38
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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.

|improve this question
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  • 3
    \$\begingroup\$ @betseq: That´s close; but this one has a variable range (and requires no modulo). \$\endgroup\$ – Titus Jan 20 '17 at 5:42
  • 4
    \$\begingroup\$ I'd recommend shortest code as winning criterion and the code-golf tag (I didn't even spot that it wasn't!). Also, you should probably should put a test case that spans 50 or 500; also maybe one that spans -50, and one that spans 0 would be a good idea. \$\endgroup\$ – Jonathan Allan Jan 20 '17 at 5:45
  • 1
    \$\begingroup\$ @JonathanAllan : I will update examples. \$\endgroup\$ – Arasuvel Jan 20 '17 at 5:51
  • 4
    \$\begingroup\$ Test case: 50, 59 -> 0. \$\endgroup\$ – Zgarb Jan 20 '17 at 12:28
  • 14
    \$\begingroup\$ You say: "The start number will always be smaller than the end number." but one of your examples (-50,-59) directly contradicts this \$\endgroup\$ – theonlygusti Jan 21 '17 at 10:36

51 Answers 51

1 2
21
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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))

|improve this answer
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  • \$\begingroup\$ (I normally prefer test over exec when you only need a boolean.) \$\endgroup\$ – Neil Jan 20 '17 at 9:01
  • \$\begingroup\$ @Neil That makes more sense indeed. Updated. \$\endgroup\$ – Arnauld Jan 20 '17 at 9:07
  • \$\begingroup\$ NB: I couldn't find any tip about ES6 currying syntax, so I wrote one. \$\endgroup\$ – Arnauld Jan 20 '17 at 10:51
  • 5
    \$\begingroup\$ @TheLethalCoder b<a is there to stop the recursion after counting through all numbers from b to a, so removing it would just cause an infinite recursion. \$\endgroup\$ – ETHproductions Jan 20 '17 at 12:37
  • 1
    \$\begingroup\$ @HristiyanDodov The unnamed outer function takes a as argument and returns the F function, which in turn takes b as argument and -- as you noticed -- is called recursively to iterate from b to a, incrementing a counter for all integers that do not contain a 5 in their decimal representation. \$\endgroup\$ – Arnauld Jan 20 '17 at 14:06
17
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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).

|improve this answer
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  • \$\begingroup\$ rAw€5¬S saves a byte. \$\endgroup\$ – Dennis Jan 20 '17 at 8:22
  • \$\begingroup\$ @Dennis thanks! Is my description "treats that number as a decimal list" accurate? \$\endgroup\$ – Jonathan Allan Jan 20 '17 at 8:36
  • 2
    \$\begingroup\$ Pretty much. w casts an integer argument to its decimal digits. \$\endgroup\$ – Dennis Jan 20 '17 at 8:37
13
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Bash + grep, 17 bytes

seq $@|grep -cv 5

Try it online!

|improve this answer
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13
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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.

|improve this answer
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  • \$\begingroup\$ You can remove the ` as it behaves the same on arrays :p. \$\endgroup\$ – Adnan Jan 20 '17 at 11:34
  • \$\begingroup\$ @Adnan: Thanks! I was gonna test that but forgot ;) \$\endgroup\$ – Emigna Jan 20 '17 at 12:45
9
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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)
|improve this answer
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  • \$\begingroup\$ Would if!`x`.count('5') work? \$\endgroup\$ – Titus Jan 20 '17 at 6:29
  • 2
    \$\begingroup\$ @Titus Python has not operator that is ! in C-like languages, but that takes 3 bytes :( \$\endgroup\$ – Yytsi Jan 20 '17 at 6:31
  • 1
    \$\begingroup\$ Think about using logical short-circuiting with and and or. \$\endgroup\$ – xnor Jan 20 '17 at 7:37
  • 1
    \$\begingroup\$ Yup, nicely done! Now think about shortening that not. \$\endgroup\$ – xnor Jan 20 '17 at 7:48
  • 1
    \$\begingroup\$ You're really close! Keep trying stuff. \$\endgroup\$ – xnor Jan 20 '17 at 8:28
6
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Bash / Unix utilities, 21 bytes

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

Try it online!

|improve this answer
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6
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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
|improve this answer
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  • \$\begingroup\$ 05AB1E also has vectorized å, which is .å, so you can do Ÿ5.å_O for 6 bytes. \$\endgroup\$ – Adnan Jan 20 '17 at 11:24
  • \$\begingroup\$ negate meaning -n, or n==0?1:0? \$\endgroup\$ – ETHproductions Jan 20 '17 at 12:38
  • \$\begingroup\$ @ETHproductions: Sorry, that was unclear. I meant logical negation, so n==0?1:0 \$\endgroup\$ – Emigna Jan 20 '17 at 12:46
6
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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!

|improve this answer
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5
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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.
|improve this answer
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5
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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
|improve this answer
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4
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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
|improve this answer
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4
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Octave, 36 bytes

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

Try it online!

|improve this answer
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4
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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!

|improve this answer
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4
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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>

|improve this answer
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  • \$\begingroup\$ Isn't this PHP7.1 syntax? \$\endgroup\$ – aross Jan 20 '17 at 10:06
  • \$\begingroup\$ @aross: It is. But PHP 7.1 is older than 5 hours (pubished on December, 1) \$\endgroup\$ – Titus Jan 20 '17 at 10:50
  • 1
    \$\begingroup\$ of course, I just asked because I'm used to specifying the version if it's 7 or up. That's also kind of the convention for Python \$\endgroup\$ – aross Jan 20 '17 at 11:00
  • 1
    \$\begingroup\$ Convention for PHP - as far as I have seen - is to use the most recent version unless specified otherwise. \$\endgroup\$ – Titus Jan 20 '17 at 16:12
  • \$\begingroup\$ I don't think many people have the latest minor version. The least common denominator at the moment would probably be 5.5. Personally I'm using FC 25 (considered pretty cutting edge), which currently distributes PHP 7.0. If you're on Windows you probably need to update manually. \$\endgroup\$ – aross Jan 20 '17 at 16:17
4
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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.

|improve this answer
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  • 1
    \$\begingroup\$ Count[Range@##,x_/;IntegerDigits@x~FreeQ~5]& \$\endgroup\$ – DavidC Jan 20 '17 at 12:59
  • 1
    \$\begingroup\$ Tr@Boole[FreeQ@5/@IntegerDigits@Range@##]& \$\endgroup\$ – alephalpha Jan 21 '17 at 3:42
3
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Ruby, 36 35 bytes

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

Thx IMP1 for -1 byte

|improve this answer
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  • 1
    \$\begingroup\$ Doesn't this return the list without the numbers containing 5, rather than the size of that list? \$\endgroup\$ – IMP1 Jan 20 '17 at 10:49
  • \$\begingroup\$ You are right, I have copy/pasted the wrong version. \$\endgroup\$ – G B Jan 20 '17 at 11:00
  • 1
    \$\begingroup\$ You can also use ?5 (the '5' character) instead of /5/ in the search to save a byte. \$\endgroup\$ – IMP1 Jan 20 '17 at 11:03
3
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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
|improve this answer
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3
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PowerShell, 42 41 bytes

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

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

|improve this answer
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  • 1
    \$\begingroup\$ You can drop the space to save a byte. With strictly numerical regex patterns, you don't need a delimiter (e.g., -replace3 or -split1 or -notmatch5). \$\endgroup\$ – AdmBorkBork Jan 20 '17 at 15:18
  • \$\begingroup\$ Ah, nice, thanks @AdmBorkBork \$\endgroup\$ – mcmurdo Jan 20 '17 at 20:46
2
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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

|improve this answer
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  • \$\begingroup\$ Don't get discouraged! Having fun and challenging yourself is what matters. You can remove two whitespaces at not "5" in :) Also, if you're using Python2, you can surround x with `` quotes, instead of doing str(x). \$\endgroup\$ – Yytsi Jan 20 '17 at 6:59
  • \$\begingroup\$ @TuukkaX Thanks! also removed space between in and `x` \$\endgroup\$ – sagiksp Jan 20 '17 at 7:52
  • \$\begingroup\$ You can remove the []. You also don't need the space before if. \$\endgroup\$ – Dennis Jan 20 '17 at 8:16
  • \$\begingroup\$ @Dennis I tried that already, but it complains that "object of type 'generator' has no len()". \$\endgroup\$ – Yytsi Jan 20 '17 at 8:26
  • \$\begingroup\$ @TuukkaX Right. lambda a,b:sum(not"5"in`n`for n in range(a,b+1)) works though. tio.run/nexus/… \$\endgroup\$ – Dennis Jan 20 '17 at 8:29
2
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Swift 52 bytes

($0...$1).filter { !String($0).contains("5") }.count
|improve this answer
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  • \$\begingroup\$ Since your challenge is a codegolf challenge, you should include your bytecount. Also, in codegolf (at least here), it's a requirement that all programs muse be actually contending (e.g. your function name can be just a single character, your actual function can probably be reduced to a single line). I don't know Swift, you might have to correct me on stuff. \$\endgroup\$ – clismique Jan 20 '17 at 5:53
2
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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.

|improve this answer
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2
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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`
|improve this answer
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  • \$\begingroup\$ Ah dang I forgot about the second trim parameter again. \$\endgroup\$ – Titus Jan 20 '17 at 10:51
2
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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.

|improve this answer
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  • \$\begingroup\$ Not "pure" but nice method. here, get a +1 :) \$\endgroup\$ – Matthew Roh Jan 23 '17 at 9:13
  • \$\begingroup\$ @MatthewRoh thanks, but what do you mean not pure? It's a solution that does fairly direct mathematical calculations on the input numbers, without iterating through the range. What else were you expecting? \$\endgroup\$ – aditsu Jan 23 '17 at 9:35
2
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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.

|improve this answer
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  • \$\begingroup\$ It is a program and not a function and it uses while instead of for. What is not different? OK, it is still looking for a string "5" inside the incremented input, agreed. Is there a better way? \$\endgroup\$ – ElPedro Jan 20 '17 at 13:10
  • \$\begingroup\$ That's exactly what it is and that's why it is deferent. Sorry, maybe should have made my comment different. \$\endgroup\$ – ElPedro Jan 20 '17 at 18:34
  • \$\begingroup\$ Same algorithm, different way of implementing. No problem with your comments. Is that better worded? \$\endgroup\$ – ElPedro Jan 20 '17 at 18:36
  • \$\begingroup\$ It is, yes :) I'll remove these comments to make the comment section look clean. \$\endgroup\$ – Yytsi Jan 20 '17 at 20:58
1
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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 !

|improve this answer
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  • \$\begingroup\$ Wellcome to PPCG ! Nice findings in an already quite nicely golfed answer. I don't know about Java that much but shouldn't (""+a).contains("5")?0:1 be replacable by !(""+a).contains("5")? \$\endgroup\$ – Christoph Jan 27 '17 at 6:56
  • 1
    \$\begingroup\$ @Christoph sadly no, since in Java a boolean really is just a boolean. So a ternary operation is the only way to go. \$\endgroup\$ – Tobias Meister Jan 27 '17 at 9:13
  • \$\begingroup\$ Hm that's sad. What about (""+a).contains("5")||r++? \$\endgroup\$ – Christoph Jan 27 '17 at 9:15
  • 1
    \$\begingroup\$ @Christoph that won't work either, because you can't have a boolean expression on its own. I've been trying to make it work in other places (like the for-loop declaration) but not with much success. Nice idea tho ;) \$\endgroup\$ – Tobias Meister Jan 27 '17 at 13:45
1
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C#, 67 bytes

a=>b=>{int c=0;for(;a<=b;)c+=(a+++"").Contains("5")?0:1;return c;};
|improve this answer
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  • \$\begingroup\$ I was hoping to use for(int c=0;...) but then it fails to compile because the return is outside the scope for c \$\endgroup\$ – TheLethalCoder Jan 20 '17 at 12:48
1
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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.

|improve this answer
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  • 1
    \$\begingroup\$ You can use c+=!/5/.test(s++) to save a few bytes :-) \$\endgroup\$ – ETHproductions Jan 20 '17 at 12:41
  • \$\begingroup\$ Thanks a lot! I had to delete my golfs, though. I was so proud of them. :( \$\endgroup\$ – Hristiyan Dodov Jan 20 '17 at 12:55
  • \$\begingroup\$ I think you can use currying i.e.` s=>e=>` \$\endgroup\$ – TheLethalCoder Jan 20 '17 at 12:58
  • \$\begingroup\$ The top answer uses currying syntax. I won't edit mine because it would become almost the same. Thanks for pointing that out, though! \$\endgroup\$ – Hristiyan Dodov Jan 20 '17 at 13:01
1
\$\begingroup\$

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
|improve this answer
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1
\$\begingroup\$

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("")).

|improve this answer
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  • \$\begingroup\$ I'm not the one downvoting, but modulo 5 isn't the correct solution for the challenge given by OP. It should exclude anything containing the digit 5 in its number, so 10 (which your answer wouldn't count) should be counted. \$\endgroup\$ – Kevin Cruijssen Jan 20 '17 at 8:41
  • \$\begingroup\$ @KevinCruijssen Fixed. \$\endgroup\$ – devRicher Jan 20 '17 at 11:02
  • \$\begingroup\$ This doesn't compile as g must be initialised when stated as it is named var so you need var g=""; and you can use currying i.e. n=>m=> \$\endgroup\$ – TheLethalCoder Jan 20 '17 at 12:30
  • \$\begingroup\$ Also this outputs the list not the count \$\endgroup\$ – TheLethalCoder Jan 20 '17 at 12:32
  • 1
    \$\begingroup\$ @KevinCruijssen With your edits this is essentially my answer... \$\endgroup\$ – TheLethalCoder Jan 20 '17 at 14:53
1
\$\begingroup\$

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
|improve this answer
\$\endgroup\$
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