

class RegexRangeNumberGenerator {

    static function parse($min, $max, $MatchWholeWord = FALSE, $MatchWholeLine = FALSE, $MatchLeadingZero = FALSE) {
        if (!is_int($min) || !is_int($max) || $min > $max || $min < 0 || $max < 0) {
            return FALSE;
        }
        if ($min == $max) {
            return self::parseIntoPattern($min, $MatchWholeWord, $MatchWholeLine, $MatchLeadingZero);
        }
        $s = [];
        $x = self::parseStartRange($min, $max);
        foreach ($x as $o) {
            $s[] = self::parseEndRange($o[0], $o[1]);
        }
        $n = self::reformatArray($s);
        $h = self::parseIntoRegex($n);
        return self::parseIntoPattern($h, $MatchWholeWord, $MatchWholeLine, $MatchLeadingZero);
    }

    static private function parseIntoPattern($t, $MatchWholeWord = FALSE, $MatchWholeLine = FALSE, $MatchLeadingZero = FALSE) {
        $r = ((is_array($t)) ? implode("|", $t) : $t);
        return (($MatchWholeLine && $MatchLeadingZero) ? "^0*(" . $r . ")$" : (($MatchLeadingZero) ? "0*(" . $r . ")" : (($MatchWholeLine) ? "^(" . $r . ")$" : (($MatchWholeWord) ? "\\b(" . $r . ")\\b" : "(" . $r . ")"))));
    }

    static private function parseIntoRegex($t) {
        if (!is_array($t)) {
            throw new Exception("Argument needs to be an array!");
        }
        $r = [];
        for ($i = 0; $i < count($t); $i++) {
            $e = str_split($t[$i][0]);
            $n = str_split($t[$i][1]);
            $s = "";
            $o = 0;
            $h = "";
            for ($a = 0; $a < count($e); $a++) {
                if ($e[$a] === $n[$a]) {
                    $h .= $e[$a];
                } else {
                    if ((intval($e[$a]) + 1) === intval($n[$a])) {
                        $h .= "[" . $e[$a] . $n[$a] . "]";
                    } else {
                        if ($s === ($e[$a] . $n[$a])) {
                            $o++;
                        }
                        $s = $e[$a] . $n[$a];
                        if ($a == (count($e) - 1)) {
                            $h .= (($o > 0) ? "{" . ($o + 1) . "}" : "[" . $e[$a] . "-" . $n[$a] . "]");
                        } else {
                            if ($o === 0) {
                                $h .= "[" . $e[$a] . "-" . $n[$a] . "]";
                            }
                        }
                    }
                }
            }
            $r[] = $h;
        }
        return $r;
    }

    static private function reformatArray($t) {
        $arrReturn = [];
        for ($i = 0; $i < count($t); $i++) {
            $page = count($t[$i]) / 2;
            for ($a = 0; $a < $page; $a++) {
                $arrReturn[] = array_slice($t[$i], (2 * $a), 2);
            }
        }
        return $arrReturn;
    }

    static private function parseStartRange($t, $r) {
        if (strlen($t) === strlen($r)) {
            return [[$t, $r]];
        }
        $break = pow(10, strlen($t)) - 1;
        return array_merge([[$t, $break]], self::parseStartRange($break + 1, $r));
    }

    static private function parseEndRange($t, $r) {
        if (strlen($t) == 1) {
            return [$t, $r];
        }
        if (str_repeat("0", strlen($t)) === "0" . substr($t, 1)) {
            if (str_repeat("0", strlen($r)) == "9" . substr($r, 1)) {
                return [$t, $r];
            }
            if ((int) substr($t, 0, 1) < (int) substr($r, 0, 1)) {
                $e = intval(substr($r, 0, 1) . str_repeat("0", strlen($r) - 1)) - 1;
                return array_merge([$t, self::strBreakPoint($e)], self::parseEndRange(self::strBreakPoint($e + 1), $r));
            }
        }
        if (str_repeat("9", strlen($r)) === "9" . substr($r, 1) && (int) substr($t, 0, 1) < (int) substr($r, 0, 1)) {
            $e = intval(intval((int) substr($t, 0, 1) + 1) . "" . str_repeat("0", strlen($r) - 1)) - 1;
            return array_merge(self::parseEndRange($t, self::strBreakPoint($e)), [self::strBreakPoint($e + 1), $r]);
        }
        if ((int) substr($t, 0, 1) < (int) substr($r, 0, 1)) {
            $e = intval(intval((int) substr($t, 0, 1) + 1) . "" . str_repeat("0", strlen($r) - 1)) - 1;
            return array_merge(self::parseEndRange($t, self::strBreakPoint($e)), self::parseEndRange(self::strBreakPoint($e + 1), $r));
        }
        $a = (int) substr($t, 0, 1);
        $o = self::parseEndRange(substr($t, 1), substr($r, 1));
        $h = [];
        for ($u = 0; $u < count($o); $u++) {
            $h[] = ($a . $o[$u]);
        }
        return $h;
    }

    static private function strBreakPoint($t) {
        return str_pad($t, strlen(($t + 1)), "0", STR_PAD_LEFT);
    }
}

Test Results

2-8         ^([2-8])$
5-35        ^([5-9]|[12][0-9]|3[0-5])$
5-100       ^([5-9]|[1-8][0-9]|9[0-9]|100)$
12-1234     ^(1[2-9]|[2-9][0-9]|[1-8][0-9]{2}|9[0-8][0-9]|99[0-9]|1[01][0-9]{2}|12[0-2][0-9]|123[0-4])$
123-123     ^(123)$
256-321     ^(25[6-9]|2[6-9][0-9]|3[01][0-9]|32[01])$
256-257     ^(25[67])$
180-195     ^(18[0-9]|19[0-5])$

Answer 4

Be careful, the excelent @Bart Kiers's code (and JS version of Travis J) in some cases it fails. For example:

12-1234    /^(?:1[2-9]|[2-9][0-9]|[1-9][0-9][0-9]|1[0-2][0-3][0-4])$/

does not match "1229", "1115", "1[0-2][0-2][5-9]"

Answer 5

Is there a reason it has to be regex? can not do some thing like this:

if ($number >= 256 && $number <= 321){
   // do something 
}

Update:

There is an easy but ugly way to do it using range:

function getRangeRegex($from, $to)
{
    $range = implode('|', range($from, $to));

    // returns: 256|257|...|321
    return $range;
}

Answer 6

That actually has been done already.

Have a look at this site. It contains a link to a python script that generates these regex's for you automagically.

Answer 7

This answer is duplicated from this question. I've also made it into a blog post

---

Using regular expressions to validate a numeric range

To be clear: When a simple if statement will suffice

if(num < -2055  ||  num > 2055)  {
   throw  new IllegalArgumentException("num (" + num + ") must be between -2055 and 2055");
}

using regular expressions for validating numeric ranges is not recommended.

In addition, since regular expressions analyze strings, numbers must first be translated to a string before they can be tested (an exception is when the number happens to already be a string, such as when getting user input from the console).

(To ensure the string is a number to begin with, you could use org.apache.commons.lang3.math.NumberUtils#isNumber(s))

Despite this, figuring out how to validate number ranges with regular expressions is interesting and instructive.

A one number range

Rule: A number must be exactly 15.

The simplest range there is. A regex to match this is

\b15\b

Word boundaries are necessary to avoid matching the 15 inside of 8215242.

A two number range

The rule: The number must be between 15 and 16. Three possible regexes:

\b(15|16)\b
\b1(5|6)\b
\b1[5-6]\b

A number range "mirrored" around zero

The rule: The number must be between -12 and 12.

Here is a regex for 0 through 12, positive-only:

\b(\d|1[0-2])\b

Free-spaced:

\b(         //The beginning of a word (or number), followed by either
   \d       //   Any digit 0 through 9
|           //Or
   1[0-2]   //   A 1 followed by any digit between 0 and 2.
)\b         //The end of a word

Making this work for both negative and positive is as simple as adding an optional dash at the start:

-?\b(\d|1[0-2])\b

(This assumes no inappropriate characters precede the dash.)

To forbid negative numbers, a negative lookbehind is necessary:

(?<!-)\b(\d|1[0-2])\b

Leaving the lookbehind out would cause the 11 in -11 to match. (The first example in this post should have this added.)

Note: \d versus [0-9]

In order to be compatible with all regex flavors, all \d-s should be changed to [0-9]. For example, .NET considers non ASCII numbers, such as those in different languages, as legal values for \d. Except for in the last example, for brevity, it's left as \d.

(With thanks to TimPietzcker at stackoverflow)

Three digits, with all but the first digit equal to zero

Rule: Must be between 0 and 400.

A possible regex:

(?<!-)\b([1-3]?\d{1,2}|400)\b

Free spaced:

   (?<!-)          //Something not preceded by a dash
   \b(             //Word-start, followed by either
      [1-3]?       //   No digit, or the digit 1, 2, or 3
         \d{1,2}   //   Followed by one or two digits (between 0 and 9)
   |               //Or
      400          //   The number 400
   )\b             //Word-end

Another possibility that should never be used:

\b(0|1|2|3|4|5|6|7|8|9|10|11|12|13|14|15|16|17|18|19|20|21|22|23|24|25|26|27|28|29|30|31|32|33|34|35|36|37|38|39|40|41|42|43|44|45|46|47|48|49|50|51|52|53|54|55|56|57|58|59|60|61|62|63|64|65|66|67|68|69|70|71|72|73|74|75|76|77|78|79|80|81|82|83|84|85|86|87|88|89|90|91|92|93|94|95|96|97|98|99|100|101|102|103|104|105|106|107|108|109|110|111|112|113|114|115|116|117|118|119|120|121|122|123|124|125|126|127|128|129|130|131|132|133|134|135|136|137|138|139|140|141|142|143|144|145|146|147|148|149|150|151|152|153|154|155|156|157|158|159|160|161|162|163|164|165|166|167|168|169|170|171|172|173|174|175|176|177|178|179|180|181|182|183|184|185|186|187|188|189|190|191|192|193|194|195|196|197|198|199|200|201|202|203|204|205|206|207|208|209|210|211|212|213|214|215|216|217|218|219|220|221|222|223|224|225|226|227|228|229|230|231|232|233|234|235|236|237|238|239|240|241|242|243|244|245|246|247|248|249|250|251|252|253|254|255|256|257|258|259|260|261|262|263|264|265|266|267|268|269|270|271|272|273|274|275|276|277|278|279|280|281|282|283|284|285|286|287|288|289|290|291|292|293|294|295|296|297|298|299|300|301|302|303|304|305|306|307|308|309|310|311|312|313|314|315|316|317|318|319|320|321|322|323|324|325|326|327|328|329|330|331|332|333|334|335|336|337|338|339|340|341|342|343|344|345|346|347|348|349|350|351|352|353|354|355|356|357|358|359|360|361|362|363|364|365|366|367|368|369|370|371|372|373|374|375|376|377|378|379|380|381|382|383|384|385|386|387|388|389|390|391|392|393|394|395|396|397|398|399|400)\b

Final example: Four digits, mirrored around zero, that does not end with zeros.

Rule: Must be between -2055 and 2055

This is from a question on stackoverflow.

Regex:

-?\b(20(5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\b

Free-spaced:

   -?                 //Optional dash
   \b(                //Followed by word boundary, followed by either of the following
      20(             //   "20", followed by either
         5[0-5]       //      A "5" followed by a digit 0-5
      |               //   or
         [0-4][0-9]   //      A digit 0-4, followed by any digit
      )
   |                  //OR
      1?[0-9]{1,3}    //   An optional "1", followed by one through three digits (0-9)
   )\b                //Followed by a word boundary.

Here is a visual representation of this regex:

And here you can try it out yourself: Debuggex demonstration

(With thanks to PlasmaPower on stackoverflow for the debugging assistance.)

Final note

Depending on what you are capturing, it is likely that all sub-groups should be made into non-capture groups. For example, this:

(-?\b(?:20(?:5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\b)

Instead of this:

-?\b(20(5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\b

Example Java implementation

  import  java.util.Scanner;
  import  java.util.regex.Matcher;
  import  java.util.regex.Pattern;
  import  org.apache.commons.lang.math.NumberUtils;
/**
  <P>Confirm a user-input number is a valid number by reading a string an testing it is numeric before converting it to an it--this loops until a valid number is provided.</P>

  <P>{@code java UserInputNumInRangeWRegex}</P>
 **/
public class UserInputNumInRangeWRegex  {
   public static final void main(String[] ignored)  {

      int num = -1;
      boolean isNum = false;

      int iRangeMax = 2055;

      //"": Dummy string, to reuse matcher
      Matcher mtchrNumNegThrPos = Pattern.compile("-?\\b(20(5[0-5]|[0-4][0-9])|1?[0-9]{1,3})\\b").matcher("");

      do  {
         System.out.print("Enter a number between -" + iRangeMax + " and " + iRangeMax + ": ");
         String strInput = (new Scanner(System.in)).next();
         if(!NumberUtils.isNumber(strInput))  {
            System.out.println("Not a number. Try again.");
         }  else if(!mtchrNumNegThrPos.reset(strInput).matches())  {
            System.out.println("Not in range. Try again.");
         }  else  {
            //Safe to convert
            num = Integer.parseInt(strInput);
            isNum = true;
         }
      }  while(!isNum);

      System.out.println("Number: " + num);
   }
}

Output

[C:\java_code\]java UserInputNumInRangeWRegex
Enter a number between -2055 and 2055: tuhet
Not a number. Try again.
Enter a number between -2055 and 2055: 283837483
Not in range. Try again.
Enter a number between -2055 and 2055: -200000
Not in range. Try again.
Enter a number between -2055 and 2055: -300
Number: -300

Answer 8

I've converted Bart Kiers's answer into C++. The function takes two integers as an input and generates the regular expression for the number range.

#include <stdio.h>
#include <iostream>
#include <vector>
#include <string>

std::string regex_range(int from, int to);

int main(int argc, char **argv)
{
    std::string regex = regex_range(1,100);

    std::cout << regex << std::endl;

    return 0;
}

std::string regex_range(int from, int to) //Credit: Bart Kiers 2011
{
    if(from < 0 || to < 0)
    {
        std::cout << "Negative values not supported. Exiting." << std::endl;
        return 0;
    }

    if(from > to)
    {
        std::cout << "Invalid range, from > to. Exiting." << std::endl;
        return 0;
    }

    std::vector<int> ranges;
    ranges.push_back(from);
    int increment = 1;
    int next = from;
    bool higher = true;

    while(true)
    {

        next += increment;

        if(next + increment > to)
        {
            if(next <= to)
            {
                ranges.push_back(next);
            }
            increment /= 10;
            higher = false;
        }
        else if(next % (increment*10) == 0)
        {
            ranges.push_back(next);
            increment = higher ? increment*10 : increment/10;
        }

        if(!higher && (increment < 10))
        {
            break;
        }
    }

    ranges.push_back(to + 1);
    std::string regex("^(?:");

    for(int i = 0; i < ranges.size() - 1; i++)
    {
        int current_from = ranges.at(i);
        std::string str_from = std::to_string(current_from);
        int current_to = ranges.at(i + 1) - 1;
        std::string str_to = std::to_string(current_to);
        for(int j = 0; j < str_from.length(); j++)
        {
            if(str_from.at(j) == str_to.at(j))
            {
                std::string str_from_at_j(&str_from.at(j));
                regex.append(str_from_at_j);
            }
            else
            {
                std::string str_from_at_j(&str_from.at(j));
                std::string str_to_at_j(&str_to.at(j));

                regex.append("[");
                regex.append(str_from_at_j);
                regex.append("-");
                regex.append(str_to_at_j);
                regex.append("]");
            }
        }
        regex.append("|");
    }
    regex = regex.substr(0, regex.length() - 1);
    regex.append(")$");
    return regex;
}

Answer 9

As I have encountered the same issue as @EmilianoT already reported, I tried to fix it, but in the end I opted for porting the PHP port of RegexNumericRangeGenerator (ported by @EmilianoT), although not in a class. I am not quite happy with this JS port, as all toString() and parseInt() methods could be still optimised (they might be somewhere unnecessary), but it works for all cases.

I thing I changed are the parameters. I replaced parse($min, $max, $MatchWholeWord = FALSE, $MatchWholeLine = FALSE, $MatchLeadingZero = FALSE) with parse(min, max, width = 0, prefix = '', suffix = ''), which gives it more options (some might want to put the regex into slashes, others want to match the line [prefix = '^'; suffix = '$'], etc). Also I wanted to be able to configure the width of the number (width = 3 → 000, 001, 052, 800, 1000, ...).

I replaced my previous answer, as it does not work all the time. If one wants to read it, they can see it in the answer history.

function parse(min, max, width = 0, prefix = '', suffix = '') {
  if (! Number.isInteger(min) || ! Number.isInteger(max) || min > max || min < 0 || max < 0) {
    return false
  }

  if (min == max) {
    return parseIntoPattern(min, prefix, suffix)
  }

  let x = parseStartRange(min, max)
  let s = []

  x.forEach(o => {
    s.push(parseEndRange(o[0], o[1]))
  })

  let n = reformatArray(s)
  let h = parseIntoRegex(n, width)

  return parseIntoPattern(h, prefix, suffix)
}

function parseIntoPattern(t, prefix = '', suffix = '') {
  let r = Array.isArray(t) ? t.join('|') : t
  return prefix + '(' + r + ')' + suffix
}

function parseIntoRegex(t, width = 0) {
  if (! Array.isArray(t)) {
    throw new Error('Argument needs to be an array!')
  }

  let r = []

  for (let i = 0; i < t.length; i++) {
    let e = t[i][0].split('')
    let n = t[i][1].split('')
    let s = ''
    let o = 0
    let h = ''

    for (let a = 0; a < e.length; a++) {
      if (e[a] === n[a]) {
        h += e[a]
      } else if (parseInt(e[a]) + 1 === parseInt(n[a])) {
        h += '[' + e[a] + n[a] + ']'
      } else {
        if (s === e[a] + n[a]) {
          o++
        }

        s = e[a] + n[a]

        if (a == e.length - 1) {
          h += o > 0 ? '{' + (o + 1) + '}' : '[' + e[a] + '-' + n[a] + ']'
        } else if (o === 0) {
          h += '[' + e[a] + '-' + n[a] + ']'
        }
      }
    }

    if (e.length < width) {
      h = '0'.repeat(width - e.length, '0') + h
    }

    r.push(h)
  }

  return r
}

function reformatArray(t) {
  let arrReturn = []

  for (let i = 0; i < t.length; i++) {
    let page = t[i].length / 2

    for (let a = 0; a < page; a++) {
      arrReturn.push(t[i].slice(2 * a))
    }
  }

  return arrReturn
}

function parseStartRange(t, r) {
  t = t.toString()
  r = r.toString()

  if (t.length === r.length) {
    return [[t, r]]
  }

  let breakOut = 10 ** t.length - 1
  return [[t, breakOut.toString()]].concat(parseStartRange(breakOut + 1, r))
}

function parseEndRange(t, r) {
  if (t.length == 1) {
    return [t, r]
  }

  if ('0'.repeat(t.length) === '0' + t.substr(1)) {
    if ('0'.repeat(r.length) == '9' + r.substr(1)) {
      return [t, r]
    }

    if (parseInt(t.toString().substr(0, 1)) < parseInt(r.toString().substr(0, 1))) {
      let e = parseInt(r.toString().substr(0, 1) + '0'.repeat(r.length - 1)) - 1
      return [t, strBreakPoint(e)].concat(parseEndRange(strBreakPoint(e + 1), r))
    }
  }

  if ('9'.repeat(r.length) === '9' + r.toString().substr(1) && parseInt(t.toString().substr(0, 1)) < parseInt(r.toString().substr(0, 1))) {
    let e = parseInt(parseInt(parseInt(t.toString().substr(0, 1)) + 1) + '0'.repeat(r.length - 1)) - 1
    return parseEndRange(t, strBreakPoint(e)).concat(strBreakPoint(e + 1), r)
  }

  if (parseInt(t.toString().substr(0, 1)) < parseInt(r.toString().substr(0, 1))) {
    let e = parseInt(parseInt(parseInt(t.toString().substr(0, 1)) + 1) + '0'.repeat(r.length - 1)) - 1
    return parseEndRange(t, strBreakPoint(e)).concat(parseEndRange(strBreakPoint(e + 1), r))
  }

  let a = parseInt(t.toString().substr(0, 1))
  let o = parseEndRange(t.toString().substr(1), r.toString().substr(1))
  let h = []

  for (let u = 0; u < o.length; u++) {
    h.push(a + o[u])
  }

  return h
}

function strBreakPoint(t) {
  return t.toString().padStart((parseInt(t) + 1).toString().length, '0')
}