How many integers contain a number in a specific range

Question

The program should take input the number, the start of the range and the end of the range, and output how many integers the number appears between the start and end of the range, inclusive. Both programs and functions are allowed.

Example Inputs

For example:

//Input example 1
3,1,100
//Input example 2
3
1
100
//Input example 3
3 1 100
//Input example 4
a(3, 1, 100);

All the above four input examples are valid and all of them mean that 3 is the number in question, 1 is the beginning of the range and 100 the the end of the range.

And then the program should output how many times 3 appears in the range from 1 to 100 inclusive. 3 appears in the integers 3, 13, 23, 30, 31, 32, 33, ..., 93 at a total of 19 times. So the program should output 19 as the output because that is how many times 3 appears in the range from 1 to 100.

Rules

• Both programs and functions are allowed.
• All numbers will be integers, meaning that there will not be any floats or doubles.
• Note: the sought number will always be in the range 0≤x≤127. There will be no cases where it will be outside this 0≤x≤127 range.
• As in the first example, with the case as 33, the number 3 will be counted as appearing only once, not twice.
• The values of the start and end of the range will be between -65536 and 65535 inclusive.
• The value of range's start will never exceed or equal to range's end. start < end
• Also the range is inclusive. For example if the input was 8 8 10, the range would be 8≤x≤10 and hence the output will be 1.
• Input can be taken in any of the ways shown in the examples. Input can be taken as a string or as a number, any way you wish.

Test Cases

3 1 100
19

3 3 93
19

12,-200,200
24          //This is because 12 appears in -129, -128, ..., -112, -12, 12, 112, 120, 121, 122, ...

123,1,3
0           //This is because all of 123's digits have to appear in the same order

3 33 34
2           //Because 3 appears in 2 numbers: 33 and 34

a(0,-1,1);
1

$ java NotVerbose 127 -12 27
0

Snack Snippet

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# Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes

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# Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the leaderboard snippet:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes

/* Configuration */

var QUESTION_ID = 98470; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 41805; // This should be the user ID of the challenge author.

/* App */

var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page;

function answersUrl(index) {
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      getComments();
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getAnswers();

var SCORE_REG = /<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

var OVERRIDE_REG = /^Override\s*header:\s*/i;

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}

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        size: +match[2],
        language: match[1],
        link: a.share_link,
      });
    
  });
  
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        bB = b.size;
    return aB - bB
  });

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  var place = 1;
  var lastSize = null;
  var lastPlace = 1;
  valid.forEach(function (a) {
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      lastPlace = place;
    lastSize = a.size;
    ++place;
    
    var answer = jQuery("#answer-template").html();
    answer = answer.replace("{{PLACE}}", lastPlace + ".")
                   .replace("{{NAME}}", a.user)
                   .replace("{{LANGUAGE}}", a.language)
                   .replace("{{SIZE}}", a.size)
                   .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
    jQuery("#answers").append(answer);

    var lang = a.language;
    if (/<a/.test(lang)) lang = jQuery(lang).text();
    
    languages[lang] = languages[lang] || {lang: a.language, user: a.user, size: a.size, link: a.link};
  });

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  for (var lang in languages)
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  langs.sort(function (a, b) {
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    if (a.lang < b.lang) return -1;
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  for (var i = 0; i < langs.length; ++i)
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    var lang = langs[i];
    language = language.replace("{{LANGUAGE}}", lang.lang)
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body { text-align: left !important}

#answer-list {
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#language-list {
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table thead {
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table td {
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}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
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<table style="display: none">
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<table style="display: none">
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    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>

Answer 1

Bash, 20 bytes

the obvious answer

seq $2 $3|grep -c $1

example

$ bash golf 3 1 100
19

Answer 2

05AB1E, 6 bytes

Input in the form: upper bound, lower bound, number.

Ÿvy³åO

Explanation:

Ÿ       # Inclusive range, [a, ..., b]
 vy     # For each element...
   ³å   # Check if the third input is a substring of the number
     O  # Sum up the results

Uses the CP-1252 encoding. Try it online!

Answer 3

Perl, 20 bytes

Saved 2 bytes by using grep as in @ardnew's answer.

Bytecount includes 18 bytes of code and -ap flags.

$_=grep/@F/,<>..<>

Give the 3 numbers on three separate lines :

perl -ape '$_=grep/@F/,<>..<>' <<< "3
1
100"

Answer 4

Python 2, 47 43 Bytes

Relatively straightforward, making use of Python 2's repr short-form.

f=lambda n,a,b:a<b and(`n`in`a`)+f(n,-~a,b)

Ouput:

f(  3,    1, 100) -> 19
f(  3,    3,  93) -> 19
f( 12, -200, 200) -> 24
f(123,    1,   3) -> 0
f(  3,   33,  34) -> 2
f(  0,   -1,   1) -> 1
f(127,   12,  27) -> 0

Answer 5

Jelly,  7  6 bytes

-1 thanks to caird coinheringaahing! (Use w and filtering to avoiding @.)

rAwƇ⁵L

TryItOnline!

Input: Start, End, ToFind

How?

rAwƇ⁵L - Main link: Start, End, ToFind
r      - range: [Start, ..., End]
 A     - absolute values
    ⁵  - third input: ToFind
   Ƈ   - filter keep those (absolute values, a) for which:
  w    -   first index of sublist (implicit digits of ToFind) in (implicit digits of a)
     L - length

The default casting of an integer to an iterable for the sublist existence check casts to a decimal list (not a character list), so negative numbers have a leading negative value (e.g. -122->[-1,2,2] which won't find a sublist of [1,2]) so taking the absolute value first seems like the golfiest solution.

Answer 6

JavaScript (ES6), 46 45 bytes

f=(n,s,e)=>s<=e&&!!`${s++}`.match(n)+f(n,s,e)

(My best nonrecursive version was 61 bytes.) Edit: Saved 1 byte thanks to @edc65.

Answer 7

PowerShell v2+, 64 62 56 bytes

param($c,$a,$b)$(for(;$a-le$b){1|?{$a++-match$c}}).count

-6 bytes thanks to mazzy

Input via command-line arguments of the form number lower_bound upper_bound. A little goofy on the notation, because of the semicolons inside the for causing parse errors if it's not surrounded in $(...) to create a script block. We basically loop upward through $a until we hit $b, using Where-Object (the |?{...}) to pull out those numbers that regex -match against $c. That's encapsulated in parens, we take the .count thereof, and that's left on the pipeline and output is implicit.

---

If, however, we guarantee that the range will be no more than 50,000 elements, we can skip the loop and just use the range operator .. directly, for 45 43 bytes. Since that's not in the challenge specifications, though, this isn't valid. Bummer.

param($c,$a,$b)($a..$b|?{$_-match$c}).count

Answer 8

Vim, 46, 41 bytes

C<C-r>=r<tab><C-r>")<cr><esc>jC0<esc>:g/<C-r>"/norm G<C-v><C-a>
kd{

Input is in this format:

1, 100
3

Answer 9

Haskell, 65 bytes

import Data.List
(s#e)i=sum[1|x<-[s..e],isInfixOf(show i)$show x]

The import ruins the score. Usage example: ((-200)#200)12 -> 24.

Answer 10

Java 7 85 bytes

int x(int a,int b,int c){int t=0;for(;b<=c;)if((b+++"").contains(a+""))t++;return t;}

Answer 11

Swift 3, 96 93 bytes

import Cocoa
func c(n:Int,s:Int,e:Int){print((s...e).filter{"\($0)".contains("\(n)")}.count)}

Edit 1:

Saved 3 bytes by using shorthand parameters

Answer 12

Scala, 50 bytes

(c:String)=>(_:Int)to(_:Int)count(""+_ contains c)

takes the first input curried; call it like this: f("12")(-200,200)

Explantion:

(c:String)=>  //define an anonymous function taking a string parameter
  (_:Int)     //create a range from an anonymous int parameter
  to          //to
  (_:Int)     //another anonymous int parameter
  count(      //and count how many...
   ""+_       //elements converted to a string
   contains c //contain c
  )

Answer 13

R, 32 bytes

Quite straightforward:

function(a,b,c)sum(grepl(a,b:c))

Answer 14

C#, 71 bytes

Beat my Java answer thanks to lambdas

(t,l,u)=>{int d=0;for(;l<=u;)if((l+++"").Contains(t+""))d++;return d;};

Answer 15

Ruby 44 bytes

m=->(n,s,f){(s..f).count{|x|x.to_s[/#{n}/]}}

Test Cases:

m.(3,1,100)     #=> 19
m.(3,3,93)      #=> 19
m.(12,-200,200) #=> 24
m.(123,1,3)     #=>  0
m.(3,33,34)     #=>  2
m.(0,-1,1)      #=>  1
m.(127,-12,27)  #=>  0

Answer 16

PHP, 62 bytes

Pretty straight forward approach:

<?=count(preg_grep('/'.($a=$argv)[1].'/',range($a[2],$a[3])));

Try it online

Answer 17

C, 143 135 bytes

Thanks to @Kritixi Lithos for helping save 8 bytes

Surely this can be done better, but its the best I've got for now. C doesn't handle strings very gracefully, so naturally it takes quite a few operations.

int C(int N,int l,int h){char b[99],n[99];int t=0,i=1;sprintf(n,"%d",N);for(;i<=h;i++){sprintf(b,"%d",i);if(strstr(b,n))++t;}return t;}

Ungolfed + program

#include <string.h>
#include <stdlib.h>
#include <stdio.h>

int C(int N,int l,int h)
{
    char b[99], n[99];
    int t=0,i=1;
    sprintf(n,"%d",N);
    for(;i<=h;i++)
    {
        sprintf(b,"%d",i);
        if(strstr(b,n))
            ++t;
    }
    return t;
}

int main()
{
  printf("%d\n", C(3, 1, 100));
}

Answer 18

JavaScript, 46 45 bytes

f=(i,s,e)=>s>e?0:RegExp(i).test(s)+f(i,++s,e)

Recursively count until start > end

Edit: Switch to RegExp test to save a byte

Answer 19

PHP, 68 63 bytes

for($a=$argv;$a[2]<=$a[3];)$o+=strstr($a[2]++,$a[1])>'';echo$o;

use like:

 php -r "for($a=$argv;$a[2]<=$a[3];)$o+=strstr($a[2]++,$a[1])>'';echo$o;" 3 1 100

edit: 5 bytes saved thanks to Titus

Answer 20

Powershell, 48 bytes

According to the rule, the range can contain more than 50,000 elements. So we can't use the range operator .. directly. Thanks AdmBorkBork.

Straightforward:

param($c,$a,$b)for(;$a-le$b){$i+=$a++-match$c}$i

Test script:

$f = {

param($c,$a,$b)for(;$a-le$b){$i+=$a++-match$c}$i

}

@(
    ,(19, 3,1,100)
    ,(19, 3,3,93)
    ,(24, 12,-200,200)
    ,(0, 123,1,3)
    ,(2, 3,33,34)
    ,(1, 0,-1,1)
    ,(0, 127,-12,27)
    ,(44175, 0,-65536,65535)
) | % {
    $e,$a = $_
    $r = &$f @a
    "$($e-eq$r): $r"
}

Output:

True: 19
True: 19
True: 24
True: 0
True: 2
True: 1
True: 0
True: 44175

Answer 21

Japt, 14 8 bytes

Takes the integer to be found as the last input value.

õV èÈsøW

Try it online

---

Explanation

             :Implicit input of integers U=start, V=end & W=number
õV           :Range [U,V]
    È        :Map
     s       :  Convert to string
      øW     :  Contains W?
   è         :Count truthy values

Answer 22

Vyxal s, 7 bytes

÷ṡƛS?hc

Try it Online!

Answer 23

Java, 92 89 71 bytes

Now with lambdas!

(t,l,u)->{int z=0;for(;l<=u;)if((l+++"").contains(t+""))z++;return z;};

Old 89 byte function solution:

int d(int t,int l,int u){int a=0,i=l;for(;i<=u;)if((i+++"").contains(t+""))a++;return a;}

Hooray for the super increment function!

Answer 24

Groovy, 48 bytes

{a,b,c->(a..b).collect{"$it".count("$c")}.sum()}

Answer 25

Racket 91 bytes

(for/sum((i(range s(+ 1 e))))(if(string-contains?(number->string i)(number->string d))1 0))

Ungolfed:

(define(f d s e)
  (for/sum ((i (range s (+ 1 e))))
    (if(string-contains?
        (number->string i)
        (number->string d))
       1 0 )))

Testing:

(f 3 1 100)
(f 3 3 93)
(f 12 -200 200)
(f 123 1 3)
(f 3 33 34)
(f 0 -1 1)

Output:

19
19
24
0
2
1

Answer 26

Axiom bytes 90

f(y,a,b)==(c:=0;for x in a..b repeat(if position(y::String,x::String,1)~=0 then c:=c+1);c)

results

(3) -> f(3,1,100)=19,f(3,3,93)=19,f(12,-200,200)=24,f(123,1,3)=0,f(3,33,34)=2
   (3)  [19= 19,19= 19,24= 24,0= 0,2= 2]
                                  Type: Tuple Equation NonNegativeInteger
(4) -> f(0,-1,1)=1, f(127,12,27)=0
   (4)  [1= 1,0= 0]
                                  Type: Tuple Equation NonNegativeInteger

Answer 27

Mathematica, 70 bytes

(w=ToString;t=0;Table[If[StringContainsQ[w@i,w@#1],t++],{i,#2,#3}];t)&

input

[12,-200,200]

output

24

Answer 28

Clojure, 65 bytes

#(count(for[i(range %2(inc %3)):when(some(set(str %))(str i))]i))

Answer 29

PHP, 56 Bytes

run as pipe Try it online

Input

$argv = [number_to_find, range_start, range_end];

Code

<?=substr_count(join(range(($a=$argv)[1],$a[2])),$a[0]);

Explanation

#substrcount, counts the aparitions of a subtring in a string
substr_count( 
           join( range(($a=$argv)[1],$a[2])), # String with the range
           $a[0]);                            # The number you are looking for

Answer 30

Perl 6, 32 bytes

{+grep *.contains($^a),$^b..$^c}

Try it online!