Don't give me five!

Question

Question :

You will be given the starting and ending integers of a sequence and should return the number of integers within it which do not contain the digit 5. The start and end numbers should be included!

Examples:

1,9 → 1,2,3,4,6,7,8,9 → Result 8

4,17 → 4,6,7,8,9,10,11,12,13,14,16,17 → Result 12

50,60 → 60 → Result 1

-59,-50 → → Result 0

The result may contain five.

The start number will always be smaller than the end number. Both numbers can be also negative!

I'm very curious for your solutions and the way you solve it. Maybe someone of you will find an easy pure mathematics solution.

Edit This is a code-golf challenge, so the shortest code wins.

Answer 1

JavaScript (ES6), 36 33 bytes

Takes input with currying syntax (a)(b).

a=>F=b=>b<a?0:!/5/.test(b)+F(b-1)

Formatted and commented

a =>                 // outer function: takes 'a' as argument, returns F
  F = b =>           // inner function F: takes 'b' as argument, returns the final result
    b < a ?          // if b is less than a
      0              //   return 0
    :                // else
      !/5/.test(b) + //   add 1 if the decimal representation of b does not contain any '5'
      F(b - 1)       //   and do a recursive call to F with b - 1

Test cases

let f =

a=>F=b=>b<a?0:!/5/.test(b)+F(b-1)

console.log(f(1)(9))
console.log(f(4)(17))
console.log(f(50)(60))
console.log(f(-50)(-59))

Answer 2

Jelly, 8 7 bytes

-1 byte thanks to Dennis (use fact that indexing into a number treats that number as a decimal list)

rAw€5¬S

TryItOnline!

How?

rAw€5¬S - Main link: from, to    e.g. -51, -44
r       - range(from, to)        e.g. [-51,-50,-49,-48,-47,-46,-45,-44]
 A      - absolute value         e.g. [51,50,49,48,47,46,45,44]
  w€    - first index of... for €ach (0 if not present)
    5   - five                   e.g. [1,1,0,0,0,0,2,0]
     ¬  - logical not            e.g. [0,0,1,1,1,1,0,1]
      S - sum                    e.g. 5

* The absolute value atom, A is necessary since a negative number cast to a decimal list has negative entries, none of which would ever be a 5 (the given example would count all eight rather than two).

Answer 3

Bash + grep, 17 bytes

seq $@|grep -cv 5

Try it online!

Answer 4

2sable, 6 5 bytes

Saved a byte thanks to Adnan

Ÿ5¢_O

Try it online!

Explanation

 Ÿ      # inclusive range
  5¢    # count 5's in each element of the range
    _   # negate
     O  # sum

Note: This works due to a bug in ¢ making the function apply itself to each element instead of counting matching elements in the list.

Answer 5

Python2, 59 55 52 51 47 43 42 bytes

f=lambda a,b:a<=b and-(`5`in`a`)-~f(a+1,b)

A recursive solution. Thanks to @xnor for giving me motivation to find a solution using logical operators! Also, thanks to @JonathanAllan and @xnor for guiding me and chopping the byte from 43 to 42!

Other attempts at 43 bytes

f=lambda a,b:a<=b and-~-(`5`in`a`)+f(a+1,b)
f=lambda a,b:a<=b and 1-(`5`in`a`)+f(a+1,b)

Answer 6

Pyth, 9 8 bytes

Saved a byte thanks to FryAmTheEggman!

lf-\5T}E

Explanation:

        Q # Input
      }E  # Form an inclusive range starting from another input
          #   order is reversed, but doesn't matter
 f-\5T    # Filter by absence of '5'
l         # Count the number of elements left

Try it online!

Answer 7

Bash / Unix utilities, 21 bytes

seq $*|sed /5/d|wc -l

Try it online!

Answer 8

05AB1E, 8 7 6 bytes

Saved a byte thanks to Adnan

Ÿ5.å_O

Try it online!

Explanation

Ÿ         # inclusive range
 5.å      # map 5 in y for each y in the list
    _     # logical negation 
     O    # sum

Answer 9

Haskell, 39 bytes

s!e=sum[1|x<-[s..e],notElem '5'$show x]

Try it online! Usage:

Prelude> 4 ! 17
12

Explanation:

             [s..e]                     -- yields the range from s to e inclusive
          x<-[s..e]                     -- for each x in this range
          x<-[s..e],notElem '5'$show x  -- if the char '5' is not in the string representation of x
       [1|x<-[s..e],notElem '5'$show x] -- then add a 1 to the resulting list      
s!e=sum[1|x<-[s..e],notElem '5'$show x] -- take the sum of the list

Answer 10

Perl 6, 23 bytes

{+grep {!/5/},$^a..$^b}

Try it online!

How it works

{                     }  # A lambda.
              $^a..$^b   # Range between the two lambda arguments.
  grep {!/5/},           # Get those whose string representation doesn't match the regex /5/.
 +                       # Return the size of this list.

Answer 11

PHP 7.1, 57 55 bytes

for([,$a,$b]=$argv;$a<=$b;)$n+=!strstr($a++,53);echo$n;

Run with php -r '<code>' <a> <b>

Answer 12

Groovy, 47 45 43 40 bytes

{a,b->(a..b).findAll{!(it=~/5/)}.size()}

This is an unnamed closure. findAll is similar to adding an if condition in a list comprehension in python.

Try it online!

Answer 13

Mathematica, 46 44 42 bytes

Thanks to alephalpha and DavidC for saving 2 bytes each!

Tr@Boole[FreeQ@5/@IntegerDigits@Range@##]&

Unnamed function taking two integer arguments and returning an integer. IntegerDigits@Range@## converts all the numbers between the inputs into lists of digits; FreeQ@5 tests those lists to decide which ones do not contain any 5. Then Boole converts booleans to zeros and ones, and Tr sums the results.

Other solutions (44 and 47 bytes):

Count[Range@##,x_/;IntegerDigits@x~FreeQ~5]&

IntegerDigits@x~FreeQ~5 determines whether the list of digits of a number is free of 5s, and Count[Range@##,x_/;...]& counts how many numbers between the inputs pass that test.

Tr[Sign[1##&@@IntegerDigits@#-5]^2&/@Range@##]&

1##&@@IntegerDigits@#-5 takes the list of digits of a number, subtracts 5 from all of them, and multplies the answers together; Sign[...]^2 then converts all nonzero numbers to 1.

Answer 14

R, 33 bytes

f=function(x,y)sum(!grepl(5,x:y))

Usage:

> f=function(x,y)sum(!grepl(5,x:y))
> f(40,60)
[1] 10
> f(1,9)
[1] 8
> f(4,17)
[1] 12

Answer 15

Octave, 36 bytes

@(m,n)sum(all(dec2base(m:n,10)'-52))

Try it online!

Answer 16

Ruby, 36 35 bytes

->a,b{(a..b).count{|x|!x.to_s[?5]}}

Thx IMP1 for -1 byte

Answer 17

Java 7, 80 78 bytes

int c(int a,int b){int r=0;for(;a<=b;)r+=(""+a++).contains("5")?0:1;return r;}

Ungolfed:

int c(int a, int b){
  int r = 0;
  for (; a <= b; ) {
    r += ("" + a++).contains("5")
          ? 0
          : 1;
  }
  return r;
}

Test code:

Try it here.

class M{
  static int c(int a,int b){int r=0;for(;a<=b;)r+=(""+a++).contains("5")?0:1;return r;}

  public static void main(String[] a){
    System.out.println(c(1, 9));
    System.out.println(c(4, 17));
  }
}

Output:

8
12

Answer 18

PowerShell, 42 41 bytes

param($a,$b)$a..$b|%{$z+=!($_-match5)};$z

Called from the command line as .\no5s.ps1 1 20

Answer 19

Swift 52 bytes

($0...$1).filter { !String($0).contains("5") }.count

Answer 20

Python 2, 61 56 bytes

lambda a,b:len([n for n in range(a,b+1) if not"5"in`n`])

-5 bytes thanks to tukkaaX

Answer 21

Batch, 95 bytes

@set/an=0,i=%1
:g
@if "%i%"=="%i:5=%" set/an+=1
@set/ai+=1
@if %i% leq %2 goto g
@echo %n%

Manually looping saves some bytes because I need the loop counter in a variable anyway.

Answer 22

Python 2, 54 bytes

i,j=input();k=0
while i<=j:k+=not"5"in`i`;i+=1
print k

Try it online!

Not the shortest Python answer Uses same algorithm but a different way of implementing with a while loop and is not a lambda function.

Answer 23

PHP, 56 bytes

for($i=$argv[1];$i<=$argv[2];)trim(5,$i++)&&$x++;echo$x;

Run like this:

php -r 'for($i=$argv[1];$i<=$argv[2];)trim(5,$i++)&&$x++;echo$x;' 1 9 2>/dev/null;echo
> 8

A version for PHP 7.1 would be 53 bytes (credits to Titus):

for([,$i,$e]=$argv;$i<=$e;)trim(5,$i++)&&$x++;echo$x;

Explanation

for(
  $i=$argv[1];          # Set iterator to first input.
  $i<=$argv[2];         # Loop until second input is reached.
)
  trim(5,$i++) && $x++; # Trim string "5" with the characters in the
                        # current number; results in empty string when
                        # `5` is present in the number. If that is not
                        # the case, increment `$x`

echo$x;                 # Output `$x`

Answer 24

CJam "easy pure mathematics solution", 60

{{Ab5+_,\_5#)<\9e]);_4f>.m9b}%}:F;q~_:z$\:*0>{((+F:-}{F:+)}?

Try it online

It takes the numbers in any order, in an array.

Explanation:

One core problem is to calculate f(n) = the number of non-5 numbers from 1 to n (inclusive) for any positive n. And the answer is: take n's decimal digits, replace all digits after the first 5 (if any) with 9, then replace all digits 5..9 with 4..8 (decrement), and convert from base 9. E.g. 1752 → 1759 → 1648 → 1*9^3+6*9^2+4*9+8=1259. Basically, each digit position has 9 acceptable values, and a 5xxxx is equivalent to a 49999 because there are no more valid numbers between them.

Once we solved this, we have a few cases: if the input numbers (say a and b, a<b) are (strictly) positive, then the result is f(b)-f(a-1). If they are negative, then we can take the absolute values, reorder them and use the same calculation. And if a<=0<=b then the result is f(-a)+f(b)+1.

The program first implements the function F as described above (but applied to each number in an array), then reads the input, converts the numbers to the absolute value and reorders them, and uses one of the 2 calculations above, based on whether a*b>0 initially.

Answer 25

C#, 77 bytes

(n,m)=>{var g=0;for(var i=n;i<m+1;i++)g+=(i+"").Contains("5")?0:1;return g;};

Anonymous lambda call.

Uses n (first number) and m (last number) as input, then checks via string containment ("".Contains("")).

Answer 26

C 143 141 132 122 bytes

f(a,b){j=0,k,f;for(;a<=b;a++){k=a;f=0;while(k!=0){if(abs(k%10)==5){f=1;break;}else f=0;k/=10;}if(f<1)j++;}printf("%d",j);}

This is a solution that adopts a mathematical approach and checks every digit for number 5. Can definitely be shortened!

Ungolfed version:

  void f(int a,int b)
  { 
       int j=0,k=0,f=0;

       for(;a<=b;a++)     
       { 
          k=a;f=0;

          while(k!=0)
          {          

            if(abs(k%10)==5) // If one of the digits in the number is 5, set a flag and break while loop, else reset the flag and continue checking.
            {
              f=1; 
              break;
            }  
            else
              f=0;

            k/=10; 

          }        

       if(f<1)
         j++; 
      }
      printf("%d ",j);       


  }

@TuukkaX Thanks for saving 2 bytes.

@nmjcman101 Thanks for saving 9 bytes.

@nmjcman101 Thanks for saving 10 bytes more, make so much sense. :)

Answer 27

TI-BASIC, 37 34 bytes

:Prompt A,B
:Delvar CFor(N,A,B
:C+not(inString(toString(N),"5→C
:End
:C

I believe count is correct, however toString() is only supported on the TI-84+ CE calculators which I do not have, so I was unable to count it or test it to make sure it runs correctly. I therefore counted it as 2 bytes in addition to the others which were counted directly on a TI-84.

Thx to Jakob Cornell for removing

Answer 28

JavaScript (ES6), 58 56 49 bytes

let f =

(s,e)=>{for(c=0;s<=e;)c+=!/5/.test(s++);return c}

console.log(f(1, 9));
console.log(f(4, 17));
console.log(f(-9, -1));

Golfed 7 bytes thanks to ETHproductions.

Answer 29

Java 7, 77 bytes

This is an improvement of Kevins Answer, but since I don't have the reputation to comment yet, this new answer will have to do.

So what I did was:

• Replace the indexOf statements with contains (-1 byte)
• Move the incrementing part of the for-loop into the conditional statement (-2 bytes)

---

for-loop (77 bytes):

int c(int a,int b){int r=1;for(;a++<b;)r+=(""+a).contains("5")?0:1;return r;}

recursive (79 bytes):

int d(int r,int a,int b){r+=(""+a).contains("5")?0:1;return a!=b?d(r,a+1,b):r;}

Output:

8
12

8
12

Test it here !

Answer 30

C#, 67 bytes

a=>b=>{int c=0;for(;a<=b;)c+=(a+++"").Contains("5")?0:1;return c;};

Answer 31

MATL, 10 bytes

&:!V53-!As

Try it online!

Explanation

        % Implicitly grab two input arguments
&:      % Create an array from [input1....input2]
!V      % Convert to a string where each number is it's own row
53-     % Subtract ASCII '5' from each character.
!A      % Detect which rows have no false values (no 5's). Returns a logical array
s       % Sum the logical array to get the # numbers without 5's
        % Implicitly display the result

Answer 32

PowerShell, 37 36 bytes

($args[0]..$args[1]-notmatch5).Count

Try it online!

Answer 33

Actually, 13 bytes

u@x`$'5íuY`░l

Try it online!

Explanation:

u@x`$'5íuY`░l
u@x            range(a, b+1)
   `$'5íuY`░   take where:
    $            string representation
     '5íuY       does not contain "5"
            l  length

Answer 34

C++ | in too many bytes, 165 125 thanks to Christoph!

int main(){int c=0;for(int i=0;i<=8;i++){int d;int n=i>=0?i:-i;while(n!=0){if(d=n%10==5){break;}n=n/10;c++;break;}}return c;}   

I took the liberty of creating a function e_ to determine if a 5(or any other number) is present in an integer instead of using to_string() and .find() so that must count for something. note: e_ is only declared as an extra function in the un-golfed version for readability.

un-golfed:

int e_(int e,int i){
    int d;
    int n=i>=0?i:-i;
    while (n != 0){
        d=n%10;
        if (d==e){
        return 1;}
        n=n/10;}
        return 0;}
int main() {
    int l = 1;int h = 8;int e = 5;int c = 0;
    for(int i=l; i<=h; i++){
        if (e_(e,i)==0)     
        c++;}
    return c;}

How e_ function works:

int n=i>=0?i:-i; Inverses our number if it is less than 0 so it's always positive. d=n%10; Divides it by 10 and gets remainder (n%base; will always return the last digit of an integer). We check that it equals 5, if it does the number can be discarded, if not n=n/10; removes the end digit and loops again.

Answer 35

CJam, 18 17 bytes

q~:S-){S+s'5&!},,

Input is end number followed by start number.

Try it online!

How it works

q~                  Read and evaluate all input. Pushes end and start.
  :S                Save the start in S.
    -)              Subtract and increment, computing L =: end - start + 1.
      {       },    Filter [0 ... L-1]; for each N in the range:
       S+             Add S to N.
         s            Cast to string.
          '5&         Intersect with '5'.
             !        Take the logical NOT.
                ,   Compute the length of the resulting array.

Answer 36

C++, 84 81 + 16 = 100 97 bytes

int f(int a,int b){int c;while(a<=b)c+=!~std::to_string(a++).find('5');return c;}

#include<string> - +16

Ungolfed:

int func(int start, int end) {
  // Defaults to 0
  int count;

  // Iterates from start to end (inclusive) and increments the count by one
  // if '5' is not found
  while (start <= end)
    count += !~std::to_string(start++).find('5');

  return count;
}

The function itself should be pretty clear. I essentially just iterate from start (or a) to end (or b). The only more complex line is this:

count += !~std::to_string(start++).find('5');

However it is also easiy explained. find returns std::npos which is the maximum value size_t can hold (exact value depends on how the compiler defines it) when the character cannot be found. Which essentially means the it is an integer value filled with binary ones. ~ performs a binary not, meaning that the value is 0 when no 5 character could be found and not 0 when it could. Then ! converts it to a bool (0 => false, everything else => true) and itverts it. So now when the 5 could be found the value is false and true if it could not. Then it gets added to the count variable. (true => 1, false => 0).

Try it online!

@Christoph Thanks for giving me the idea to save 3 bytes

Answer 37

Perl 5, 29 bytes

perl -le 'print 0+grep!/5/,shift..shift' 4 17               #prints 12

Answer 38

C#, 82 81 bytes

using System.Linq;a=>b=>Enumerable.Range(a,++b-a).Count(n=>(n+"").All(d=>d!=53));

Added as a separate answer as it's logic is different.

Answer 39

SimpleTemplate, 63 bytes

This was harder than expected.

Expects each number as a single parameter in the class, outputting the number of element without 5.

{@eachargv as_}{@if"~5~"is not matches_}{@incX}{@/}{@/}{@echoX}

Ungolfed

{@each argv as argument}
    {@if "~5~"is not matches argument}
        {@inc result}
    {@/}
{@/}
{@echo result}

Answer 40

Perl, 35 20+1 bytes

$\+=!/5/ for$_..<>}{

After considerable help from @Dada in the comment below, this turned into a beautiful 20 bytes - plus 1 for the p. I've left my original answer below, for posterity.

for(shift..shift){$z+=!/5/}print$z;

Call from command line as no5s.pl 1 20.

First Perl golf, so hopefully someone can improve it :)

Also using Strawberry Perl on Windows, so I can run no5s.pl 1 20 directly on the command line and it works - might try using perl -e ...

Answer 41

Clojure, 73 bytes

(fn[i a](count(for[n(range i(inc a)):when(every? #(not=\5 %)(str n))]n)))

I had to rollback my previous "improvement", as somehow count was dropped from the new version, and it ended up being longer when I fixed it.

Filters the range of numbers; selecting only the numbers where every digit is not a 5.

(defn count-minus-5 [mi ma]
  (count ; Get the length of the resulting list
    ; Comprehension over the range mi(n) to ma(x).
    (for [n (range mi (inc ma))
          ; Only allow the number when every digit isn't a 5.
          :when (every? #(not= \5 %) (str n))]
      n)))

Answer 42

QBIC, 33 32 bytes

Saved a byte by reversing the conditional and moving the increment to ELSE

::[a,b|~instr(!c$,@5`)|\d=d+1}?d

Explanation:

::       get a and b from the command line
[a,b|    FOR c = a; c <= b; c++
~instr(  IF indexOf ( instr is a QBasic function that doesn't have a QBIC equivalent)
    !c$  our loop iterator cast to string
    ,@5` a literal 5
)        instr returns 0 if it didn't find '5', which is truthy.
|        THEN --> Don't do anything for values with a '5'
\d=d+1   ELSE count this number into the total
}        Close all constructs (END IF, NEXT c)
?d       Print 'd': the total number of numbers without a '5'

Answer 43

Perl 6, 21 bytes

+(+*..+*).grep:{!/5/}

Basically equivalent to smls's solution, but I was able to shave off a couple of bytes by expressing the range as (+*..+*) and eliminating the brackets. (The plusses are necessary, otherwise the stars would be interpreted as negative or positive infinity rather than as arguments to the WhateverCode lambda.)

Answer 44

Scala, 51 bytes

(a:Int,b:Int)=>a to b map{_+""indexOf "5"}count(0>)

I haven't been able to concisely combine the map and count.

Answer 45

Clojure, 64 bytes

#(count(for[i(range % (inc %2)):when(not(some #{\5}(str i)))]i))

Answer 46

C++ (function), 115

int f(int a,int b){int r=b-a+1,t;while((t=a++)<=b)while(t!=0)if(abs(t)%10==5){r--;break;}else t/=10;std::cout<<r;}

C++ (full), 143

#include<iostream> 
int main(){int a=0,b=4,r=b-a+1,t;while((t=a++)<=b)while(t!=0)if(abs(t)%10==5){r--;break;}else t/=10;std::cout<<r;return 0;}

Ungolfed version:

#include<iostream> 
int main()
{
    int a=0, b=4, r = b-a+1, t;
    while((t = a++) <= b)
        while(t != 0)
            if(abs(t) % 10 == 5)
            {
                r--;
                break;
            }
            else 
                t /= 10;
    std::cout<<r;
    return 0;
}

Answer 47

Java 8, 68 bytes

(a,b)->IntStream.range(a,b+1).map(i->(""+i).contains("5")?0:1).sum()

This is basically just a rewrite of the Java7 solution by @Kevin Cruijssen to Java8. And I hope we don't care about imports here, otherwise it would be longer by 17 bytes due to the package).

Ungolfed:

(a,b)->IntStream.range(a,b+1)
  .map(i->(""+i).contains("5")?0:1)
  .sum()

Test code:

Try it here.

import java.util.function.BiFunction;
import java.util.stream.IntStream;

class M {
  public static void main (String[] args) {
    BiFunction<Integer, Integer, Integer> dontGimme5 = 
      (a,b)->IntStream.range(a,b+1).map(i->(""+i).contains("5")?0:1).sum();
    System.out.println(dontGimme5.apply(1,9));
    System.out.println(dontGimme5.apply(4,17));
    System.out.println(dontGimme5.apply(-50,-59));
  }
}

Output:

8
12
0

Answer 48

C++, 187 bytes

int f(int a,int b){int r=0;for(int i=a;i<=b;i++){r++;for(int j=0;j<floor(log(abs(i)));j++){int k=floor(abs(i)%int(pow(10,j+1))/pow(10,j));if(k==5){r--;break;}}}return r;}

#include <cmath> - 16 bytes

I couldn't think of a purely mathematical way, so I did this.

1. Loop through a to b
2. Calculate the length with floor(log(abs(i)))
3. Get numbers of each digit with floor(abs(i)%int(pow(10, j + 1)) / pow(10,j))
4. If it is 5, subtract then move to next number

Still, isn't practical for code-golf, but used much math as I could.

Try it online!

Answer 49

TI-Basic, 43 bytes

Since linear regression to string is so costly, let's try a different approach.

DelVar AInput 
For(I,X,Y
1
For(J,1,9
Ans and 5≠int(10fPart(I/10^(J
End
A+Ans→A
End

Old method, 55 bytes

Majority of the program size is converting number to string through linear regression... why wasn't there a built-in for this?

DelVar AInput 
For(I,X,Y
{0,1→L₁
{0,A→L₂
LinReg(ax+b) Y₁
Equ►String(Y₁,Str1
A+not(inString(Str1,"5→A
End

P.S. (For both methods:) Since the last statement evaluated by the program is that seventh line, it will return the value as normal through Ans. Also, Input stores to X and Y similar to Prompt X,Y.

Answer 50

SmileBASIC, 55 bytes

INPUT S,E
FOR I=S TO E
INC N,INSTR(STR$(I),"5")<0NEXT?N

Nothing special, just uses INSTR and STR$ to check for 5.

Answer 51

Hoon, 110 94 bytes

Hoon's range function, gulf, doesn't work for signed integers, which increases the length by a bit :(

=+
si^f=:(curr lien test 53)
|=
{a/@s b/@s}
|-
?:
=(a b)
(f <b>)
(add (f <a>) $(a (sum -1 a)))

Use the signed integer library. Create a function f: :(a b c d) is a macro that expands into (a b (a c d)) so this is (curr lien (curr test 53)), aka create a curried function that tests if any element of a list is 53 ('5')

Create a function that takes a and b. Create a loop: if a==b return f(tostring(b)), else return add(f(tostring(a)) recurse(a=a+1))

> =f =+
  si^f=:(curr lien test 53)
  |=
  {a/@s b/@s}
  |-
  ?:
  =(a b)
  (f <b>)
  (add (f <a>) $(a (sum -1 a)))
> (f -1 -9)
8
> (f -4 -17)
12
> (f -50 -60)
1
> (f --59 --50)
0

(Hoon's signed integers use - as a prefix, so --5 is negative 5 and -5 is positive 5)