# Zeroes in interval

Your task is to write a function or program that takes two non-negative integers i and k (ik), and figure out how many zeroes you'd write if you wrote all whole numbers from i to k (inclusive) in your base of choice on a piece of paper. Output this integer, the number of zeroes, to stdout or similar.

-30% if you also accept a third argument b, the integer base to write down the numbers in. At least two bases must be handled to achieve this bonus.

• You may accept the input in any base you like, and you may change the base between test cases.
• You may accept the arguments i, k and optionally b in any order you like.
• Answers must handle at least one base that is not unary.

Test cases (in base 10):

i k -> output
10 10 -> 1
0 27 -> 3
100 200 -> 22
0 500 -> 92


This is code-golf; fewest bytes win.

• If you can go with whatever base you'd like from case to case, couldn't you do each in base k and print 0 or 1, depending on whether i = 0? – StephenTG Jan 29 '16 at 20:23
• You might want to exclude unary as a base, or else this problem is trivial: get inputs, print 0. – user45941 Jan 29 '16 at 20:29
• Can you add some test cases for other bases? – Morgan Thrapp Jan 29 '16 at 20:34
• I think this would be more interesting if the base argument were required. "Base of your choice" is weird to me. – Alex A. Jan 29 '16 at 20:34
• Yes, @AlexA. but too late to change that now, 10 answers in. – Filip Haglund Jan 29 '16 at 20:53

## Jelly, 1 byte

¬


This uses base k+2, in which case there's a single 0 iff i is 0. It takes two arguments, but applies the logical NOT to only the first one.

If we don't want to cheat:

### 7 bytes - 30% = 4.9

-1.1 points by @Dennis

rb⁵$¬SS  This gets the bonus.  dyadic link: r inclusive range b⁵$           Convert all to base input.
¬          Vectorized logical NOT
S         Sum up 0th digits, 1st digits, etc.
S        Sum all values

• This is the second Jelly program I've written on my phone. – lirtosiast Jan 29 '16 at 20:44
• Damn, 1 byte? Give us a chance. – Rɪᴋᴇʀ Jan 29 '16 at 21:09
• This can be easily done in very few bytes in any other language. I say stick to the non-cheating version. – ETHproductions Jan 29 '16 at 21:13
• @ETHproductions The rules of the question explicitly allow doing this. Cheaty or not, it's the answer the rules call for. – Dennis Jan 29 '16 at 21:17

# Python 2, 36 bytes

lambda a,b:range(a,b+1).count('0')


Credit to muddyfish for the  trick.

• Welcome to Programming Puzzles & Code Golf! This is a nice first answer. :) – Alex A. Jan 30 '16 at 3:58
• Wow! I didn't know it's working! – Dantal Jan 31 '16 at 2:23

# 05AB1E, 3 1 byte

Uses base k+2 like the Jelly answer, Code:

_


Explanation:

_  # Logical NOT operator


3 byte non-cheating version:

Code:

Ÿ0¢


Explanation:

Ÿ    # Inclusive range
0¢  # Count zeroes


The bonus gives me 3.5 bytes due to a bug:

ŸB)0¢


Explanation:

Ÿ      # Inclusive range
B     # Convert to base input
)    # Wrap into an array (which should not be needed)
0¢  # Count zeroes


Uses CP-1252 encoding.

• How does this work? – lirtosiast Jan 29 '16 at 21:17

# Japt, 3 bytes

+!U


Uses base k+2, as the Jelly answer. There is a zero iff i==0. Test it online!

### Better version, 10 8 bytes

UòV ¬è'0


This one uses base 10. Test it online!

### Bonus version, 14 12 bytes - 30% = 8.4

UòV msW ¬è'0


Sadly, with the golfing I did, the bonus is no longer worth it... Test it online!

### How it works

UòV msW ¬è'0   // Implicit: U = start int, V = end int, W = base
UòV            // Create the inclusive range [U..V].
msW        // Map each item by turning it into a base-W string.
¬      // Join into a string.
è'0   // Count the number of occurances of the string "0".


## ES6, 91 86 - 30% = 60.2 bytes

(i,k,b=10)=>([...Array(k+1-i)].map((_,n)=>(i+n).toString(b))+'0').match(/0/g).length-1


Or save 3 (2.1) bytes if b doesn't need to default to 10.

Best non-bonus version I could do was 65 bytes:

(i,k)=>([...Array(k+1).keys()].slice(i)+'0').match(/0/g).length-1


Edit: Saved 5 bytes by using @edc65's zero-counting trick.

• As I don't manage to get votes for my answer, I'll upvote yours (at least there my name inside) – edc65 Jan 30 '16 at 21:11

## Seriously, 10 bytes

'0,,u@xεjc


Explanation:

'0,,u@xεjc
'0,,u       push "0", i, k+1
@x     swap i and k+1, range(i, k+1)
εjc  join on empty string and count 0s


Try it online!

With bonus: 11.9 bytes

'0,,u@x,╗╜@¡Mεjc


Try it online!

Explanation:

'0,,u@x,╗╜@¡MΣc
'0,,u@x             push "0", range(i, k+1)
,╗           push b to register 0
   M     map:
╜@¡         push b, push string of a written in base b
Σc   sum (concat for strings), count 0s


# CJam, 1210 3 bytes

li!


This uses the shortcut @ThomasKwa does.

If this is not allowed, then here is a 10 byte answer.

q~),>s'0e=


Nice and short! Works like @Mego's Seriously answer.

Thanks @Dennis!

Try it here!

# T-SQL, 394 Bytes (No bonus)

I figure 'why not', right?

DECLARE @i INT, @k INT SET @i = 100 SET @k = 200  WITH g AS (SELECT @i AS n UNION ALL SELECT n+1 FROM g WHERE n+1<=@k ) SELECT LEN(n) AS c FROM (SELECT STUFF((SELECT REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(n, 1, ''), 2, ''), 3, ''), 4, ''), 5, ''), 6, ''), 7, ''), 8, ''), 9, ''), ' ', '') FROM g FOR XML PATH ('')) ,1,0,'') n ) a OPTION (maxrecursion 0)


And the friendly one:

-- CG!

DECLARE @i INT, @k INT
SET @i = 100
SET @k = 200

WITH g AS
(
SELECT @i AS n
UNION ALL
SELECT n+1 FROM g WHERE n+1<=@k
)

SELECT LEN(n) AS c FROM
(
SELECT
STUFF((SELECT REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(REPLACE(n, 1, ''), 2, ''), 3, ''), 4, ''), 5, ''), 6, ''), 7, ''), 8, ''), 9, ''), ' ', '')
FROM g FOR XML PATH ('')) ,1,0,'') n
) a

OPTION (maxrecursion 0)

• that's dedication. – colsw Mar 23 '17 at 16:31

# Ruby, 46 - 30% = 32.2 bytes

You could probably golf this more, but at least I get the 30% bonus!

->i,k,b{((i..k).map{|a|a.to_s b}*"").count ?0}


...or without the bonus (27 bytes.)

->i,k{([*i..k]*"").count ?0}


Tips are welcome, still learning this whole "Ruby" thing.

• Nice answer, you don't need the splat operator when using map, this could save 1 byte. (i..k) is as good as [*i..k] in the first case. – G B Mar 23 '17 at 13:11

# Brachylog, 26 bytes

,{,.e?}?:1frcS:0xlI,Sl-I=.


Takes input as a list [i,k].

### Explanation

,{    }?:1f                § Unify the output with a list of all inputs which verify the
§ predicate between brackets {...} with output set as the input
§ of the main predicate

,.e?                     § Unify the input with a number between i and k with the ouput
§ being the list [i,k]

rcS             § Reverse the list and concatenate everything into a single
§ number (we reverse it to not lose the leading 0 if i = 0 when
§ we concatenate into a single number). Call this number S.

:0xlI        § Remove all occurences of 0 from S, call I the length of this new
§ number with no zeros

,Sl-I=. § Output the length of S minus I.


# Julia, 48 bytes - 30% = 33.6

f(i,k,b)=sum(j->sum(c->c<49,[base(b,j)...]),i:k)


This is a function that accepts three integers and returns an integer. One of the arguments specifies the base, so this qualifies for the bonus.

Ungolfed:

function f(i, k, b)
# For each j in the inclusive range i to k, convert j to base
# b as a string, splat the string into a character array, and
# compare each character to the ASCII code 49 (i.e. '1'). The
# condition will only be true if the character is '0'. We sum
# these booleans to get the number of zeros in that number,
# then we sum over the set of sums to get the result.
sum(j -> sum(c -> c < 49, [base(b, j)...]), i:k)
end


Implementing the bonus yields a score just barely better than the not implementing it (34 bytes):

f(i,k)=sum(c->c<49,[join(i:k)...])


# Seriously, 2 bytes

This might be taking the Jelly answer trick to the limit, but here is a simple 2 byte Seriously answer.

,Y


Try it online!

# Pyth, 6.3 bytes, with bonus (9 bytes - 30%)

/sjRQ}EE0


Explanation:

  jRQ     - [conv_base(Q, d) for d in V]
}EE  - inclusive_range(eval(input), eval(input))
s        - sum(^, [])
/       0 - ^.count(0)


Try it here

Or 7 bytes without the bonus:

/}EE\0


Explanation:

  }EE   - inclusive_range(eval(input), eval(input))
- repr(^)
/    \0 - ^.count("0")


Try it here

Or use a test suite

• I think getting the bonus is worth it: /sjRQ}EE0 – FryAmTheEggman Jan 29 '16 at 20:48
• Ehh, it's the same code with a base conversion, I'm pretty sure you know what you're doing, just the problem of a bonus forcing you to try different stuff and count... :P – FryAmTheEggman Jan 29 '16 at 21:17

# PHP, 50 Bytes

supports decimal only

<?=substr_count(join(range($argv,$argv)),0);


<?=substr_count(join(array_map([2=>decbin,10=>""][$argv],range($argv,$argv))),0);  ## supports decimal,hexadecimal, octal and binary with Bonus 77.7 <?=substr_count(join(array_map([2=>decbin,8=>decoct,10=>"",16=>dechex][$argv],range($argv,$argv))),0);


<?=substr_count(join(array_map(function($i){return base_convert($i,10,$_GET);},range($_GET,$_GET))),0);  • Nice collection! Care to do a version 3a including base 64? :D – Titus Mar 23 '17 at 11:12 • @Titus How is the order of base 64? Why not en.wikipedia.org/wiki/Ascii85 or make a little more with all printable ascii chars – Jörg Hülsermann Mar 23 '17 at 12:11 # JavaScript (ES6), 50 (71 - 30%) (n,k,b)=>eval("for(o=0;n<=k;++n)o+=n.toString(b)").match(/0/g).length-1  No bonus, base k+2 is 10 bytes (i,k)=>+!i No bonus, unary is 8 bytes (i,k)=>0 TEST f=(n,k,b)=>eval("for(o=0;n<=k;++n)o+=n.toString(b)").match(/0/g).length-1 function go() { var i=I.value.match(/\d+/g) R.textContent = f(i,i,i) } go() i,k,b:<input id=I value='0,500,10' oninput="go()"> <span id=R></span> • If you move the o='0' before the loop your code continues to work even when k<i. – Neil Jan 30 '16 at 19:58 • @Neil nice, but the spec says (i ≤ k). Update I tried this but in fact it does not work for k<i – edc65 Jan 30 '16 at 20:32 • Well, it worked for me (and I know the spec guarantees that i <= k, but your code crashes when k < i; by comparison my code only crashes when k < i - 1 !) – Neil Jan 30 '16 at 21:02 • @Neil uh ok now I get it. It does not give a sensible answer but at least does not crash – edc65 Jan 30 '16 at 21:05 • @ForcentVintier anyway after your input I revised the code saving some bytes – edc65 Mar 24 '17 at 10:42 # Jolf, 7 bytes Replace ♂ with \x11. Try it here! Zl♂sjJ0 sjJ inclusive range between two numeric inputs ♂ chopped into single-length elements Zl 0 and count the number of zeroes implicitly printed  ## Lua 74 bytes z,c=io.read,""for a=z(),z()do c=c..a end o,b=string.gsub(c,"0","")print(b)  There's gotta be a more effective way to do this... I thought I was really onto something here: c,m,z=0,math,io.read for a=z(),1+z()do c=c+((m.floor(a/10))%10==0 and 1 or a%100==0 and 1 or a%10==0 and 1 or 0) end print(c)  But alas... It keeps getting longer and longer as I realize there's more and more zeroes I forgot about... # APL, 22 bytes {+/'0'⍷∊0⍕¨(⍺-1)↓⍳⍵}  This is a monadic function that accepts the range boundaries on the left and right and returns an integer. Ungolfed:  (⍺-1)↓⍳⍵} ⍝ Construct the range ⍺..⍵ by dropping the first ⍝ ⍺-1 values in the range 1..⍵ ∊0⍕¨ ⍝ Convert each number to a string {+/'0'⍷ ⍝ Count the occurrences of '0' in the string  Try it here ## Haskell, 29 bytes i#k=sum[1|'0'<-show=<<[i..k]]  I'm using base 10. Usage example: 100 # 200 -> 22 How it works: turn each element in the list from i to k into it's string representation, concatenate into a single string, take a 1 for every char '0' and sum those 1s. # MATL, 7 (10 bytes − 30% bonus) 2$:i:qYA~z


Try it online!

This works in release 11.0.2, which is earlier than this challenge.

2$: % implicitly input two numbers and generate inclusive range i:q % input base b and generate vector [0,1,...,b-1] YA % convert range to base b using symbols 0,1,...,b-1. Gives 2D array ~ % logical negation. Zeros become 1, rest of symbols become 0 z % number of nonzero elements in array  # Matlab: 27 bytes @(q,w)nnz(num2str(q:w)==48)  creates a vector from lower number to larger one, then converts all numbers to string and counts all the '0' symbols. # Python 3, 52. Tried to implement the bonus, but it doesn't seem to be worth it. lambda a,b:''.join(map(str,range(a,b+1))).count('0')  With test cases: assert f(10, 10) == 1 assert f(0, 27) == 3 assert f(100, 200) == 22 assert f(0, 500) == 92  • I literally never heard about the assert statement before this comment. Thanks mate! – sagiksp Mar 23 '17 at 6:04 # Perl 6, 23 bytes {+($^i..$^k).comb(/0/)}  1. creates a Range ( $^i..$^k ) 2. joins the values with spaces implicitly ( .comb is a Str method ) 3. creates a list of just the zeros ( .comb(/0/) ) 4. returns the number of elems in that list ( + ) ### Usage: my &zero-count = {…} for (10,10), (0,27), (100,200), (0,500), (0,100000) { say zero-count |@_ }  1 3 22 92 38895  • You know, that comment at the end of your code makes it seem longer... – ETHproductions Jan 31 '16 at 22:35 • @ETHproductions I usually do that so that if I come up with more than one way to do things that I can see if it is shorter than others. I just keep adding more ways to do it until I come up with what I think is the shortest way. – Brad Gilbert b2gills Jan 31 '16 at 22:38 # Mathematica, 39 bytes, 27.3 with bonus Count[#~Range~#2~IntegerDigits~#3,0,2]&  # C# 112 Bytes int z(int i,int k)=>String.Join("",Enumerable.Range(i,k-i+1)).Count(c=>c=='0')  1. Create a string with numbers from the first number up to the last number 2. Count the zero characters in the string • Welcome to PPCG! I'm not super familiar with C# but I think you could probably save a few bytes if you removed some of the spaces. – 0 ' Mar 23 '17 at 1:05 • thank you 0, you are right but only a couple bytes. I believe my edited answer removes all the spaces I can. :) – lee Mar 23 '17 at 2:32 # PHP, 84 bytes *.7=58.8 (bases 2 to 36) for(;($v=$argv)>$a=$v++;)$n+=substr_count(base_convert($a,10,$v),0);echo$n;  or for(;($v=$argv)>$v;)$n+=substr_count(base_convert($v++,10,$v),0);echo$n;


takes decimal input from command line arguments; run with -r.

• For fun: <?=0 supports unary and alphabetic. ;) – Titus Mar 23 '17 at 11:13

## PowerShell, 56545148 42 bytes

param($i,$k)(-join($i..$k)-split0).count-1


Takes input, creates a range with $i..$k then -joins that together into a string, followed by a regex -split command that separates the string into an array by slicing at the 0s. We encapsulate that with ().count-1 to measure how many zeros. That's left on the pipeline, and output is implicit.

Saved 6 bytes thanks to @ConnorLSW

Try it online!

Base-handling in PowerShell is limited and doesn't support arbitrary bases, so I'm not going for the bonus.

• param($i,$k)(-join($i..$k)-split'0').Length-1 works for me, -3, or use .Count-1 to save even more, haven't tested that yet though. – colsw Mar 23 '17 at 16:29
• @ConnorLSW Thanks! Don't need the quotes around '0', so that trimmed off a few more. – AdmBorkBork Mar 24 '17 at 13:13
• nice one, I always forget powershell handles numbers like that. – colsw Mar 24 '17 at 14:21

# Java 8, 102 bytes - 30% = 71.4

Why not.

(i,k,b)->{int j=0;for(;i<=k;i++)for(char c:Integer.toString(i,b).toCharArray())if(c==48)j++;return j;}


Without the bonus, 96 bytes (so the bonus actually improves my score!):

(i,k)->{int j=0;for(;i<=k;i++)for(char c:String.valueOf(i).toCharArray())if(c==48)j++;return j;}


This implements the following:

interface Function {
public int apply(int i, int k, int b);
}

• @mbomb007 The problem is that formatting it this way renders the answer as 102 bytes on the Leaderboard in the question. – HyperNeutrino Mar 23 '17 at 13:36
• That's a flaw with the leaderboard, not the post. Look at how most of the other answers are doing it the same way. – mbomb007 Mar 23 '17 at 13:37
• @mbomb007 I'm looking at the answers and I see a ton of different formats, some of which work with the leaderboard, some of which don't. – HyperNeutrino Mar 23 '17 at 13:39

## Clojure, 50 49 bytes

#(count(re-seq #"0"(apply str(range %(inc %2)))))


Oh regex is shorter than filtering. Original:

#(count(filter #{\0}(apply str(range %(inc %2)))))


Very basic, uses the set of character \0 to remove others and counts how many were found.