# Output the largest number with the fewest digits

Given a nonempty list of positive decimal integers, output the largest number from the set of numbers with the fewest digits.

The input list will not be in any particular order and may contain repeated values.

Examples:

 -> 1
 -> 9
 -> 1729
[1, 1] -> 1
[34, 3] -> 3
[38, 39] -> 39
[409, 12, 13] -> 13
[11, 11, 11, 1] -> 1
[11, 11, 11, 11] -> 11
[78, 99, 620, 1] -> 1
[78, 99, 620, 10] -> 99
[78, 99, 620, 100] -> 99
[1, 5, 9, 12, 63, 102] -> 9
[3451, 29820, 2983, 1223, 1337] -> 3451
[738, 2383, 281, 938, 212, 1010] -> 938


The shortest code in bytes wins.

• Can the input numbers be on separate lines? – seshoumara Sep 16 '16 at 5:48
• @seshoumara That sounds reasonable, yes. – Calvin's Hobbies Sep 16 '16 at 5:48

# Pyth, 73 6 bytes

eS.ml


Test Suite

Explanation:

e      Still grab the last element
S      Still sort
.ml   But prefilter the list for those with the (m)inimum length.


7 byte solution:

eSh.gl


Test Suite

Explanation:

   .g   Group items in (implicit) input by:
l  The length of
their representation
h     Get those with the shortest length
S      Sort the resulting list
e       and grab the last (i.e. largest) element


# Python 2, 48 42 bytes

-6 bytes thanks to @Dennis (use min rather than sorted)

lambda l:min(l,key=lambda x:(len(x),-x))


All test cases are at ideone

Take the minimum of the list by (length, -value)

• min should work instead of sorted. – Dennis Sep 16 '16 at 3:48
• @Dennis, oh jeez - thanks! Probably different enough to have posted that yourself though. – Jonathan Allan Sep 16 '16 at 4:14
• Swapping sorted() for min? I consider that a trivial modification of your original code. – Dennis Sep 16 '16 at 4:35
• There's also len(x)+1./x for the same length. Too bad you need the 1.. – xnor Sep 16 '16 at 6:37
• Well, that's shorter than what I came up with. Good job! – mbomb007 Sep 16 '16 at 19:39

# Jelly, 7 bytes

DL,NµÞḢ


Test it at TryItOnline
Or see all test cases also at TryItOnline

How?

DL,NµÞḢ - Main link takes one argument, the list, e.g. [738, 2383, 281, 938, 212, 1010]
D       - convert to decimal, e.g. [[7,3,8],[2,3,8,3],[2,8,1],[9,3,8],[2,1,2],[1,0,1,0]]
L      - length, e.g. [3,4,3,3,3,4]
N    - negate, e.g [-738, -2383, -281, -938, -212, -1010]
,     - pair, e.g. [[3,-738],[4,-2383],[3,-281],[3,-938],[3,-212],[4,-1010]]
µ   - make a monadic chain
Þ  - sort the input by that monadic function, e.g [938,738,281,212,2383,1010]
(the lists in the example are not created, but we sort over the values shown)
Ḣ - pop and return the first element, e.g. 938

• Great use of sort! – miles Sep 16 '16 at 3:21
• @miles your way was still inspired :) – Jonathan Allan Sep 16 '16 at 3:28

# 05AB1E, 5 bytes

Code:

({é¬(


Explanation:

(      # Negate the list, e.g. [22, 33, 4] -> [-22, -33, -4]
{     # Sort, e.g. [-22, -33, -4] -> [-33, -22, -4]
é    # Sort by length, e.g. [-33, -22, -4] -> [-4, -22, -33]
¬   # Get the first element.
(  # And negate that.


Uses the CP-1252 encoding. Try it online!

# Ruby, 34 bytes

->a{a.max_by{|n|[-n.to_s.size,n]}}


See it on eval.in: https://eval.in/643153

# MATL, 14 bytes

10&YlktX<=G*X>


Try it online!

Explanation:

  &Yl           % Log
10              % Base 10
kt         % Floor and duplicate
X<       % Find the smallest element
=      % Filter out elements that do not equal the smallest element
G     % Push the input again
*    % Multiply (this sets numbers that do not have the fewest digits to 0)
X>  % And take the maximum


# Retina, 24 16 bytes

O^
O$#$.0
G1


Saved 8 bytes thanks to Martin!

The all test is using a slightly older version of the code, but the algorithm is identical. I'll update it to be closer when I get more time.

The trailing newline is significant. Sorts the numbers by reverse numeric value, then sorts them by number of digits. This leaves us with the largest number with the fewest digits in the first position, so we can just delete the remaining digits.

• If you make the input linefeed-separated you can omit the regex from both sorting stages and then use G1 for the last stage. – Martin Ender Sep 16 '16 at 5:56
• Also, the first stage doesn't need #. You only care about relative order for a given integer length, and within one length lexicographic sorting of numbers is correct. – Martin Ender Sep 16 '16 at 6:21
• @MartinEnder Thanks! I've added both your tips. I should have suggested \w+ as the default for sorting, that way I wouldn't need to struggle as much to make the test suites ;) – FryAmTheEggman Sep 16 '16 at 18:12
• Here is another 16, in case it gives you any ideas for further golfing: retina.tryitonline.net/… – Martin Ender Sep 16 '16 at 18:44

# Mathematica, 33 31 bytes

Max@MinimalBy[#,IntegerLength]&


MinimalBy selects all the elements of the original input list with the smallest score according to IntegerLength, i.e., with the smallest number of digits; and then Max outputs the largest one.

Thanks to Martin Ender for finding, and then saving, 2 bytes for me :)

# Perl 6, 18 bytes

*.min:{.chars,-$_}  ## Explanation: *\ # Whatever lambda .min: # find the minimum using { # bare block lambda with implicit parameter ｢$_｣

.chars, # number of characters first ( implicit method call on ｢$_｣ ) -$_     # then negative of the value in case of a tie
}


## Usage:

say [738, 2383, 281, 938, 212, 1010].&( *.min:{.chars,-$_} ); # 938 my &code = *.min:{.chars,-$_}

say code [78, 99, 620, 10]; # 99


# Jelly, 8 bytes

DL€İMị¹Ṁ


## Explanation

DL€İMị¹Ṁ  Input: list A
D         Convert each integer to a list of base 10 digits
L€       Get the length of each list (number of digits of each)
İ      Take the reciprocal of each
M     Get the indices of the maximal values
¹   Get A
ị    Select the values at those indices from A
Ṁ  Find the maximum and return

• How is this 8 bytes? Do all of these characters fit in ASCII? – Federico Poloni Sep 16 '16 at 14:29
• @FedericoPoloni Yes, they do fit, although in another codepage. – Erik the Outgolfer Sep 16 '16 at 14:52

# JavaScript (ES6), 51

l=>l.sort((a,b)=>(a+l).length-(b+l).length||b-a)


Test

f=l=>l.sort((a,b)=>(a+l).length-(b+l).length||b-a)

;[
[, 1]
,[, 9]
,[, 1729]
,[[1, 1], 1]
,[[34, 3], 3]
,[[38, 39], 39]
,[[409, 12, 13], 13]
,[[11, 11, 11, 1], 1]
,[[11, 11, 11, 11], 11]
,[[78, 99, 620, 1], 1]
,[[78, 99, 620, 10], 99]
,[[78, 99, 620, 100], 99]
,[[1, 5, 9, 12, 63, 102], 9]
,[[3451, 29820, 2983, 1223, 1337], 3451]
,[[738, 2383, 281, 938, 212, 1010], 938]
].forEach(([l,x])=>{
var r=f(l)
console.log(r==x?'OK':'KO',l+' -> '+r)
})  

# J, 21 14 bytes

Saved 7 bytes thanks to miles and (indirectly) Jonathan!

{.@/:#@":"0,.-


This is a four-chain:

{.@/: (#@":"0 ,. -)


Let's walk over the input 10 27 232 1000. The inner fork consists of three tines. #@":"0 calculates the sizes, ,. concats each size with its negated (-) member. For input 10 27 232 1000, we are left with this:

   (#@":"0 ,. -) 10 27 232 1000
2   _10
2   _27
3  _232
4 _1000


Now, we have {.@/: as the outer tine. This is monadic first ({.) over dyadic sort (/:). That is, we'll be taking the first element of the result of dyadic /:. This sorts its right argument according to its left argument, which gives us for our input:

   (/: #@":"0 ,. -) 10 27 232 1000
27 10 232 1000


Then, using {. gives us the first element of that list, and we are done:

   ({.@/: #@":"0 ,. -) 10 27 232 1000
27


## Old version

>./@(#~]=<./@])#@":"0


Still working on improvements. I golfed it down from 30, and I think this is good enough. I'm going to first break it down into basic parts:

   size =: #@":"0
max =: >./
min =: <./
over =: @
right =: ]
left =: [
selectMin =: #~ right = min over right

f =: max over selectMin size
f 3 4 5
5
f 3 4 53
4
f 343 42 53
53


Here's how this works.

>./@(#~ ] = <./@]) #@":"0


This is a monadic train, but this part is a hook. The verb >./@(#~ ] = <./@]) is called with left argument as the input to the main chain and the sizes, defined as #@":"0, as the right argument. This is computed as length (#) over (@) default format (":), that is, numeric stringification, which is made to apply to the 0-cells (i.e. members) of the input ("0).

Let's walk over the example input 409 12 13.

   (#@":"0) 409 12 13
3 2 2


Now for the inner verb, >./@(#~ ] = <./@]). It looks like >./@(...), which effectively means maximum value (>./) of (@) what's inside (...). As for the inside, this is a four-train, equivalent to this five-train:

[ #~ ] = <./@]


[ refers to the original argument, and ] refers to the size array; 409 12 13 and 3 2 2 respectively in this example. The right tine, <./@], computes the minimum size, 2 in this case. ] = <./@] is a boolean array of values equal to the minimum, 0 1 1 in this case. Finally, [ #~ ... takes values from the left argument according the right-argument mask. This means that elements that correspond to 0 are dropped and 1 retained. So we are left with 12 13. Finally, according to the above, the max is taken, giving us the correct result of 13, and we are done.

• Some shuffling plus a hook can save a byte >./@#~[:(=<./)#@":"0. I think there might be a bit more to save – miles Sep 16 '16 at 2:05
• @miles XD I just finished writing explanation. Ah well, let me take a look at this beauty... – Conor O'Brien Sep 16 '16 at 2:06
• Jonathan found a better method. If we convert it to J, its 14 bytes {.@/:#@":"0,.- but the input has to be shaped as a list – miles Sep 16 '16 at 3:29
• @miles "shaped as a list"? You mean, like 400 12 13? – Conor O'Brien Sep 16 '16 at 10:49

# JavaScript (ES6), 62 bytes

var solution =

a=>a.map(n=>(l=${n}.length)>a?l>a+1|n<r?0:r=n:(a=l-1,r=n))|r ;document.write('<pre>' +   -> 1  -> 9  -> 1729 [1, 1] -> 1 [34, 3] -> 3 [38, 39] -> 39 [409, 12, 13] -> 13 [11, 11, 11, 1] -> 1 [11, 11, 11, 11] -> 11 [78, 99, 620, 1] -> 1 [78, 99, 620, 10] -> 99 [78, 99, 620, 100] -> 99 [1, 5, 9, 12, 63, 102] -> 9 [3451, 29820, 2983, 1223, 1337] -> 3451 [738, 2383, 281, 938, 212, 1010] -> 938 .split('\n').slice(1, -1).map(c => c + ', result: ' + solution(eval(c.slice(0, c.indexOf('->')))) ).join('\n')) ## dc, 54 bytes ?dZsL0sN[dsNdZsL]su[dlN<u]sU[dZlL=UdZlL>ukz0<R]dsRxlNp  Explanation: ?dZsL0sN # read input, initialize L (length) and N (number) [dsNdZsL]su # macro (function) 'u' updates the values of L and N [dlN<u]sU # macro 'U' calls 'u' if N < curr_nr [dZlL=U dZlL>ukz0<R]dsR # macro 'R' is a loop that calls 'U' if L == curr_nr_len #or 'u' if L > curr_nr_len xlNp # the main: call 'R' and print N at the end  Run example: 'input.txt' contains all the test cases in the question's statement while read list;do echo "$list -> "$(dc -f program.dc <<<$list);done < input.txt


Output:

1 -> 1
9 -> 9
1729 -> 1729
1 1 -> 1
34 3 -> 3
38 39 -> 39
409 12 13 -> 13
11 11 11 1 -> 1
11 11 11 11 -> 11
78 99 620 1 -> 1
78 99 620 10 -> 99
78 99 620 100 -> 99
1 5 9 12 63 102 -> 9
3451 29820 2983 1223 1337 -> 3451
738 2383 281 938 212 1010 -> 938


# Java 7, 112 104 bytes

int c(int[]a){int i=a,j;for(int b:a)i=(j=(i+"").length()-(b+"").length())>0?b:b>i&j==0?b:i;return i;}


Different approach to save multiple bytes thanks to @Barteks2x.

Ungolfed & test cases:

Try it here.

class M{
static int c(int[] a){
int i = a,
j;
for(int b : a){
i = (j = (i+"").length() - (b+"").length()) > 0
? b
: b > i & j == 0
? b
: i;
}
return i;
}

public static void main(String[] a){
System.out.println(c(new int[]{ 1 }));
System.out.println(c(new int[]{ 9 }));
System.out.println(c(new int[]{ 1729 }));
System.out.println(c(new int[]{ 1, 1 }));
System.out.println(c(new int[]{ 34, 3 }));
System.out.println(c(new int[]{ 409, 12, 13 }));
System.out.println(c(new int[]{ 11, 11, 11, 1 }));
System.out.println(c(new int[]{ 11, 11, 11, 11 }));
System.out.println(c(new int[]{ 78, 99, 620, 1 }));
System.out.println(c(new int[]{ 78, 99, 620, 100 }));
System.out.println(c(new int[]{ 1, 5, 9, 12, 63, 102 }));
System.out.println(c(new int[]{ 3451, 29820, 2983, 1223, 1337 }));
System.out.println(c(new int[]{ 738, 2383, 281, 938, 212, 1010 }));
}
}


Output:

1
9
1729
1
3
13
1
11
1
99
9
3451
938

• shorter version: int c(int[]a){int i=a,j;for(int b:a)i=(j=(i+"").length()-(b+"").length())>0?b:b>i&j==0?b:i;return i;} – barteks2x Sep 17 '16 at 20:40
• @Barteks2x Thanks, I've edited it. – Kevin Cruijssen Sep 17 '16 at 21:29

set awk '{print $0,length($0)}'|sort -rnk2n;echo $1  Read input from stdin, one value per line # bash and sort, 58 57 bytes set sort -n;while((${#2}==${#1}));do shift;done;echo$1


• doesn't work for last sample gave 2383 instead of 938 – Archemar Sep 16 '16 at 14:47
• @Archemar sorry I misread the question, it is corrected now – Emmanuel Sep 16 '16 at 16:54
• You can remove the space between while and ((. – seshoumara Sep 17 '16 at 11:26

# JavaScript ES6, 8077 70 bytes

a=>Math.max(...a.filter(l=>l.length==Math.min(...a.map(i=>i.length))))


I hope I am going in the right direction...

• Could you replace a.map(i=>i.length).sort((a,b)=>a-b) with Math.min(...a.map(i=>i.length))? – user81655 Sep 16 '16 at 4:58
• @user81655 yup, I can. I thought I had made that edit but apparently I did not – Downgoat Sep 16 '16 at 4:59
• You could also try negating the minimum so that you can reuse the Math.max: a=>(m=Math.max)(...a.filter(l=>l.length==-m(...a.map(i=>-i.length)))) It seems to save only 1 byte though. – user81655 Sep 16 '16 at 5:03
• For another byte the filter can be replaced with a map that returns 0 for values that do not pass the test: a=>(m=Math.max)(...a.map(l=>l.length+m(...a.map(i=>-i.length))?0:l)) – user81655 Sep 16 '16 at 5:12

# Brachylog, 16 bytes

or:@]feL:la#=,Lh


Try it online!

### Explanation

or                 Sort the list in descending order.
:@]f             Find all suffixes of the list.
eL           Take one suffix L of the list.
:la        Apply length to all numbers in that suffix.
#=,     All lengths must be equal.
Lh   Output is the first element of L.


snd.maximum.map((0-).length.show>>=(,))

• This doesn't work, it prefers 34 to 2. – xnor Sep 16 '16 at 6:39
• oh, thanks. I have to rethink it.. – Damien Sep 16 '16 at 6:42
• Works better now! – Damien Sep 16 '16 at 6:52

## Javascript (ES6), 5754 53 bytes

l=>l.sort((a,b)=>(s=a=>1/a+${a}.length)(a)-s(b))  For the record, my previous version was more math-oriented but 1 byte bigger: l=>l.sort((a,b)=>(s=a=>1/a-~Math.log10(a))(a)-s(b))  ### Test cases let f = l=>l.sort((a,b)=>(s=a=>1/a+${a}.length)(a)-s(b))

console.log(f());                              //  -> 1
console.log(f());                              //  -> 9
console.log(f());                           //  -> 1729
console.log(f([1, 1]));                           //  -> 1
console.log(f([34, 3]));                          //  -> 3
console.log(f([38, 39]));                         //  -> 39
console.log(f([409, 12, 13]));                    //  -> 13
console.log(f([11, 11, 11, 1]));                  //  -> 1
console.log(f([11, 11, 11, 11]));                 //  -> 11
console.log(f([78, 99, 620, 1]));                 //  -> 1
console.log(f([78, 99, 620, 10]));                //  -> 99
console.log(f([78, 99, 620, 100]));               //  -> 99
console.log(f([1, 5, 9, 12, 63, 102]));           //  -> 9
console.log(f([3451, 29820, 2983, 1223, 1337]));  //  -> 3451
console.log(f([738, 2383, 281, 938, 212, 1010])); //  -> 938

# MATL, 11 bytes

tV48\&XS0))


Input is a column vector (using ; as separator), such as

[78; 99; 620; 100]


### Explanation

Let's use input [78; 99; 620; 100] as an example.

t      % Input column vector implicitly. Duplicate
%   STACK: [78; 99; 620; 100], [78; 99; 620; 100]
V      % Convert to string. Each number is a row, left-padded with spaces
%   STACK: [78; 99; 620; 100], [' 78'; ' 99'; '620'; '100']
48\    % Modulo 48. This transforms each digit into the corresponding number,
% and space into 32. Thus space becomes the largest "digit"
%   STACK: [78; 99; 620; 100], [32 7 8; 32 9 9; 6 2 0; 1 0 0]
&XS    % Sort rows in lexicographical order, and push the indices of the sorting
%   STACK: [78; 99; 620; 100], [4; 3; 1; 2]
0)     % Get last value
%   STACK: [78; 99; 620; 100], 2
)      % Index
%   STACK: 99
% Implicitly display

• Nice to see the stack states in your explanation! – flawr Sep 16 '16 at 23:18

# Perl, 38 37 bytes

Includes +1 for -a

Give input on STDIN:

perl -M5.010 maxmin.pl <<< "3451 29820 2983 1223 1337"


maxmin.pl:

#!/usr/bin/perl -a
\$G[99-y///c][$_]for@F;say$#{$G[-1]}


Uses memory linear in the largest number, so don't try this on too large numbers. A solution without that flaw is 38 bytes:

#!/usr/bin/perl -p
$.++until$\=(sort/\b\S{$.}\b/g)[-1]}{  All of these are very awkward and don't feel optimal at all... # R, 7241 36 bytes Rewrote the function with a new approach. Golfed 5 bytes thanks to a suggestion from @bouncyball. n=nchar(i<-scan());max(i[n==min(n)])  Explained:  i<-scan() # Read input from stdin n=nchar( ); # Count the number of characters in each number in i max( ) # Return the maximum of the set where i[n==min(n)] # the number of characters is the minimum number of characters.  function(i){while(1){if(length(o<-i[nchar(i)==T]))return(max(o));T=T+1}}  Indented/explained: function(i){ # Take an input i while(1){ # Do the following continuously: if(length( o<-i[nchar(i)==T]) # Define o to be the subset of i with numbers of length T, ) # where T is 1 (a built-in!). # We take the length of this subset (its size), and then pass # it to if(). Thanks to weak typing, this numeric is converted # to a logical value. When this occurs, zero evaluates to FALSE # and any non-zero number evaluates to TRUE. Therefore, the if() # is TRUE iff the subset is not empty. return(max(o)); # If it's true, then we just return the largest element of the # subset, breaking out of our loop. T=T+1 # Otherwise, increment our counter and continue. } }  • Save 4 bytes by not defining function: i=scan();n=nchar(i);max(i[n==min(n)]) – bouncyball Sep 16 '16 at 14:13 • @bouncyball Thanks! And 1 further byte saved by n=nchar(i<-scan()). – rturnbull Sep 16 '16 at 19:32 ## Bash + coreutils, 58 bytes d=sort -n;egrep ^.{sed q<<<"$d"|wc -L}$<<<"$d"|tail -1


Input format is one value per line. Golfing suggestions are welcomed.

Explanation:

d=sort -n                             #save the list in ascending numerical order
egrep ^.{                    }$<<<"$d"  #print only list lines having as many chars
sed q<<<"\$d"|wc -L                 #as the first sorted line does
|tail -1                                #and then get the last one (the answer)

• +1 thank you now I know that sed q = head -1 – Emmanuel Sep 19 '16 at 0:58

# Python 2 - 41 bytes

lambda l:max((-len(x),x) for x in l)


# APL (Dyalog Unicode), 36 bytes

-4 thanks to @rak1507.

f←⌈/⊢×((⌊/=⊢)≢∘⍕¨)


Try it online!

• ⌈/⊢×((⌊/=⊢)≢∘⍕¨) works I think – rak1507 Dec 20 '20 at 21:05

## Python 2, 58 bytes

def F(x):l={len(i):i for i in sorted(x)};print l[min(l)]


# Python 3, 56 bytes

lambda a:sorted(sorted(a),key=lambda x:-len(str(x)))[-1]


Uses a lambda in a lambda!

## Python 2, 53 bytes

s=lambda a:sorted(sorted(a),key=lambda x:-len(x))[-1]


Same but with backticks

## Pip, 11 bytes

(SNgSK-#_v)


Takes input as command-line args. Try it online!

First time using the Sort-Keyed operator! Like Python's sorted(), it takes a function that is applied to each item of the iterable and the result used as a sort key. Here's how this program works:

 SNg         List of cmdline args, sorted numerically in increasing order
SK       Sort with key function...
-#_    ... negative length(x), thus putting the shortest numbers at the end but not
affecting the relative ordering among numbers with the same length
(        v)  Get the last element (index -1) and auto-print


# Clojure, 63 bytes

(reduce #(if(=(quot %1 10)(quot %2 10))(max %1 %2) %1)(sort x))


as in:

(reduce #(if(=(quot %1 10)(quot %2 10))(max %1 %2) %1)(sort[3 7 121 11 8 2 10 9]))
=> 9


Though I'm sure there's a way to make it smaller.