How many 1's we get

Question

Description

Count how many occurrences there are of the digit 1 between two given numbers \$[a, b]\$, inclusive.

For example, from 1 to 100 it should be 21:

1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91, 100

The number 1 is repeated 21 times.

Rules

1. Each number in the input list is guaranteed is an integer in the range \$0 \leq a \leq b < 2^{32}\$.
2. The shortest answer in bytes wins.

Test cases

[1, 100] -> 21
[11, 200] -> 138
[123, 678] -> 182

Example

Here is my code using bash

eval echo {$1..$2}|grep -o 1|wc -l

Answer 1

05AB1E, 5 4 bytes

Thanks a lot to OP! Now I can save a byte

ŸSΘO

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Explanation

Ÿ    Inclusive range
 S   Split the string into individual chars
  Θ  (Vectorizes) Does this character == "1"?
   O Sum the resulting list

Answer 2

Python 2, ³⁹ 36 bytes

-3 bytes thanks to @SurculoseSputum

lambda a,b:`range(a,b+1)`.count('1')

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Python 3, ⁴² 40 bytes

-2 bytes thanks to @JoKing

lambda a,b:f"{*range(a,b),b}".count('1')

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Answer 3

Rust+itertools, 42 bytes

|a,b|(a..=b).join("").matches('1').count()

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Answer 4

APL (Dyalog Extended), 9 bytes

+/'1'=⍕⍤…

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+/ sum

'1'= where the character is equal to

⍕ the string representation

⍤ of

… the range

Answer 5

Octave, ^{29 28} 27 bytes

@(x,y)sum(mat2str(x:y)==49)

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Answer 6

Java 8, 71 67 66 bytes

a->b->{var s="";for(;b>=a;)s+=b--;return~-s.split("1",-1).length;}

-4 bytes thanks to @OlivierGrégoire.

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Explanation:

a->b->{                // Method with two integer inputs and integer return-type
  var s="";            //  String `s`, starting empty
  for(;b>=a;)          //  Loop `b` downwards in the range [`b`, `a`]:
    s+=b--;            //   And append `b` to to String `s`
  return~-s.split("1", //  Split String `s` on "1",
                  -1)  //  and keep empty trailing items
           .length;    //  Then get the amount of parts of this array
                       //  And decrease it by 1 with `~-`, before turning it as result

Answer 7

R, 43 bytes

function(x,y)sum(unlist(gregexpr(1,x:y))>0)

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Answer 8

Ruby, 62 46 28 bytes

->c,d{[*c..d].join.count ?1}

-18 bytes, courtesy of Dingus.

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Answer 9

JavaScript (ES6), 44 bytes

Expects (a)(b).

a=>g=b=>b<a?0:(b+g).split(1).length-3+g(b-1)

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How?

We use a recursive function g to count how many 1's we have in b and decrement b until it's lower than a.

In order to count the 1's, we have to coerce b to a string. We could do b+'' but it's shorter to use b+g. Because the source code of g itself contains two 1's, we subtract 3 instead of just 1 from the result of (b+g).split(1).length.

Answer 10

Retina 0.8.2, 36 32 bytes

\d+
$*_
(?<=(_+) _*)(?=\1)
$.'
1

Try it online! Link includes test cases. Explanation:

\d+
$*_

Convert a and b to unary, but use _ instead of 1 to avoid confusion. (In Retina 1 this would just be *, saving 2 bytes.)

(?<=(_+) _*)(?=\1)
$.'

At each boundary in b up to and including a from the end, insert the distance to the end in decimal, thus generating the range from b down to a.

1

Count the resulting number of 1s.

Answer 11

Google Sheets, 68

=SUM(ArrayFormula(LEN(REGEXREPLACE(""&SEQUENCE(A2-A1+1,1,A1),"[^1]",

Sheets auto-closes parens.

This is super slow at large ranges, but as this is Code Golf, we're here to optimize character count. My first attempt was to use JOIN(SEQUENCE(...)) to make one long string then count the 1's, but as it turns out, Sheets has a limit of 50000 characters, so that didn't work.

Answer 12

Kotlin, 49 bytes

{x:Int,y:Int->(x..y).sumBy{"$it".count{it=='1'}}}

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Answer 13

Haskell, 26 bytes

a!b=sum[1|'1'<-show[a..b]]

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Answer 14

MathGolf, 8 bytes

↨æ▒m┴Σ]Σ

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Explanation:

↨         # Loop in the range [a,b] using the two implicit inputs a,b,
 æ        # and execute the following four commands:
  ▒       #  Convert the integer to a list of digits
   m      #  Map over each digit:
    ┴     #   And check which are equal to 1 (1 if 1; 0 otherwise)
     Σ    #  Get the sum of those checks
      ]   # After the loop, wrap all values on the stack into a list
       Σ  # And sum this list
          # (after which the entire stack joined together is output implicitly as result)

Answer 15

Perl 5 -p, 21 bytes

map$\+=y/1//,$_..<>}{

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Answer 16

PowerShell, 40 bytes

$a,$b=$args
($a..$b|sls 1 -a|% m*).Count

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---

PowerShell, 40 bytes

($args-join'..'|iex|sls 1 -a|% m*).Count

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Answer 17

PHP, 46 45 bytes

fn($a,$b)=>substr_count(join(range($a,$b)),1)

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Answer 18

C (gcc), 58 bytes

i;c;f(a,b){for(c=0;b/a;)for(i=a++;i;i/=10)c+=i%10==1;a=c;}

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Answer 19

Factor, 65 bytes

: f ( a b -- n ) [a,b] [ number>string [ 49 = ] count ] map sum ;

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Answer 20

K (oK), 19 17 bytes

Solution:

{+/"1"=,/$x_!1+y}

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Explanation:

Range generation is inefficient (i.e. generate range 0..Y and then drop from front, rather than generating range X..Y) but saves 2 bytes.

{+/"1"=,/$x_!1+y} / the solution
{               } / lambda taking implicit x & y args
             1+y  / add 1 to y
            !     / range 0..N
          x_      / drop (_) x items from front
         $        / convert to string
       ,/         / flatten 
   "1"=           / is string equal to "1"?
 +/               / sum

Answer 21

JavaScript (ES6), 63 62 bytes

a=>b=>([...Array(b-a+1)].map((_,i)=>i+a)+'').split(1).length-1

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[...Array(b-a+1)] // an array of length b-a+1
map((_,i)=>i+a)   // fill it with numbers from a to b
+''               // convert it to a string with each number separated by a comma
.split(1)         // split at each 1
.length-1         // count the chunks and subtract 1

-1 byte thanks to @Jo King

Answer 22

J, 19 16 bytes

-3 bytes thanks to Jonah!

1#.1=/&":[,-.&i.

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Answer 23

Japt, 5 bytes

The spec contradicts itself as to whether the range should be inclusive or not. If it shouldn't then replace õ with o.

õ ¬è1

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Answer 24

Rust, 63 bytes

|a,b|(a..=b).map(|x|format!("{}",x).matches('1').count()).sum()

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Answer 25

MATL, 7 bytes

&:V1V=s

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Explanation

&:V1V=s
     =s  % Count occurrences
   1V    % of '1' in
  V      % string of
&:       % inclusive range of input

Answer 26

Java (OpenJDK 8), 80 bytes

a->b->IntStream.range(a,b+1).flatMap(i->(""+i).chars()).filter(x->x==49).count()

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Doesn't need much of an explanation, but here's one anyways:

Function<Integer, Function<Integer, Long>> f = 
    a -> b-> 
      IntStream
        .range(a,b+1) //Create an IntStream going from a to b
        .flatMap(i -> //Map every int i in that stream
          (""+i)      //Make it a string
          .chars()    //Turn that string to an IntStream
        )    //Flatten that
        .filter(x -> x == 49)  //Keep all the '1's
        .count();              //Find out how many '1's there are

Answer 27

Jelly, 5 bytes

rDFċ1

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Explanation:

rDFċ1
r      range of the inputs
 DF    make decimal and flatten; list of all the digits in the range
   ċ1  count occurences of 1

Answer 28

Pyth, ⁸ 7 bytes

-1 byte thanks to @FryAmTheEggman

/`}QE"1

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Explanation

/`}QE"1
  }QE     # Inclusive range on input
 `        # string of the range ([1, 2, 3] -> "[1, 2, 3]")
/    "1   # count ones

Answer 29

Wolfram Language (Mathematica), 29 bytes

Tr@DigitCount[Range@##,10,1]&

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Answer 30

Dart, 56 bytes

f(a,b)=>'1'.allMatches([for(;a<=b;a++)a].join()).length;

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