Find how many alphanumeric characters can be made up of a single number

Question

The alphanumeric characters have ASCII-values:

0-9  ->  48-57
A-Z  ->  65-90
a-z  ->  97-122

Your challenge is to take an integer as input, and output how many characters can be made using consecutive digits of that number. The character codes may be overlapping. 666 should result in 2, since you have 66 twice.

Test cases:

Input: 5698
Possible characters: '8' (56), 'E' (69), 'b' (98)
Output: 3

Input: 564693
Possible characters: '8' (56), 'E' (69)
Output: 2

Input: 530923864209124521
Possible characters: '5' (53), 'V' (86), '4' (52)  
Output: 3

Input: 1111111
Possible characters: 'ooooo' (5*111)
Output: 5

Input: 5115643141276343
Possible characters: '3' (51), '8' (56), 'L' (76), 's' (115)
Output: 4

Input: 56789
Possible characters: '8' (56), 'C' (67), 'N' (78), 'Y' (89)
Output: 4

Input: 94
Possible characters: ''
Output: 0

Input: 1
Output: 0

Input and output formats are optional (yes, you may take the integer as a string).

Answer 1

05AB1E, 8 7 bytes

žKÇIŒÃg

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Explanation

žK       # push [a-zA-Z0-9]
  Ç      # convert to list of ascii codes
   IΠ   # push all substrings of input
     Ã   # keep only the subtrings which exist in the list of acsii codes
      g  # push length of resulting list

Answer 2

MATL, 17 13 bytes

8Y2"G@oVXf]vn

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Explanation

8Y2     % Predefined literal: string with all letters, uppercase and lowercase,
        % and digits
"       % For each character in that string
  G     %   Push input, for example the string '5115643141276343'
  @     %   Push current character, such as 'A'
  o     %   Convert to its ASCII code, such as 65
  V     %   String representation, such as '65'
  Xf    %   Find string '65' within string '5115643141276343'. This gives a vector
        %   (possibly empty) with indices of occurrences
]       % End
v       % Concatenate all stack contents vertically
n       % Number of entries. Implicitly display

Answer 3

Brachylog, 22 bytes

∧Ạụ:Ạ:Ịcạ:?{tT&h∋~sT}ᶜ

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Explanation

       c                 Concatenate together:
∧Ạụ:                       "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
    Ạ:                     "abcdefghijklmnopqrstuvwxyz"
      Ị                    "0123456789"
        ạ                Get the list of ASCII codes of that string
         :?{        }ᶜ   Count the number of results, for input [list of codes, Input], of:
            tT             Call the Input T
              &h∋          Take one ASCII code
                 ~sT       It is a substring of T

Answer 4

Java 8, 204 197 195 186 174 bytes

n->{int r=0,i=0,e=n.length()-1,t;for(;i<e;r+=t>47&t<57|t>64&t<91|t>96&t<100|(t=new Short(n.substring(i,i++>e-2?i:i+2)))>99&t<123?1:0)t=new Byte(n.substring(i,i+2));return r;}

-9 bytes thanks to @ceilingcat.

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Explanation:

n->{                   // Method with String parameter and integer return-type
  int r=0,             //  Result, starting at 0
      i=0,             //  Index
      e=n.length()-1,  //  Length of String -1
      t;               //  Temp integer
  for(;i<e             //  Loop over the String using the index
      ;                //    After every iteration:
       r+=             //     Increase the result-sum by:
          t>47&t<57|t>64&t<91|t>96&t<100
                       //      If two adjacent digits are a digit or letter
          |            //      or
          (t=new Short(n.substring(i,i++>e-2?i:i+2)))>99&t<123?
                       //      if three adjacent digits are a letter:
           1           //       Increase the sum by 1
          :            //      Else:
           0)          //       Keep the sum the same by adding 0
    t=new Byte(n.substring(i,i+2));
                       //   Set `t` to the substring of [i, i+2) converted to integer
  return r;}           //  Return the result-sum

Answer 5

JavaScript (ES6), 71 70 bytes

f=([a,...b])=>a?(a&(a+=b[0])+b[1]<123|a>47&a<58|a>64&a<91|a>96)+f(b):0

Test cases

f=([a,...b])=>a?(a&(a+=b[0])+b[1]<123|a>47&a<58|a>64&a<91|a>96)+f(b):0

console.log(f("5698"))                // 3
console.log(f("564693"))              // 2
console.log(f("530923864209124521"))  // 3
console.log(f("1111111"))             // 5
console.log(f("5115643141276343"))    // 4
console.log(f("56789"))               // 4
console.log(f("94"))                  // 0
console.log(f("1"))                   // 0

Answer 6

Perl 5, 47 bytes

46 bytes of code + -p flag.

$"="|";$_=()=/(?=@{[48..57,65..90,97..122]})/g

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I couldn't find any shorter way to write that 48..57,65..90,97..122: map{ord}0..9,a..z,A..Z (getting the ascii value of the characters) is one byte longer. And doing for$c(0..122){$\+=chr($c)=~/\pl|\d/ for/(?=$c)/g}}{ (looking for all numbers, but keeping only those whose numbers corresponds to the ascii value of letters (\pl) or digits (\d)) will be 5 bytes longer (note that \pl|\d can't be replaced by \w as the latter also includes underscores)).

---

Previous approach (49 bytes):

for$@(48..57,65..90,97..122){$\+=()=/(?=$@)/g}}{

Answer 7

PHP, 68 Bytes

for(;a&($a=$argn)[$i];)$d+=ctype_alnum(chr($a[$i].$a[++$i]));echo$d;

Online Version

Answer 8

JavaScript (ES), 165 161 156 154 153 bytes

Yeah, RegEx definitely wasn't the right tool for the job here!

n=>[/\d{2}/g,/\d{3}/g].map(e=>eval("while(x=e.exec(n)){a.push(m=x[0]);e.lastIndex-=m.length-1}"),a=[])|a.filter(x=>x>47&x<58|x>64&x<91|x>96&x<123).length

---

Try It

f=

n=>[/\d{2}/g,/\d{3}/g].map(e=>eval("while(x=e.exec(n)){a.push(m=x[0]);e.lastIndex-=m.length-1}"),a=[])|a.filter(x=>x>47&x<58|x>64&x<91|x>96&x<123).length

console.log(f(5698))//3
console.log(f(564693))//2
console.log(f(530923864209124521))//3
console.log(f(1111111))//5
console.log(f(5115643141276343))//4
console.log(f(56789))//4
console.log(f(94))//0
console.log(f(1))//0

Answer 9

Retina, 52 bytes

&`1[01]\d|12[012]|4[89]|5[0-7]|6[5-9]|[78]\d|9[0789]

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Answer 10

Pyth, 19 17 14 bytes

l@jGUTmr0Csd.:

takes a string.

-3 Bytes thanks to @LeakyNun

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Explanation

l@jGUTmr0Csd.:
    UT                # the list of digits [0,1,2,...,9]
  jG                  # join that on the lowercase alphabet (repetition doesn't matter)
              Q       # implicit input
            .:        # all substrings of the input
      m               # for each of those substrings
          sd          # Convert the string to a base 10 integer
         C            # convert that integer to the character with that number
       r0             # make that character lowercase
l@                    # length of the intersection of those two list of chars we generated
 

Answer 11

Python 2, 74 64 62 bytes

f=lambda n:n and chr(n%10**(3-(n%1000>122))).isalnum()+f(n/10)

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Answer 12

Haskell, 161 157 138 129 126 bytes

import Data.List
f x=sum[1|y<-nub$concat$words.concat<$>mapM(\c->[[c],c:" "])x,any(elem$read y)[[48..57],[65..90],[97..122]]]

I wonder if there is a better way to remove dupes of the list than importing Data.List for nub?

Answer 13

Jelly, 8 bytes

ØBODf@ẆL

Input is a digit array.

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How it works

ØBODf@ẆL  Main link. Argument: A (digit array)

ØB        Base 62; yield all alphanumeric ASCII characters.
  O       Ordinal; get their code points.
   D      Decimal; convert the code points into digit arrays.
      Ẇ   Window; yield all contiguous subarrays of A.
    f@    Filter swapped; keep all elements of the result to the right that appear
          in the result to the left.
       L  Length; count the matches.

Answer 14

Ruby, 50 bytes

p (0..~/$/).any?{|n|$_[n,2].to_i.chr=~/\p{Alnum}/}

Reads from standard input; requires the Ruby interpreter be invoked with the -n option (implicit while gets loop).

Could be reduced to 43 bytes if it were allowed to match underscores.

p (0..~/$/).any?{|n|$_[n,2].to_i.chr=~/\w/}

Answer 15

Japt, 24 bytes

;è"(?={9ò ¬+B+C ¬mc q|})

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Alternate 27-byte version:

è"(?={#~o f_d f"%d|%l"Ãq|})

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