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

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Answer 4

2sable, 6 5 bytes

Saved a byte thanks to Adnan

Ÿ5¢_O

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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

Bash / Unix utilities, 21 bytes

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

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Answer 7

05AB1E, 8 7 6 bytes

Saved a byte thanks to Adnan

Ÿ5.å_O

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Explanation

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

Answer 8

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 9

Perl 6, 23 bytes

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

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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 10

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 11

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 12

Octave, 36 bytes

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

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Answer 13

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 14

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 15

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 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

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 20

Swift 52 bytes

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

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

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 23

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 24

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 25

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 26

C#, 67 bytes

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

Answer 27

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 28

MATL, 10 bytes

&:!V53-!As

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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 29

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 30

Actually, 13 bytes

u@x`$'5íuY`░l

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Explanation:

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