Largest int from concatenated ints: Difference between revisions
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=={{header|Quackery}}== |
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<lang quackery>[ sortwith |
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[ 2dup swap join |
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dip join $< ] |
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[] swap witheach join ] is largest-int ( [ --> $ ) |
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$ '1 34 3 98 9 76 45 4' nest$ largest-int echo$ cr |
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$ '54 546 548 60' nest$ largest-int echo$</lang> |
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{{out}} |
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<pre> |
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998764543431 |
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6054854654 |
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</pre> |
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=={{header|Racket}}== |
=={{header|Racket}}== |
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Revision as of 19:30, 13 March 2021
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Given a set of positive integers, write a function to order the integers in such a way that the concatenation of the numbers forms the largest possible integer and return this integer.
Use the following two sets of integers as tests and show your program output here.
- {1, 34, 3, 98, 9, 76, 45, 4}
- {54, 546, 548, 60}
- Possible algorithms
- A solution could be found by trying all combinations and return the best.
- Another way to solve this is to note that in the best arrangement, for any two adjacent original integers X and Y, the concatenation X followed by Y will be numerically greater than or equal to the concatenation Y followed by X.
- Yet another way to solve this is to pad the integers to the same size by repeating the digits then sort using these repeated integers as a sort key.
- See also
- Algorithms: What is the most efficient way to arrange the given numbers to form the biggest number?
- Constructing the largest number possible by rearranging a list
11l
<lang 11l>F maxnum(x)
V maxlen = String(max(x)).len R sorted((x.map(v -> String(v))), key' i -> i * (@maxlen * 2 I/ i.len), reverse' 1B).join(‘’)
L(numbers) [[212, 21221], [1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]]
print("Numbers: #.\n Largest integer: #15".format(numbers, maxnum(numbers)))</lang>
- Output:
Numbers: [212, 21221] Largest integer: 21221221 Numbers: [1, 34, 3, 98, 9, 76, 45, 4] Largest integer: 998764543431 Numbers: [54, 546, 548, 60] Largest integer: 6054854654
Ada
The algorithmic idea is to apply a twisted comparison function:
<lang Ada>function Order(Left, Right: Natural) return Boolean is
( (Img(Left) & Img(Right)) > (Img(Right) & Img(Left)) );</lang>
This function converts the parameters Left and Right to strings and returns True if (Left before Right) exceeds (Right before Left). It needs Ada 2012 -- the code for older versions of Ada would be more verbose.
The rest is straightforward: Run your favourite sorting subprogram that allows to use the function "Order" instead of standard comparison operators ("<" or ">" or so) and print the results:
<lang Ada>with Ada.Text_IO, Ada.Containers.Generic_Array_Sort;
procedure Largest_Int_From_List is
function Img(N: Natural) return String is
S: String := Integer'Image(N);
begin
return S(S'First+1 .. S'Last); -- First character is ' '
end Img;
function Order(Left, Right: Natural) return Boolean is
( (Img(Left) & Img(Right)) > (Img(Right) & Img(Left)) );
type Arr_T is array(Positive range <>) of Natural;
procedure Sort is new Ada.Containers.Generic_Array_Sort
(Positive, Natural, Arr_T, Order);
procedure Print_Sorted(A: Arr_T) is
B: Arr_T := A;
begin
Sort(B);
for Number of B loop
Ada.Text_IO.Put(Img(Number));
end loop;
Ada.Text_IO.New_Line;
end Print_Sorted;
begin
Print_Sorted((1, 34, 3, 98, 9, 76, 45, 4)); Print_Sorted((54, 546, 548, 60));
end Largest_Int_From_List;</lang>
Aime
<lang aime>largest(...) {
index x;
for (, integer e in xcall(list).__list) {
x[999999999 - 9.times(b_, data(), e).b_size(9).atoi] = e;
}
x.ucall(o_, 0);
o_newline();
}
main(void) {
largest(1, 34, 3, 98, 9, 76, 45, 4); largest(54, 546, 548, 60); 0;
}</lang> works for input up to 999999999.
- Output:
998764543431 6054854654
ALGOL 68
Using method 2 - first sorting into first digit order and then comparing concatenated pairs. <lang algol68>BEGIN
# returns the integer value of s #
OP TOINT = ( STRING s)INT:
BEGIN
INT result := 0;
FOR s pos FROM LWB s TO UPB s DO
result *:= 10 +:= ( ABS s[ s pos ] - ABS "0" )
OD;
result
END # TOINT # ;
# returns the first digit of n #
OP FIRSTDIGIT = ( INT n )INT:
BEGIN
INT result := ABS n;
WHILE result > 9 DO result OVERAB 10 OD;
result
END # FIRSTDIGIT # ;
# returns a string representaton of n #
OP TOSTRING = ( INT n )STRING: whole( n, 0 );
# returns an array containing the values of a sorted such that concatenating the values would result in the largest value #
OP CONCATSORT = ( []INT a )[]INT:
IF LWB a >= UPB a THEN
# 0 or 1 element(s) #
a
ELSE
# 2 or more elements #
[ 1 : ( UPB a - LWB a ) + 1 ]INT result := a[ AT 1 ];
# sort the numbers into reverse first digit order #
FOR o pos FROM UPB result - 1 BY -1 TO 1
WHILE BOOL swapped := FALSE;
FOR i pos TO o pos DO
IF FIRSTDIGIT result[ i pos ] < FIRSTDIGIT result[ i pos + 1 ] THEN
INT t = result[ i pos + 1 ];
result[ i pos + 1 ] := result[ i pos ];
result[ i pos ] := t;
swapped := TRUE
FI
OD;
swapped
DO SKIP OD;
# now re-order adjacent numbers so they have the highest concatenated value #
WHILE BOOL swapped := FALSE;
FOR i pos TO UPB result - 1 DO
STRING l := TOSTRING result[ i pos ];
STRING r := TOSTRING result[ i pos + 1 ];
IF TOINT ( l + r ) < TOINT ( r + l ) THEN
INT t = result[ i pos + 1 ];
result[ i pos + 1 ] := result[ i pos ];
result[ i pos ] := t;
swapped := TRUE
FI
OD;
swapped
DO SKIP OD;
result
FI # CONCATSORT # ;
# prints the array a #
OP PRINT = ( []INT a )VOID:
FOR a pos FROM LWB a TO UPB a DO
print( ( TOSTRING a[ a pos ] ) )
OD # PRINT # ;
# task test cases # PRINT CONCATSORT []INT( 1, 34, 3, 98, 9, 76, 45, 4 ); print( ( newline ) ); PRINT CONCATSORT []INT( 54, 546, 548, 60 ); print( ( newline ) )
END</lang>
- Output:
998764543431 6054854654
Arturo
<lang rebol>largestConcInt: function [arr]->
max map permutate arr 's [
to :integer join map s => [to :string]
]
loop [[1 34 3 98 9 76 45 4] [54 546 548 60]] 'a ->
print largestConcInt a</lang>
AutoHotkey
<lang AutoHotkey>LargestConcatenatedInts(var){ StringReplace, var, A_LoopField,%A_Space%,, all Sort, var, D`, fConcSort StringReplace, var, var, `,,, all return var }
ConcSort(a, b){ m := a . b , n := b . a
return m < n ? 1 : m > n ? -1 : 0
}</lang> Examples:<lang AutoHotkey>d = ( 1, 34, 3, 98, 9, 76, 45, 4 54, 546, 548, 60 4 , 45, 54, 5 ) loop, parse, d, `n MsgBox % LargestConcatenatedInts(A_LoopField)</lang>
- Output:
998764543431 6054854654 554454
AWK
<lang awk> function cmp(i1, v1, i2, v2, u1, u2) { u1 = v1""v2; u2 = v2""v1;
return (u2 - u1)
} function largest_int_from_concatenated_ints(X) {
PROCINFO["sorted_in"]="cmp";
u=""; for (i in X) u=u""X[i]; return u }
BEGIN { split("1 34 3 98 9 76 45 4",X); print largest_int_from_concatenated_ints(X)
split("54 546 548 60",X); print largest_int_from_concatenated_ints(X) } </lang>
- Output:
998764543431 6054854654
BBC BASIC
<lang bbcbasic> DIM Nums%(10)
Nums%()=1,34,3,98,9,76,45,4
PRINT FNlargestint(8)
Nums%()=54,546,548,60
PRINT FNlargestint(4)
END
DEF FNlargestint(len%)
LOCAL i%,l$,a$,b$,sorted%
REPEAT
sorted%=TRUE
FOR i%=0 TO len%-2
a$=STR$Nums%(i%)
b$=STR$Nums%(i%+1)
IF a$+b$<b$+a$ SWAP Nums%(i%),Nums%(i%+1):sorted%=FALSE
NEXT
UNTIL sorted%
FOR i%=0 TO len%-1
l$+=STR$Nums%(i%)
NEXT
=l$</lang>
- Output:
998764543431 6054854654
Bracmat
<lang bracmat>( ( maxnum
= A Z F C
. !arg:#
| !arg
: %@?F
?
( #%@?C
& ( str$(!F !C)+-1*str$(!C !F):~<0
| !C:?F
)
& ~
)
?
| !arg:?A !F ?Z&!F maxnum$(!A !Z)
)
& out$(str$(maxnum$(1 34 3 98 9 76 45 4))) & out$(str$(maxnum$(54 546 548 60))) );</lang>
- Output:
998764543431 6054854654
C
<lang C>#include <stdio.h>
- include <stdlib.h>
- include <string.h>
int catcmp(const void *a, const void *b) { char ab[32], ba[32]; sprintf(ab, "%d%d", *(int*)a, *(int*)b); sprintf(ba, "%d%d", *(int*)b, *(int*)a); return strcmp(ba, ab); }
void maxcat(int *a, int len) { int i; qsort(a, len, sizeof(int), catcmp); for (i = 0; i < len; i++) printf("%d", a[i]); putchar('\n'); }
int main(void) { int x[] = {1, 34, 3, 98, 9, 76, 45, 4}; int y[] = {54, 546, 548, 60};
maxcat(x, sizeof(x)/sizeof(x[0])); maxcat(y, sizeof(y)/sizeof(y[0]));
return 0; }</lang>
- Output:
998764543431 6054854654
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq;
class Program {
static void Main(string[] args)
{
var source1 = new int[] { 1, 34, 3, 98, 9, 76, 45, 4 };
var source2 = new int[] { 54, 546, 548, 60 };
var largest1 = LargestPossibleSequence(source1);
var largest2 = LargestPossibleSequence(source2);
Console.WriteLine("The largest possible integer from set 1 is: {0}", largest1);
Console.WriteLine("The largest possible integer from set 2 is: {0}", largest2);
}
static long LargestPossibleSequence(int[] ints)
{
return long.Parse(string.Join("", ints.OrderBy(i => i, new IntConcatenationComparer()).Reverse()));
}
}
class IntConcatenationComparer : IComparer<int> {
public int Compare(int x, int y)
{
var xy = int.Parse(x.ToString() + y.ToString());
var yx = int.Parse(y.ToString() + x.ToString());
return xy - yx; }
} </lang>
- Output:
The largest possible integer from set 1 is: 998764543431 The largest possible integer from set 2 is: 6054854654
C++
<lang cpp>#include <iostream>
- include <sstream>
- include <algorithm>
- include <vector>
- include <string>
std::string findLargestConcat ( std::vector< int > & mynumbers ) {
std::vector<std::string> concatnumbers ;
std::sort ( mynumbers.begin( ) , mynumbers.end( ) ) ;
do {
std::ostringstream numberstream ;
for ( int i : mynumbers )
numberstream << i ;
concatnumbers.push_back( numberstream.str( ) ) ; } while ( std::next_permutation( mynumbers.begin( ) ,
mynumbers.end( ) )) ;
return *( std::max_element( concatnumbers.begin( ) ,
concatnumbers.end( ) ) ) ; }
int main( ) {
std::vector<int> mynumbers = { 98, 76 , 45 , 34, 9 , 4 , 3 , 1 } ;
std::vector<int> othernumbers = { 54 , 546 , 548 , 60 } ;
std::cout << "The largest concatenated int is " <<
findLargestConcat( mynumbers ) << " !\n" ;
std::cout << "And here it is " << findLargestConcat( othernumbers )
<< " !\n" ;
return 0 ;
}</lang>
- Output:
The largest concatenated int is 998764543431 ! And here it is 6054854654 !
Ceylon
<lang ceylon>shared void run() {
function comparator(Integer x, Integer y) { assert (is Integer xy = Integer.parse(x.string + y.string), is Integer yx = Integer.parse(y.string + x.string)); return yx <=> xy; }
function biggestConcatenation({Integer*} ints) => "".join(ints.sort(comparator));
value test1 = {1, 34, 3, 98, 9, 76, 45, 4}; print(biggestConcatenation(test1));
value test2 = {54, 546, 548, 60}; print(biggestConcatenation(test2)); }</lang>
Clojure
<lang Clojure>(defn maxcat [coll]
(read-string
(apply str
(sort (fn [x y]
(apply compare
(map read-string [(str y x) (str x y)])))
coll))))
(prn (map maxcat [[1 34 3 98 9 76 45 4] [54 546 548 60]]))</lang>
- Output:
(998764543431 6054854654)
Common Lisp
Sort by two-by-two comparison of largest concatenated result
<lang lisp> (defun int-concat (ints)
(read-from-string (format nil "~{~a~}" ints)))
(defun by-biggest-result (first second)
(> (int-concat (list first second)) (int-concat (list second first))))
(defun make-largest-int (ints)
(int-concat (sort ints #'by-biggest-result)))
</lang>
- Output:
> (make-largest-int '(1 34 3 98 9 76 45 4)) 998764543431 > (make-largest-int '(54 546 548 60)) 6054854654
Variation around the sort with padded most significant digit
<lang lisp>
- Sort criteria is by most significant digit with least digits used as a tie
- breaker
(defun largest-msd-with-less-digits (x y)
(flet ((first-digit (x)
(digit-char-p (aref x 0))))
(cond ((> (first-digit x)
(first-digit y))
t)
((> (first-digit y)
(first-digit x))
nil)
((and (= (first-digit x)
(first-digit y))
(> (length x)
(length y)))
nil)
(t t))))
(loop
:for input :in '((54 546 548 60) (1 34 3 98 9 76 45 4))
:do (format t "~{~A~}~%"
(sort (mapcar #'write-to-string input)
#'largest-msd-with-less-digits)))
</lang>
- Output:
6054548546 998764453341
D
The three algorithms. Uses the second module from the Permutations Task. <lang d>import std.stdio, std.algorithm, std.conv, std.array, permutations2;
auto maxCat1(in int[] arr) pure @safe {
return arr.to!(string[]).permutations.map!join.reduce!max;
}
auto maxCat2(in int[] arr) pure nothrow @safe {
return arr.to!(string[]).sort!q{b ~ a < a ~ b}.join;
}
auto maxCat3(in int[] arr) /*pure nothrow @safe*/ {
immutable maxL = arr.reduce!max.text.length;
return arr.to!(string[])
.schwartzSort!(s => s.replicate(maxL/s.length + 1), "a > b")
.join;
}
void main() {
const lists = [[1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]]; [&maxCat1, &maxCat2, &maxCat3].map!(cat => lists.map!cat).writeln;
}</lang>
- Output:
[["998764543431", "6054854654"], ["998764543431", "6054854654"], ["998764543431", "6054854654"]]
Elixir
<lang elixir>defmodule RC do
def largest_int(list) do
sorted = Enum.sort(list, fn x,y -> "#{x}#{y}" >= "#{y}#{x}" end)
Enum.join(sorted)
end
end
IO.inspect RC.largest_int [1, 34, 3, 98, 9, 76, 45, 4] IO.inspect RC.largest_int [54, 546, 548, 60]</lang>
- Output:
"998764543431" "6054854654"
Erlang
<lang Erlang> -module( largest_int_from_concatenated ).
-export( [ints/1, task/0] ).
ints( Ints ) -> Int_strings = [erlang:integer_to_list(X) || X <- Ints], Pad_ints = [{X ++ X, X} || X <- Int_strings], erlang:list_to_integer( lists:append([Int || {_Pad, Int} <- lists:reverse(lists:sort(Pad_ints))]) ).
task() -> [io:fwrite("Largest ~p from ~p~n", [ints(X), X]) || X <- [[1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]]]. </lang>
- Output:
8> largest_int_from_concatenated:task(). Largest 998764543431 from [1,34,3,98,9,76,45,4] Largest 6054854654 from [54,546,548,60]
F#
<lang fsharp> // Form largest integer which is a permutation from a list of integers. Nigel Galloway: March 21st., 2018 let fN g = List.map (string) g |> List.sortWith(fun n g->if n+g<g+n then 1 else -1) |> System.String.Concat </lang>
- Output:
fN [1; 34; 3; 98; 9; 76; 45; 4] -> "998764543431" fN [54; 546; 548; 60] -> "6054854654"
Factor
Using algorithm 3: <lang factor>USING: assocs io kernel math qw sequences sorting ; IN: rosetta-code.largest-int
- pad ( target seq -- padded )
2dup length / swap <repetition> concat swap head ;
- largest-int ( seq -- )
dup dup [ length ] map supremum ! find longest length so we know how much to pad [ swap pad ] curry map ! pad the integers <enum> sort-values ! sort the padded integers keys ! find the original indices of the sorted integers swap nths ! order non-padded integers according to their sorted order reverse concat print ;
qw{ 1 34 3 98 9 76 45 4 } qw{ 54 546 548 60 } [ largest-int ] bi@</lang>
- Output:
998764543431 6054854654
Or alternatively, a translation of F#.
<lang factor>USING: kernel math.order qw sequences sorting ;
- fn ( seq -- str )
[ 2dup swap [ append ] 2bi@ after? +lt+ +gt+ ? ] sort concat ;</lang>
- Output:
qw{ 1 34 3 98 9 76 45 4 } qw{ 54 546 548 60 } [ fn ] bi@
--- Data stack:
"998764543431"
"6054854654"
Fortran
There is often a potential ambiguity when reading numbers. While three definitely names the Platonic number notion, 3 might instead be regarded as being a text that happens to have the glyph of a number but is not a number. This sort of discussion arises when a spreadsheet has read in a text file and behold! numbers are on the display and they look just like what is displayed when numbers are being shown, but, they are not numbers, they are only drawn that way. Within the spreadsheet they are parts of some text, and the notion that takes over is one of a "blunt, heavy object", not alas close to hand.
So, the plan is to regard the numbers as being text sequences aligned to the left, containing only digit characters of course - except for the fact that CHARACTER variables often end up having trailing spaces. F2003 formalised a scheme whereby such variables can be "cut-to-fit" as execution proceeds but with earlier Fortrans the standard method is to pay attention to the number of characters in use. F90 introduced a function LEN_TRIM(text) to return the index of the last non-blank character in a text so the only problem now is to decide on how long might the largest number be (and by representing numbers as text strings, there is no difficulty with the limits of INTEGER*2 or INTEGER*4 etc.), and what will be the maximum number of numbers. By devising a subroutine to do the work, these issues can be handled by the caller that is providing the data. The subroutine however intends to sort the collection of texts. This could be done by damaging its parameter which might be regarded as impolite or even unwanted so instead the sort is effected via an array XLAT and juggling its values. This has the advantage that the possibly large elements of the text array are not being moved about, but means that the subroutine must be able to have an XLAT array that is "large enough". F90 standardised the ability for a routine to declare such an array at run-time; previously, arrays within a subroutine (or indeed anywhere) had to have a size fixed at compilation time. In the past this might have been handled by the caller supplying such an array as an additional parameter.
Passing arrays as parameters can be tricky, especially for multi-dimensional arrays. This uses the old style whereby the size is left unstated via the * in TEXT(*), though one could use TEXT(N) instead - but at the risk that the actual value of N is wrong and array index checking might be confused thereby. Still earlier one would simply place some integer value there, any valid integer, as in TEXT(666), and not worry about bound checking at all because old-style compilers did not produce checking code even if it was wanted. F90 standardised the MODULE protocol, within which the size is specified as TEXT(:) whereby secret additional parameters are supplied that contain the actual bound information and bound checking will be correct, possibly not so if the TEXT(N) form is used instead and N is wrong. This extra overhead in every use is possibly better than undetected errors in some uses...
The sorting of the text array was to be by the notorious BubbleSort, taking advantage of the fact that each pass delivers the maximum value of the unsorted portion to its final position: the output could thereby be produced as the sort worked. Rather than mess about with early termination (no element being swapped) or attention to the bounds within which swapping took place, attention concentrated upon the comparison. Because of the left-alignment of the texts, a simple comparison seemed sufficient until I thought of unequal text lengths and then the following example. Suppose there are two numbers, 5, and one of 54, 55, or 56 as the other. Via normal comparisons, the 5 would always be first (because short texts are considered expanded with trailing spaces when compared against longer texts, and a space precedes every digit) however the biggest ordering is 5 54 for the first case but 56 5 for the last. This possibility is not exemplified in the specified trial sets. So, a more complex comparison is required. One could of course write a suitable function and consider the issue there but instead the comparison forms the compound text in the same manner as the result will be, in the two ways AB and BA, and looks to see which yields the bigger sequence. This need only be done for unequal length text pairs.
The source is F77 style, except for the declaration of XLAT(N), the use of <N> in the FORMAT statements instead of some large constant or similar, and the ability to declare an array via constants as in (/"5","54"/) rather than mess about declaring arrays and initialising them separately. The I0 format code to convert a number (an actual number) into a digit string aligned leftwards in a CHARACTER variable of sufficient size is also a F90 introduction, though the B6700 compiler allowed a code J instead. This last is to demonstrate usage of actual numbers for those unpersuaded by the argument for ambiguity that allows for texts. If the I0 format code is unavailable then I9 (or some suitable size) could be used, followed by text = ADJUSTL(text), except that this became an intrinsic function only in F90, so perhaps you will have to write a simple alignment routine. <lang Fortran> SUBROUTINE SWAP(A,B) !Why can't the compiler supply these!
INTEGER A,B,T
T = B
B = A
A = T
END
SUBROUTINE BIGUP(TEXT,N) !Outputs the numbers in TEXT to give the biggest number.
CHARACTER*(*) TEXT(*) !The numbers as text, aligned left.
INTEGER N !The number of them.
INTEGER XLAT(N),L(N) !An index and a set of lengths.
INTEGER I,J,M !Assorted steppers.
INTEGER TI,TJ !Fingers to a text.
INTEGER LI,LJ !Lengths of the fingered texts.
INTEGER MSG !I/O unit number.
COMMON /IODEV/ MSG !Old style.
DO I = 1,N !Step through my supply of texts.
XLAT(I) = I !Preparing a finger to them.
L(I) = LEN_TRIM(TEXT(I)) !And noting their last non-blank.
END DO !On to the next.
WRITE (MSG,1) "Supplied",(TEXT(I)(1:L(I)), I = 1,N) !Show the grist.
1 FORMAT (A12,":",<N>(A,",")) !Instead of <N>, 666 might suffice.
Crude bubblesort. No attempt at noting the bounds of swaps made.
DO M = N,1,-1 !Just for fun, go backwards.
DO I = 2,M !Start a scan.
J = I - 1 !Comparing element I to element I - 1.
TI = XLAT(I) !Thus finger the I'th text in XLAT order.
TJ = XLAT(J) !And its supposed predecessor.
LI = L(TI) !The length of the fingered text.
LJ = L(TJ) !All this to save on typing below.
IF (LI .EQ. LJ) THEN !If the texts are equal lengths,
IF (TEXT(TI).LT.TEXT(TJ)) CALL SWAP(XLAT(I),XLAT(J)) !A simple comparison.
ELSE !But if not, construct the actual candidate texts for comparison.
IF (TEXT(TI)(1:LI)//TEXT(TJ)(1:LJ) !These two will be the same length.
1 .LT.TEXT(TJ)(1:LJ)//TEXT(TI)(1:LI)) !Just as above.
2 CALL SWAP(XLAT(I),XLAT(J)) !J shall now follow I.
END IF !So much for that comparison.
END DO !On to the next.
END DO !The original plan was to reveal element XLAT(M) as found.
WRITE (MSG,2) "Biggest",(TEXT(XLAT(I))(1:L(XLAT(I))),I = N,1,-1) !But, all at once is good too.
2 FORMAT (A12,":",<N>(A," ")) !The space maintains identity.
END !That was fun.
PROGRAM POKE
CHARACTER*4 T1(10) !Prepare some example arrays.
CHARACTER*4 T2(4) !To hold the specified examples.
INTEGER MSG
COMMON /IODEV/ MSG
DATA T1(1:8)/"1","34","3","98","9","76","45","4"/
DATA T2/"54","546","548","60"/
MSG = 6 !Standard output.
WRITE (MSG,1)
1 FORMAT ("Takes a list of integers and concatenates them so as ",
1 "to produce the biggest possible number.",/,
2 "The result is shown with spaces between the parts ",
3 "to show provenance. Ignore them otherwise."/)
CALL BIGUP(T1,8)
WRITE (MSG,*)
CALL BIGUP(T2,4)
WRITE (MSG,*) "These are supplied in lexicographical order..."
CALL BIGUP((/"5","54"/),2)
WRITE (MSG,*) "But this is not necessarily the biggest order."
CALL BIGUP((/"5","56"/),2)
WRITE (MSG,*) "And for those who count..."
DO I = 1,10
WRITE (T1(I),"(I0)") I !This format code produces only the necessary text.
END DO !Thus, the numbers are aligned left in the text field.
CALL BIGUP(T1,10)
END </lang>
Output: the Fortran compiler ignores spaces when reading fortran source, so, hard-core fortranners should have no difficulty doing likewise for the output...
Takes a list of integers and concatenates them so as to produce the biggest possible number.
The result is shown with spaces between the parts to show provenance. Ignore them otherwise.
Supplied:1,34,3,98,9,76,45,4,
Biggest:9 98 76 45 4 34 3 1
Supplied:54,546,548,60,
Biggest:60 548 546 54
These are supplied in lexicographical order...
Supplied:5,54,
Biggest:5 54
But this is not necessarily the biggest order.
Supplied:5,56,
Biggest:56 5
And for those who count...
Supplied:1,2,3,4,5,6,7,8,9,10,
Biggest:9 8 7 6 5 4 3 2 1 10
FreeBASIC
<lang freebasic>#define MAXDIGITS 8
function catint( a as string, b as string ) as uinteger
return valint(a+b)
end function
function grt( a as string, b as string ) as boolean
return catint(a, b)>catint(b, a)
end function
sub shellsort( a() as string )
'quick and dirty shellsort, not the focus of this exercise
dim as uinteger gap = ubound(a), i, j, n=ubound(a)
dim as string temp
do
gap = int(gap / 2.2)
for i=gap to n
temp = a(i)
j=i
while j>=gap andalso grt( a(j-gap), temp )
a(j) = a(j - gap)
j -= gap
wend
a(j) = temp
next i
loop until gap = 1
end sub
sub sort_and_print( a() as string )
dim as uinteger i
dim as string outstring = ""
shellsort(a())
for i=0 to ubound(a)
outstring = a(i)+outstring
next i
print outstring
end sub
dim as string set1(8) = {"1", "34", "3", "98", "9", "76", "45", "4"} dim as string set2(4) = {"54", "546", "548", "60"}
sort_and_print(set1()) sort_and_print(set2())</lang>
- Output:
998764543431 6054854654
Gambas
Click this link to run this code <lang gambas>'Largest int from concatenated ints
Public Sub Main() Dim iList1 As Integer[] = [1, 34, 3, 98, 9, 76, 45, 4] 'Integer list 1 Dim iList2 As Integer[] = [54, 546, 548, 60] 'Integer list 2
Calc(iList1) 'Send List 1 to Calc routine Calc(iList2) 'Send List 2 to Calc routine
End '_________________________________________________________________________________________
Public Sub Calc(iList As Integer[]) Dim siCount1, siCount2, siCounter As Short 'Counters Dim sList As New String[] 'To hold converted integers Dim bTrigger As Boolean 'To trigger a found match
For Each siCount1 In iList 'For each integer in the list..
sList.Add(Str(siCount1)) 'Convert to a string and add to sList If Len(Str(siCount1)) > siCounter Then 'If the length of the string is greater than siCounter then.. siCounter = Len(Str(siCount1)) 'siCounter = length of the string End If
Next
For siCount1 = 0 To sList.Max 'For each item in sList
If Len(sList[siCount1]) < siCounter Then 'If the length of the string is less that siCounter then.. sList[siCount1] &= Right(sList[siCount1], 1) 'Add the same digit to the string e.g. in list 1 "9" becomes "99", list 2 "54" becomes "544" End If
Next
sList.Sort(gb.Descent) 'Sort the list in decending order
For siCount1 = 0 To sList.Max 'For each item in sList
bTrigger = False 'Set bTrigger to False
For siCount2 = 0 To iList.Max 'Loop through each item in iList
If Val(sList[siCount1]) = iList[siCount2] Then 'If the value of each is the same e.g. "98" = 98 then
bTrigger = True 'Set bTrigger to True
Continue 'Exit the loop
Endif
Next
If Not bTrigger Then 'If there was no match e.g. there is no "99" then..
sList[siCount1] = Left(sList[siCount1], siCounter - 1) 'Strip out the end digit e.g. "99" becomes 9 again
End If
Next
Print Val(sList.Join("")) 'Join all items in sList together and print
End</lang> Output:
998764543431 6054854654
Go
<lang go>// Variation of method 3. Repeat digits to at least the size of the longest, // then sort as strings. package main
import (
"fmt" "math/big" "sort" "strconv" "strings"
)
type c struct {
i int s, rs string
}
type cc []*c
func (c cc) Len() int { return len(c) } func (c cc) Less(i, j int) bool { return c[j].rs < c[i].rs } func (c cc) Swap(i, j int) { c[i], c[j] = c[j], c[i] }
// Function required by task. Takes a list of integers, returns big int. func li(is ...int) *big.Int {
ps := make(cc, len(is))
ss := make([]c, len(is))
ml := 0
for j, i := range is {
p := &ss[j]
ps[j] = p
p.i = i
p.s = strconv.Itoa(i)
if len(p.s) > ml {
ml = len(p.s)
}
}
for _, p := range ps {
p.rs = strings.Repeat(p.s, (ml+len(p.s)-1)/len(p.s))
}
sort.Sort(ps)
s := make([]string, len(ps))
for i, p := range ps {
s[i] = p.s
}
b, _ := new(big.Int).SetString(strings.Join(s, ""), 10)
return b
}
func main() {
fmt.Println(li(1, 34, 3, 98, 9, 76, 45, 4)) fmt.Println(li(54, 546, 548, 60))
}</lang>
- Output:
998764543431 6054854654
Groovy
<lang groovy>def largestInt = { c -> c.sort { v2, v1 -> "$v1$v2" <=> "$v2$v1" }.join() as BigInteger }</lang> Testing: <lang groovy>assert largestInt([1, 34, 3, 98, 9, 76, 45, 4]) == 998764543431 assert largestInt([54, 546, 548, 60]) == 6054854654</lang>
Haskell
Compare repeated string method
<lang Haskell>import Data.List (sortBy) import Data.Ord (comparing)
main = print (map maxcat [[1,34,3,98,9,76,45,4], [54,546,548,60]] :: [Integer])
where
sorted xs = let pad x = concat $ replicate (maxLen `div` length x + 1) x
maxLen = maximum $ map length xs
in sortBy (flip $ comparing pad) xs
maxcat = read . concat . sorted . map show</lang>
- Output:
[998764543431,6054854654]
Since repeating numerical string "1234" is the same as taking all the digits of 1234/9999 after the decimal point, the following does essentially the same as above: <lang haskell>import Data.List (sortBy) import Data.Ord (comparing) import Data.Ratio ((%))
nines = iterate ((+9).(*10)) 9
maxcat = foldl (\a (n,d)->a * (1 + d) + n) 0 .
sortBy (flip $ comparing $ uncurry (%)) . map (\a->(a, head $ dropWhile (<a) nines))
main = mapM_ (print.maxcat) [[1,34,3,98,9,76,45,4], [54,546,548,60]]</lang>
Sort on comparison of concatenated ints method
<lang Haskell>import Data.List (sortBy)
main = print (map maxcat [[1,34,3,98,9,76,45,4], [54,546,548,60]] :: [Integer])
where sorted = sortBy (\a b -> compare (b++a) (a++b))
maxcat = read . concat . sorted . map show</lang>
- Output as above.
Try all permutations method
<lang Haskell>import Data.List (permutations)
main :: IO () main =
print (maxcat <$> [[1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]] :: [Integer]) where maxcat = read . maximum . fmap (concatMap show) . permutations</lang>
- Output as above.
Icon and Unicon
This solution only works in Unicon as it uses a Heap class to do the heavy lifting.
<lang unicon>import Collections # For the Heap (dense priority queue) class
procedure main(a)
write(lici(a))
end
procedure lici(a)
every (result := "") ||:= Heap(a,,cmp).gen() return result
end
procedure cmp(a,b)
return (a||b) > (b||a)
end</lang>
Sample runs:
->lici 1 34 3 98 9 76 45 4 998764543431 ->lici 54 546 548 60 6054854654 ->
J
Solution: <lang j>maxlen=: [: >./ #&> maxnum=: (0 ". ;)@(\: maxlen $&> ])@(8!:0)</lang> Usage: <lang j> maxnum&> 1 34 3 98 9 76 45 4 ; 54 546 548 60 998764543431 6054854654</lang>
Java
This example sets up a comparator to order the numbers using Collections.sort as described in method #3 (padding and reverse sorting).
It was also necessary to make a join method to meet the output requirements.
<lang java5>import java.util.*;
public class IntConcat {
private static Comparator<Integer> sorter = new Comparator<Integer>(){
@Override
public int compare(Integer o1, Integer o2){
String o1s = o1.toString();
String o2s = o2.toString();
if(o1s.length() == o2s.length()){
return o2s.compareTo(o1s);
}
int mlen = Math.max(o1s.length(), o2s.length());
while(o1s.length() < mlen * 2) o1s += o1s;
while(o2s.length() < mlen * 2) o2s += o2s;
return o2s.compareTo(o1s);
}
};
public static String join(List<?> things){
String output = "";
for(Object obj:things){
output += obj;
}
return output;
}
public static void main(String[] args){
List<Integer> ints1 = new ArrayList<Integer>(Arrays.asList(1, 34, 3, 98, 9, 76, 45, 4));
Collections.sort(ints1, sorter);
System.out.println(join(ints1));
List<Integer> ints2 = new ArrayList<Integer>(Arrays.asList(54, 546, 548, 60));
Collections.sort(ints2, sorter);
System.out.println(join(ints2));
}
}</lang>
<lang java5>import java.util.Comparator; import java.util.stream.Collectors; import java.util.stream.Stream;
public interface IntConcat {
public static Comparator<Integer> SORTER = (o1, o2) -> {
String o1s = o1.toString();
String o2s = o2.toString();
if (o1s.length() == o2s.length()) {
return o2s.compareTo(o1s);
}
int mlen = Math.max(o1s.length(), o2s.length());
while (o1s.length() < mlen * 2) {
o1s += o1s;
}
while (o2s.length() < mlen * 2) {
o2s += o2s;
}
return o2s.compareTo(o1s);
};
public static void main(String[] args) {
Stream<Integer> ints1 = Stream.of(
1, 34, 3, 98, 9, 76, 45, 4
);
System.out.println(ints1
.parallel()
.sorted(SORTER)
.map(String::valueOf)
.collect(Collectors.joining())
);
Stream<Integer> ints2 = Stream.of(
54, 546, 548, 60
);
System.out.println(ints2
.parallel()
.sorted(SORTER)
.map(String::valueOf)
.collect(Collectors.joining())
);
}
}</lang>
- Output:
998764543431 6054854654
JavaScript
ES5
<lang JavaScript> (function () {
'use strict';
// maxCombine :: [Int] -> Int
function maxCombine(xs) {
return parseInt(
xs.sort(
function (x, y) {
var a = x.toString(),
b = y.toString(),
ab = parseInt(a + b),
ba = parseInt(b + a);
return ab > ba ? -1 : (ab < ba ? 1 : 0);
}
)
.join(), 10
);
}
return [
[1, 34, 3, 98, 9, 76, 45, 4],
[54, 546, 548, 60]
].map(maxCombine);
})();
</lang>
- Output:
[998764543431, 6054854654]
ES6
<lang JavaScript>var maxCombine = (a) => +(a.sort((x, y) => +("" + y + x) - +("" + x + y)).join());
// test & output console.log([
[1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]
].map(maxCombine));</lang>
jq
Padding
For jq versions greater than 1.4, it may be necessary to change "sort_by" to "sort". <lang jq>def largest_int:
def pad(n): . + (n - length) * .[length-1:];
map(tostring)
| (map(length) | max) as $max
| map([., pad($max)])
| sort_by( .[1] )
| map( .[0] ) | reverse | join("") ;
- Examples:
([1, 34, 3, 98, 9, 76, 45, 4],
[54, 546, 548, 60]) | largest_int
</lang>
- Output:
$ /usr/local/bin/jq -n -M -r -f Largest_int_from_concatenated_ints.jq 998764543431 6054854654
Custom Sort
The following uses quicksort/1: <lang jq>def largest_int:
map(tostring)
| quicksort( .[0] + .[1] < .[1] + .[0] )
| reverse | join("") ;</lang>
Julia
Perhaps algorithm 3 is more efficient, but algorithm 2 is decent and very easy to implement in Julia. So this solution uses algorithm 2.
<lang julia>function maxconcat(arr::Vector{<:Integer})
b = sort(string.(arr); lt=(x, y) -> x * y < y * x, rev=true) |> join return try parse(Int, b) catch parse(BigInt, b) end
end
tests = ([1, 34, 3, 98, 9, 76, 45, 4],
[54, 546, 548, 60],
[1, 34, 3, 98, 9, 76, 45, 4, 54, 546, 548, 60])
for arr in tests
println("Max concatenating in $arr:\n -> ", maxconcat(arr))
end</lang>
- Output:
Max concatenating in [1, 34, 3, 98, 9, 76, 45, 4]: -> 998764543431 Max concatenating in [54, 546, 548, 60]: -> 6054854654 Max concatenating in [1, 34, 3, 98, 9, 76, 45, 4, 54, 546, 548, 60]: -> 9987660548546544543431
Kotlin
<lang scala>import java.util.Comparator
fun main(args: Array<String>) {
val comparator = Comparator<Int> { x, y ->
val xy = (x.toString() + y).toInt()
val yx = (y.toString() + x).toInt()
xy.compareTo(yx)
}
fun findLargestSequence(array: IntArray): String {
return array.sortedWith(comparator).reversed().map { it.toString() }.joinToString("")
}
val source1 = intArrayOf(1, 34, 3, 98, 9, 76, 45, 4) println(findLargestSequence(source1))
val source2 = intArrayOf(54, 546, 548, 60) println(findLargestSequence(source2))
}</lang>
- Output:
998764543431 6054854654
Lua
<lang Lua>function icsort(numbers) table.sort(numbers,function(x,y) return (x..y) > (y..x) end) return numbers end
for _,numbers in pairs({{1, 34, 3, 98, 9, 76, 45, 4}, {54, 546, 548, 60}}) do print(('Numbers: {%s}\n Largest integer: %s'):format( table.concat(numbers,","),table.concat(icsort(numbers)) )) end</lang>
- Output:
Numbers: {1,34,3,98,9,76,45,4}
Largest integer: 998764543431
Numbers: {54,546,548,60}
Largest integer: 6054854654
Mathematica
<lang Mathematica>makeLargestInt[list_] := Module[{sortedlist},
sortedlist = Sort[list, Order[ToString[#1] <> ToString[#2], ToString[#2] <> ToString[#1]] < 0 &]; Map[ToString, sortedlist] // StringJoin // FromDigits ]
(* testing with two examples *) makeLargestInt[{1, 34, 3, 98, 9, 76, 45, 4}] makeLargestInt[{54, 546, 548, 60}]</lang>
- Output:
998764543431 6054854654
min
<lang min>(quote cons "" join) :s+ ('string map (over over swap s+ 's+ dip <) sort "" join int) :fn
(1 34 3 98 9 76 45 4) fn puts! (54 546 548 60) fn puts!</lang>
- Output:
998764543431 6054854654
NetRexx
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary
runSample(arg) return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method largestInt(il) public static
ri = wa = -- put the list into an indexed string wa[0] = il.words loop ww = 1 to wa[0] wa[ww] = il.word(ww) end ww
-- order the list
loop wx = 1 to wa[0] - 1
loop wy = wx + 1 to wa[0]
xx = wa[wx]
yy = wa[wy]
xy = xx || yy
yx = yy || xx
if xy < yx then do
-- swap xx and yy
wa[wx] = yy
wa[wy] = xx
end
end wy
end wx
-- rebuild list from indexed string loop ww = 1 to wa[0] ri = ri wa[ww] end ww return ri.space(0) -- concatenate the list elements into a single numeric
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method runSample(arg) private static
ints = [ - '1 34 3 98 9 76 45 4', - '54 546 548 60' - ] loop il over ints say largestInt(il).right(20) ':' il.space(1, ',') end il return
</lang>
- Output:
998764543431 : 1,34,3,98,9,76,45,4
6054854654 : 54,546,548,60
Nim
<lang nim>import algorithm, sequtils, strutils, sugar
proc maxNum(x: seq[int]): string =
var c = x.mapIt($it) c.sort((x, y) => cmp(y&x, x&y)) c.join()
echo maxNum(@[1, 34, 3, 98, 9, 76, 45, 4]) echo maxNum(@[54, 546, 548, 60])</lang>
- Output:
998764543431 6054854654
OCaml
<lang ocaml>let myCompare a b = compare (b ^ a) (a ^ b) let icsort nums = String.concat "" (List.sort myCompare (List.map string_of_int nums))</lang>
- testing
# icsort [1;34;3;98;9;76;45;4];; - : string = "998764543431" # icsort [54;546;548;60];; - : string = "6054854654"
Oforth
<lang Oforth>: largestInt map(#asString) sortWith(#[ 2dup + -rot swap + > ]) sum asInteger ;</lang>
- Output:
[ [1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60] ] map(#largestInt) . [998764543431, 6054854654]
PARI/GP
Sorts then joins. Most of the noise comes from converting a vector of integers into a concatenated integer: eval(concat(apply(n->Str(n),v))). Note that the short form eval(concat(apply(Str,v))) is not valid here because Str is variadic.
<lang parigp>large(v)=eval(concat(apply(n->Str(n),vecsort(v,(x,y)->eval(Str(y,x,"-",x,y)))))); large([1, 34, 3, 98, 9, 76, 45, 4]) large([54, 546, 548, 60])</lang>
- Output:
%1 = 998764543431 %2 = 6054854654
Pascal
tested with freepascal.Used a more extreme example 3.
algorithm 3
<lang pascal>const
base = 10; MaxDigitCnt = 11; source1 : array[0..7] of integer = (1, 34, 3, 98, 9, 76, 45, 4); source2 : array[0..3] of integer = (54,546,548,60); source3 : array[0..3] of integer = (60, 54,545454546,0);
type
tdata = record
datOrg,
datMod : LongWord;
datStrOrg : string[MaxDigitCnt];
end;
tArrData = array of tData;
procedure DigitCount(var n: tdata); begin
with n do //InttoStr is very fast str(datOrg,datStrOrg);
end;
procedure InsertData(var n: tdata;data:LongWord); begin
n.datOrg := data; DigitCount(n);
end;
function FindMaxLen(const ArrData:tArrData): LongWord; var
cnt : longInt; res,t : LongWord;
begin
res := 0;// 1 is minimum
for cnt := High(ArrData) downto Low(ArrData) do
begin
t := length(ArrData[cnt].datStrOrg);
IF res < t then
res := t;
end;
FindMaxLen := res;
end;
procedure ExtendCount(var ArrData:tArrData;newLen: integer); var
cnt, i,k : integer;
begin
For cnt := High(ArrData) downto Low(ArrData) do
with ArrData[cnt] do
begin
datMod := datOrg;
i := newlen-length(datStrOrg);
k := 1;
while i > 0 do
begin
datMod := datMod *Base+Ord(datStrOrg[k])-Ord('0');
inc(k);
IF k >length(datStrOrg) then
k := 1;
dec(i);
end;
end;
end;
procedure SortArrData(var ArrData:tArrData); var
i, j,idx : integer; tmpData : tData;
begin
For i := High(ArrData) downto Low(ArrData)+1 do
begin
idx := i;
j := i-1;
For j := j downto Low(ArrData) do
IF ArrData[idx].datMod < ArrData[j].datMod then
idx := j;
IF idx <> i then
begin
tmpData := ArrData[idx];
ArrData[idx]:= ArrData[i];
ArrData[i] := tmpData;
end;
end;
end;
procedure ArrDataOutput(const ArrData:tArrData); var
i,l : integer; s : string;
begin { the easy way
For i := High(ArrData) downto Low(ArrData) do
write(ArrData[i].datStrOrg);
writeln;
*}
l := 0;
For i := High(ArrData) downto Low(ArrData) do
inc(l,length(ArrData[i].datStrOrg));
setlength(s,l);
l:= 1;
For i := High(ArrData) downto Low(ArrData) do
with ArrData[i] do
begin
move(datStrOrg[1],s[l],length(datStrOrg));
inc(l,length(datStrOrg));
end;
writeln(s);
end;
procedure HighestInt(var ArrData:tArrData); begin
ExtendCount(ArrData,FindMaxLen(ArrData)); SortArrData(ArrData); ArrDataOutput(ArrData);
end;
var
i : integer; tmpData : tArrData;
begin
// Source1 setlength(tmpData,length(source1)); For i := low(tmpData) to high(tmpData) do InsertData(tmpData[i],source1[i]); HighestInt(tmpData); // Source2 setlength(tmpData,length(source2)); For i := low(tmpData) to high(tmpData) do InsertData(tmpData[i],source2[i]); HighestInt(tmpData); // Source3 setlength(tmpData,length(source3)); For i := low(tmpData) to high(tmpData) do InsertData(tmpData[i],source3[i]); HighestInt(tmpData);
end.</lang>
- Output:
998764543431 6054854654 60545454546540
Inspired by Haskell
generate the repetition by dividing /(10^CountDigits-1) http://rosettacode.org/wiki/Largest_int_from_concatenated_ints#Compare_repeated_string_method
<lang pascal>const
base = 10; MaxDigitCnt = 11; source1 : array[0..7] of LongInt = (10 , 34, 3, 98, 9, 76, 45, 4); source2 : array[0..3] of LongInt = (54,546,548,60); source3 : array[0..3] of LongInt = (0,2121212122,21,60);
type
tdata = record
datMod : double;
datOrg : LongInt;
//InttoStr is very fast and the string is always needed
datStrOrg : string[MaxDigitCnt];
end;
tArrData = array of tData;
procedure InsertData(var n: tdata;data:LongWord); begin
with n do begin datOrg := data; str(datOrg,datStrOrg); end;
end;
function FindMaxLen(const ArrData:tArrData): LongWord; var
cnt : longInt; res,t : LongWord;
begin
res := 0;// 1 is minimum
for cnt := High(ArrData) downto Low(ArrData) do
begin
t := length(ArrData[cnt].datStrOrg);
IF res < t then
res := t;
end;
FindMaxLen := res;
end;
procedure ExtendData(var ArrData:tArrData;newLen: integer); var
cnt, i : integer;
begin
For cnt := High(ArrData) downto Low(ArrData) do
with ArrData[cnt] do
begin
//generating 10^length(datStrOrg)
datMod := 1;
i := length(datStrOrg);
// i always >= 1
repeat
datMod := base*datMod;
dec(i);
until i <= 0;
// 1/(datMod-1.0) = 1/(9...9)
datMod := datOrg/(datMod-1.0)+datOrg;
i := newlen-length(datStrOrg);
For i := i downto 1 do
datMod := datMod*Base;
end;
end;
procedure SortArrData(var ArrData:tArrData); //selection sort var
i, j,idx : integer; tmpData : tData;
begin
For i := High(ArrData) downto Low(ArrData)+1 do
begin
idx := i;
j := i-1;
//select max
For j := j downto Low(ArrData) do
IF ArrData[idx].datMod < ArrData[j].datMod then
idx := j;
//finally swap
IF idx <> i then
begin
tmpData := ArrData[idx];
ArrData[idx]:= ArrData[i];
ArrData[i] := tmpData;
end;
end;
end;
procedure ArrDataOutput(const ArrData:tArrData); var
i : integer;
begin { the easy way}
For i := High(ArrData) downto Low(ArrData) do write(ArrData[i].datStrOrg); writeln;
end;
procedure HighestInt(var ArrData:tArrData); begin
ExtendData(ArrData,FindMaxLen(ArrData)); SortArrData(ArrData); ArrDataOutput(ArrData);
end;
var
i : integer; tmpData : tArrData;
begin
// Source1 setlength(tmpData,length(source1)); For i := low(tmpData) to high(tmpData) do InsertData(tmpData[i],source1[i]); HighestInt(tmpData); // Source2 setlength(tmpData,length(source2)); For i := low(tmpData) to high(tmpData) do InsertData(tmpData[i],source2[i]); HighestInt(tmpData); // Source3 setlength(tmpData,length(source3)); For i := low(tmpData) to high(tmpData) do InsertData(tmpData[i],source3[i]); HighestInt(tmpData);
end.</lang>
- Output:
9987645434310 6054854654 602121212122210>
Perl
<lang perl>sub maxnum {
join , sort { "$b$a" cmp "$a$b" } @_
}
print maxnum(1, 34, 3, 98, 9, 76, 45, 4), "\n"; print maxnum(54, 546, 548, 60), "\n";</lang>
- Output:
998764543431 6054854654
Phix
<lang Phix>function catcmp(string a, string b)
return compare(b&a,a&b)
end function
function method2(sequence s)
for i=1 to length(s) do
s[i] = sprintf("%d",s[i])
end for
s = custom_sort(routine_id("catcmp"),s)
return join(s,"")
end function
? method2({1,34,3,98,9,76,45,4}) ? method2({54,546,548,60})</lang>
- Output:
"998764543431" "6054854654"
PHP
<lang php>function maxnum($nums) {
usort($nums, function ($x, $y) { return strcmp("$y$x", "$x$y"); });
return implode(, $nums);
}
echo maxnum(array(1, 34, 3, 98, 9, 76, 45, 4)), "\n"; echo maxnum(array(54, 546, 548, 60)), "\n";</lang>
- Output:
998764543431 6054854654
PicoLisp
Here are solutions for all three algorithms.
The third solution actually avoids padding the numbers, by converting them into circular lists and comparing these. As a drawback, however, this works only for unique lists (as the comparison of identical numbers would not terminate), so a better solution might involve additional checks. <lang PicoLisp>(load "@lib/simul.l") # For 'permute'</lang>
Algorithm 1
<lang PicoLisp>(for L '((1 34 3 98 9 76 45 4) (54 546 548 60))
(prinl (maxi format (permute L))) )</lang>
Algorithm 2
<lang PicoLisp>(for L '((1 34 3 98 9 76 45 4) (54 546 548 60))
(prinl
(sort L
'((A B)
(>
(format (pack A B))
(format (pack B A)) ) ) ) ) )</lang>
Algorithm 3
<lang PicoLisp>(for L '((1 34 3 98 9 76 45 4) (54 546 548 60))
(prinl
(flip
(by '((N) (apply circ (chop N))) sort L) ) ) )</lang>
- Output:
in all three cases
998764543431 6054854654
PL/I
<lang pli> /* Largest catenation of integers 16 October 2013 */ /* Sort using method 2, comparing pairs of adjacent integers. */
Largest: procedure options (main);
declare s(*) char (20) varying controlled, n fixed binary;
get (n);
allocate s(n);
get list (s);
s = trim(s);
put skip edit (s) (a, x(1));
put skip list ('Largest integer=', Largest_integer());
largest_integer: procedure () returns (char(100) varying);
declare sorted bit (1);
declare (true value ('1'b), false value ('0'b)) bit (1);
declare i fixed binary;
declare temp character(20) varying;
do until (sorted);
sorted = true;
do i = 1 to n-1;
if char(s(i)) || char(s(i+1)) < char(s(i+1)) || char(s(i)) then
do;
temp = s(i); s(i) = s(i+1); s(i+1) = temp; sorted = false;
end;
end;
end;
return (string(s));
end largest_integer; end Largest; </lang>
54 546 548 60 Largest integer= 6054854654 1 34 3 98 9 76 45 4 Largest integer= 998764543431
PowerShell
Using algorithm 3 <lang PowerShell>Function Get-LargestConcatenation ( [int[]]$Integers )
{
# Get the length of the largest integer
$Length = ( $Integers | Sort -Descending | Select -First 1 ).ToString().Length
# Convert to an array of strings,
# sort by each number repeated Length times and truncated to Length,
# and concatenate (join)
$Concat = ( [string[]]$Integers | Sort { ( $_ * $Length ).Substring( 0, $Length ) } -Descending ) -join
# Convert to integer (upsizing type if needed)
try { $Integer = [ int32]$Concat }
catch { try { $Integer = [ int64]$Concat }
catch { $Integer = [bigint]$Concat } }
return $Integer
}</lang>
<lang PowerShell>Get-LargestConcatenation 1, 34, 3, 98, 9, 76, 45, 4 Get-LargestConcatenation 54, 546, 548, 60 Get-LargestConcatenation 54, 546, 548, 60, 54, 546, 548, 60</lang>
- Output:
998764543431 6054854654 60605485485465465454
Prolog
Works with SWI-Prolog 6.5.3.
All permutations method
<lang Prolog>largest_int_v1(In, Out) :- maplist(name, In, LC), aggregate(max(V), get_int(LC, V), Out).
get_int(LC, V) :-
permutation(LC, P),
append(P, LV),
name(V, LV).
</lang>
- Output:
?- largest_int_v1([1, 34, 3, 98, 9, 76, 45, 4], Out). Out = 998764543431. ?- largest_int_v1([54, 546, 548, 60], Out). Out = 6054854654.
Method 2
<lang Prolog>largest_int_v2(In, Out) :- maplist(name, In, LC), predsort(my_sort,LC, LCS), append(LCS, LC1), name(Out, LC1).
my_sort(R, L1, L2) :-
append(L1, L2, V1), name(I1, V1),
append(L2, L1, V2), name(I2, V2),
( I1 < I2, R = >; I1 = I2, R = '='; R = <).
% particular case 95 958 my_sort(>, [H1], [H1, H2 | _]) :- H1 > H2.
my_sort(<, [H1], [H1, H2 | _]) :- H1 < H2.
my_sort(R, [H1], [H1, H1 | T]) :- my_sort(R, [H1], [H1 | T]).
% particular case 958 95 my_sort(>, [H1, H2 | _], [H1]) :- H1 > H2.
my_sort(<, [H1, H2 | _], [H1]) :- H1 < H2.
my_sort(R, [H1, H1 | T], [H1]) :- my_sort(R, [H1 | T], [H1]) . </lang>
- Output:
?- largest_int_v2([1, 34, 3, 98, 9, 76, 45, 4], Out). Out = 998764543431 . ?- largest_int_v2([54, 546, 548, 60], Out). Out = 5486054654 .
Python
Python: Sort on comparison of concatenated ints method
This also shows one of the few times where cmp= is better than key= on sorted()
<lang python>try:
cmp # Python 2 OK or NameError in Python 3
def maxnum(x):
return .join(sorted((str(n) for n in x),
cmp=lambda x,y:cmp(y+x, x+y)))
except NameError:
# Python 3
from functools import cmp_to_key
def cmp(x, y):
return -1 if x<y else ( 0 if x==y else 1)
def maxnum(x):
return .join(sorted((str(n) for n in x),
key=cmp_to_key(lambda x,y:cmp(y+x, x+y))))
for numbers in [(1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:
print('Numbers: %r\n Largest integer: %15s' % (numbers, maxnum(numbers)))</lang>
- Output:
Numbers: (1, 34, 3, 98, 9, 76, 45, 4) Largest integer: 998764543431 Numbers: (54, 546, 548, 60) Largest integer: 6054854654
Python: Compare repeated string method
<lang python>def maxnum(x):
maxlen = len(str(max(x)))
return .join(sorted((str(v) for v in x), reverse=True,
key=lambda i: i*(maxlen * 2 // len(i))))
for numbers in [(212, 21221), (1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:
print('Numbers: %r\n Largest integer: %15s' % (numbers, maxnum(numbers)))</lang>
- Output:
Numbers: (212, 21221) Largest integer: 21221221 Numbers: (1, 34, 3, 98, 9, 76, 45, 4) Largest integer: 998764543431 Numbers: (54, 546, 548, 60) Largest integer: 6054854654
<lang python>from fractions import Fraction from math import log10
def maxnum(x):
return .join(str(n) for n in sorted(x, reverse=True,
key=lambda i: Fraction(i, 10**(int(log10(i))+1)-1)))
for numbers in [(1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:
print('Numbers: %r\n Largest integer: %15s' % (numbers, maxnum(numbers)))</lang>
- Output as first Python example, above.
Python: Try all permutations method
<lang python>from itertools import permutations def maxnum(x):
return max(int(.join(n) for n in permutations(str(i) for i in x)))
for numbers in [(1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:
print('Numbers: %r\n Largest integer: %15s' % (numbers, maxnum(numbers)))</lang>
- Output as above.
Quackery
<lang quackery>[ sortwith
[ 2dup swap join
dip join $< ]
[] swap witheach join ] is largest-int ( [ --> $ )
$ '1 34 3 98 9 76 45 4' nest$ largest-int echo$ cr $ '54 546 548 60' nest$ largest-int echo$</lang>
- Output:
998764543431 6054854654
Racket
<lang Racket>
- lang racket
(define (largest-int ns)
(string->number (apply ~a (sort ns (λ(x y) (string>? (~a x y) (~a y x)))))))
(map largest-int '((1 34 3 98 9 76 45 4) (54 546 548 60)))
- -> '(998764543431 6054854654)
</lang>
Raku
(formerly Perl 6) <lang perl6>sub maxnum(*@x) {
[~] @x.sort: -> $a, $b { $b ~ $a leg $a ~ $b }
}
say maxnum <1 34 3 98 9 76 45 4>; say maxnum <54 546 548 60>;</lang>
- Output:
998764543431 6054854654
Red
<lang Rebol>Red []
foreach seq [[1 34 3 98 9 76 45 4] [54 546 548 60]] [
print rejoin sort/compare seq function [a b] [ (rejoin [a b]) > rejoin [b a] ]
] </lang>
- Output:
998764543431 6054854654
REXX
The algorithm used is based on exact comparisons (left to right) with right digit fill of the left digit.
This allows the integers to be of any size.
This REXX version works with any size integer (negative, zero, positive), and does some basic error checking to
verify that the numbers are indeed integers (and it also normalizes the integers).
The absolute value is used for negative numbers. No sorting of the numbers is required for the 1st two examples.
simple integers
<lang rexx>/*REXX program constructs the largest integer from an integer list using concatenation.*/ @.=.; @.1 = 1 34 3 98 9 76 45 4 /*the 1st integer list to be used. */
@.2 = 54 546 548 60 /* " 2nd " " " " " */
@.3 = 4 45 54 5 /* " 3rd " " " " " */
w=0 /* [↓] process all the integer lists.*/
do j=1 while @.j\==.; z= space(@.j) /*keep truckin' until lists exhausted. */
w=max(w, length(z) ); $= /*obtain maximum width to align output.*/
do while z\=; idx= 1; big= norm(1) /*keep examining the list until done.*/
do k=2 to words(z); #= norm(k) /*obtain an a number from the list. */
L= max(length(big), length(#) ) /*get the maximum length of the integer*/
if left(#, L, left(#, 1) ) <<= left(big, L, left(big, 1) ) then iterate
big= #; idx= k /*we found a new biggie (and the index)*/
end /*k*/ /* [↑] find max concatenated integer. */
z= delword(z, idx, 1) /*delete this maximum integer from list*/
$= $ || big /*append " " " ───► $. */
end /*while z*/ /* [↑] process all integers in a list.*/
say 'largest concatenatated integer from ' left( space(@.j), w) " is ─────► " $
end /*j*/ /* [↑] process each list of integers. */
exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ norm: arg i; #= word(z, i); er= '***error***'; if left(#, 1)=="-" then #= substr(#, 2)
if \datatype(#,'W') then do; say er # "isn't an integer."; exit 13; end; return #/1</lang>
- output when using the default (internal) integer lists:
largest concatenatated integer from 1 34 3 98 9 76 45 4 is ─────► 998764543431 largest concatenatated integer from 54 546 548 60 is ─────► 6054854654 largest concatenatated integer from 4 45 54 5 is ─────► 554454
exponentiated integers
In REXX, a number such as 6.6e77 would be considered an integer if the (current) numeric digits is
large enough to express that number as an integer without the exponent.
The default for REXX is 9 decimal digits, but the norm function automatically uses enough decimal digits to
express the number as an integer.
This REXX version can handle any sized integer (most REXXes can handle up to around eight million decimal
digits, but displaying the result would be problematic for results wider than the display area).
<lang rexx>/*REXX program constructs the largest integer from an integer list using concatenation.*/
@.=.; @.1 = 1 34 3 98 9 76 45 4 /*the 1st integer list to be used. */
@.2 = 54 546 548 60 /* " 2nd " " " " " */
@.3 = 4 45 54 5 /* " 3rd " " " " " */
@.4 = 4 45 54 5 6.6e77 /* " 4th " " " " " */
w= 0 /* [↓] process all the integer lists.*/
do j=1 while @.j\==.; z= space(@.j) /*keep truckin' until lists exhausted. */
w=max(w, length(z) ); $= /*obtain maximum width to align output.*/
do while z\=; idx=1; big= norm(1) /*keep examining the list until done.*/
do k=2 to words(z); #= norm(k) /*obtain an a number from the list. */
L= max(length(big), length(#) ) /*get the maximum length of the integer*/
if left(#, L, left(#, 1) ) <<= left(big, L, left(big, 1) ) then iterate
big=#; idx= k /*we found a new biggie (and the index)*/
end /*k*/ /* [↑] find max concatenated integer. */
z= delword(z, idx, 1) /*delete this maximum integer from list*/
$= $ || big /*append " " " ───► $. */
end /*while z*/ /* [↑] process all integers in a list.*/
say 'largest concatenatated integer from ' left( space(@.j), w) " is " $
end /*j*/ /* [↑] process each list of integers. */
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ norm: arg i; #= word(z, i); er= '***error***'; if left(#, 1)=="-" then #= substr(#, 2)
if \datatype(#, 'N') then signal er13 /*go and tell err msg.*/
else #= # / 1 /*a #, so normalize it*/
if pos('E',#)>0 then do; parse var # mant "E" pow /*Has exponent? Expand*/
numeric digits pow + length(mand) /*expand digs, adjust#*/
end
if datatype(#, 'W') then return # / 1
er13: say er # "isn't an integer."; exit 13</lang>
- output when using the default (internal) integer lists:
(Output shown at three-quarter size.)
largest concatenatated integer from 1 34 3 98 9 76 45 4 is 998764543431 largest concatenatated integer from 54 546 548 60 is 6054854654 largest concatenatated integer from 4 45 54 5 is 554454 largest concatenatated integer from 4 45 54 5 6.6e77 is 660000000000000000000000000000000000000000000000000000000000000000000000000000554454
Alternate Version
Inspired by the previous versions. <lang>/*REXX program constructs the largest integer from an integer list using concatenation.*/ l.=; l.1 = '1 34 3 98 9 76 45 4' /*the 1st integer list to be used. */
l.2 = '54 546 548 60' /* " 2nd " " " " " */
l.3 = ' 4 45 54 5' /* " 3rd " " " " " */
l.4 = ' 4 45 54 5 6.6e77' /* " 4th " " " " " */
l.5 = ' 3 3 .2' /* " 5th " " " " " */
/* soll.1=998764543431 soll.2=6054854654 soll.3=554454 soll.4=660000000000000000000000000000000000000000000000000000000000000000000000000000545454
- /
l_length=0 Do li=1 By 1 While l.li<>
l_length=max(l_length,length(space(l.li))) End
Do li=1 By 1 While l.li<>
z=
Do j=1 To words(l.li)
int=integer(word(l.li,j))
If int='?' Then Do
Say left(space(l.li),l_length) '-> ** invalid ** bad integer' word(l.li,j)
Iterate li
End
Else
z=z int
End
/*Say copies(' ',l_length) ' ' soll.li */
Say left(space(l.li),l_length) '->' largeint(l.li) End
Exit
integer: Procedure Numeric Digits 1000 Parse Arg z If Datatype(z,'W') Then
Return z+0
Else
Return '?'
largeint: result= Do While z<> /* [?] check the rest of the integers.*/
big=word(z,1); index=1; LB=length(big) /*assume that first integer is biggest.*/ do k=2 to words(z); n=word(z,k) /*obtain an integer from the list. */ L=max(LB,length(n)) /*get the maximum length of the integer*/ if left(n,L,left(n,1))<<=left(big,L,left(big,1)) then iterate big=n; index=k /*we found a new biggie (and the index)*/ LB=length(big) End /*k*/ z=delword(z,index,1) /*delete this maximum integer from list*/ result=result||big /*append " " " ---? $. */ end /*while z*/ /* [?] process all integers in a list.*/
Return result</lang>
- Output:
1 34 3 98 9 76 45 4 -> 998764543431 54 546 548 60 -> 6054854654 4 45 54 5 -> 554454 4 45 54 5 6.6e77 -> 660000000000000000000000000000000000000000000000000000000000000000000000000000554454 3 3 .2 -> ** invalid ** bad integer .2
Version 4
<lang rexx>/*REXX program constructs the largest integer from an integer list using concatenation.*/ l.=; l.1 = '1 34 3 98 9 76 45 4' /*the 1st integer list to be used. */
l.2 = '54 546 548 60' /* " 2nd " " " " " */
l.3 = ' 4 45 54 5' /* " 3rd " " " " " */
l.4 = ' 4 45 54 5 6.6e77' /* " 4th " " " " " */
l.5 = ' 3 3 .2' /* " 5th " " " " " */
l.6 = ' 4 45 54 5 6.6e1001' /* " 6th " " " " " */
l.7 = ' 4.0000 45 54 5.00' /* " 7th " " " " " */
l.8 = ' 10e999999999 5' /* " 8th " " " " " */
l_length=0 Do li=1 By 1 While l.li<>
l_length=max(l_length,length(space(l.li))) End
Do li=1 By 1 While l.li<>
z=
msg=
Do j=1 To words(l.li)
int=integer(word(l.li,j))
If int='?' Then Do
Say left(space(l.li),l_length) '-> ** invalid ** bad list item:' word(l.li,j) msg
Iterate li
End
Else
z=z int
End
zz=largeint(z)
If length(zz)<60 Then
Say left(space(l.li),l_length) '->' zz
Else
Say left(space(l.li),l_length) '->' left(zz,5)'...'right(zz,5)
End
Exit
integer: Procedure Expose msg Numeric Digits 1000 Parse Arg z If Datatype(z,'W') Then
Return z/1
Else Do
If Datatype(z,'NUM') Then Do
Do i=1 To 6 Until dig>=999999999
dig= digits()*10
dig=min(dig,999999999)
Numeric Digits dig
If Datatype(z,'W') Then
Return z/1
End
msg='cannot convert it to an integer'
Return '?'
End
Else Do
msg='not a number (larger than what this REXX can handle)'
Return '?'
End
End
largeint: Procedure Parse Arg list w.0=words(list) Do i=1 To w.0
w.i=word(list,i) End
Do wx=1 To w.0-1
Do wy=wx+1 To w.0
xx=w.wx
yy=w.wy
xy=xx||yy
yx=yy||xx
if xy < yx then do
/* swap xx and yy */
w.wx = yy
w.wy = xx
end
End
End
list= Do ww=1 To w.0
list=list w.ww End
Return space(list,0)</lang>
- Output:
1 34 3 98 9 76 45 4 -> 998764543431 54 546 548 60 -> 6054854654 4 45 54 5 -> 554454 4 45 54 5 6.6e77 -> 66000...54454 3 3 .2 -> ** invalid ** bad list item: .2 cannot convert it to an integer 4 45 54 5 6.6e1001 -> 66000...54454 4.0000 45 54 5.00 -> 554454 10e999999999 5 -> ** invalid ** bad list item: 10e999999999 not a number (larger than what this REXX can handle)
Ring
<lang ring> nums=[1,34,3,98,9,76,45,4] see largestInt(8) + nl nums=[54,546,548,60] see largestInt(4) + nl
func largestInt len l = "" sorted = false while not sorted
sorted=true
for i=1 to len-1
a=string(nums[i])
b=string(nums[i+1])
if a+b<b+a
temp = nums[i]
nums[i] = nums[i+1]
nums[i+1] = temp
sorted=false ok
next
end for i=1 to len
l+=string(nums[i])
next return l </lang> Output:
998764543431 6054854654
Ruby
Sort on comparison of concatenated ints method
<lang Ruby>def icsort nums
nums.sort { |x, y| "#{y}#{x}" <=> "#{x}#{y}" }
end
[[54, 546, 548, 60], [1, 34, 3, 98, 9, 76, 45, 4]].each do |c|
p c # prints nicer in Ruby 1.8 puts icsort(c).join
end</lang>
- Output:
[54, 546, 548, 60] 6054854654 [1, 34, 3, 98, 9, 76, 45, 4] 998764543431
Compare repeated string method
<lang ruby>def icsort nums
maxlen = nums.max.to_s.length
nums.map{ |x| x.to_s }.sort_by { |x| x * (maxlen * 2 / x.length) }.reverse
end
[[54, 546, 548, 60], [1, 34, 3, 98, 9, 76, 45, 4]].each do |c|
p c # prints nicer in Ruby 1.8 puts icsort(c).join
end</lang>
- Output as above.
<lang ruby>require 'rational' #Only needed in Ruby < 1.9
def icsort nums
nums.sort_by { |i| Rational(i, 10**(Math.log10(i).to_i+1)-1) }.reverse
end
[[54, 546, 548, 60], [1, 34, 3, 98, 9, 76, 45, 4]].each do |c|
p c # prints nicer in Ruby 1.8 puts icsort(c).join
end</lang>
- Output as above.
Run BASIC
<lang runbasic>a1$ = "1, 34, 3, 98, 9, 76, 45, 4" a2$ = "54,546,548,60"
print "Max Num ";a1$;" = ";maxNum$(a1$) print "Max Num ";a2$;" = ";maxNum$(a2$)
function maxNum$(a1$) while word$(a1$,i+1,",") <> ""
i = i + 1 a$(i) = trim$(word$(a1$,i,","))
wend
s = 1 while s = 1
s = 0 for j = 1 to i -1 if a$(j)+a$(j+1) < a$(j+1)+a$(j) then h$ = a$(j) a$(j) = a$(j+1) a$(j+1) = h$ s = 1 end if next j
wend
for j = 1 to i
maxNum$ = maxNum$ ; a$(j)
next j end function</lang>
- Output:
Max Num 1, 34, 3, 98, 9, 76, 45, 4 = 998764543431 Max Num 54,546,548,60 = 6054854654
Rust
<lang Rust>fn maxcat(a: &mut [u32]) {
a.sort_by(|x, y| {
let xy = format!("{}{}", x, y);
let yx = format!("{}{}", y, x);
xy.cmp(&yx).reverse()
});
for x in a {
print!("{}", x);
}
println!();
}
fn main() {
maxcat(&mut [1, 34, 3, 98, 9, 76, 45, 4]); maxcat(&mut [54, 546, 548, 60]);
}</lang>
- Output:
998764543431 6054854654
S-lang
<lang S-lang>define catcmp(a, b) {
a = string(a); b = string(b); return strcmp(b+a, a+b);
}
define maxcat(arr) {
arr = arr[array_sort(arr, &catcmp)];
variable result = "", elem;
foreach elem (arr)
result += string(elem);
return result;
}
print("max of series 1 is " + maxcat([1, 34, 3, 98, 9, 76, 45, 4])); print("max of series 2 is " + maxcat([54, 546, 548, 60])); </lang>
- Output:
"max of series 1 is 998764543431" "max of series 2 is 6054854654"
Scala
<lang Scala>object LIFCI extends App {
def lifci(list: List[Long]) = list.permutations.map(_.mkString).max
println(lifci(List(1, 34, 3, 98, 9, 76, 45, 4))) println(lifci(List(54, 546, 548, 60)))
}</lang>
- Output:
998764543431 6054854654
Scheme
<lang Scheme>(define (cat . nums) (apply string-append (map number->string nums)))
(define (my-compare a b) (string>? (cat a b) (cat b a)))
(map (lambda (xs) (string->number (apply cat (sort xs my-compare))))
'((1 34 3 98 9 76 45 4) (54 546 548 60)))</lang>
- Output:
(998764543431 6054854654)
Sidef
<lang ruby>func maxnum(nums) {
nums.sort {|x,y| "#{y}#{x}" <=> "#{x}#{y}" };
}
[[54, 546, 548, 60], [1, 34, 3, 98, 9, 76, 45, 4]].each { |c|
say maxnum(c).join.to_num;
}</lang>
- Output:
6054854654 998764543431
Smalltalk
Version 1) sort by padded print strings:
<lang smalltalk>#(
(54 546 548 60) (1 34 3 98 9 76 45 4)
) do:[:ints |
|resultString|
"sort ints by padded strings (sort a copy - literals are immudatble),
then collect their strings, then concatenate"
resultString :=
((ints copy sort:[:a :b |
|pad|
pad := (a integerLog10) max:(b integerLog10).
(a printString paddedTo:pad with:$0) > (b printString paddedTo:pad with:$0)])
collect:#printString) asStringWith:.
Stdout printCR: resultString.
].</lang> Version 2) alternative: sort by concatenated pair's strings: <lang smalltalk>#(
(54 546 548 60) (1 34 3 98 9 76 45 4)
) do:[:ints |
|resultString|
resultString :=
((ints copy sort:[:a :b | e'{a}{b}' > e'{b}{a}']) "(1)"
collect:#printString) asStringWith:.
Stdout printCR: resultString.
].</lang> Note ¹ replace "e'{a}{b}'" by "(a printString,b printString)" in dialects, which do not support embedded expression strings.
Version 3) no need to collect the resultString; simply print the sorted list (ok, if printing is all we want): <lang smalltalk>#(
(54 546 548 60) (1 34 3 98 9 76 45 4)
) do:[:ints |
(ints copy sort:[:a :b | e'{a}{b}' > e'{b}{a}'])
do:[:eachNr | eachNr printOn:Stdout].
Stdout cr.
]</lang>
Version 4) no need to generate any intermediate strings; the following will do as well: <lang smalltalk>#(
(54 546 548 60) (1 34 3 98 9 76 45 4)
) do:[:ints |
(ints copy sortByApplying:[:i | i log10 fractionPart]) reverseDo:#print. Stdout cr.
]</lang>
- Output:
6054854654 989764543431
Tcl
<lang tcl>proc intcatsort {nums} {
lsort -command {apply {{x y} {expr {"$y$x" - "$x$y"}}}} $nums
}</lang> Demonstrating: <lang tcl>foreach collection {
{1 34 3 98 9 76 45 4}
{54 546 548 60}
} {
set sorted [intcatsort $collection] puts "\[$collection\] => \[$sorted\] (concatenated: [join $sorted ""])"
}</lang>
- Output:
[1 34 3 98 9 76 45 4] => [9 98 76 45 4 34 3 1] (concatenated: 998764543431) [54 546 548 60] => [60 548 546 54] (concatenated: 6054854654)
VBScript
<lang vb> Function largestint(list) nums = Split(list,",") Do Until IsSorted = True IsSorted = True For i = 0 To UBound(nums) If i <> UBound(nums) Then a = nums(i) b = nums(i+1) If CLng(a&b) < CLng(b&a) Then tmpnum = nums(i) nums(i) = nums(i+1) nums(i+1) = tmpnum IsSorted = False End If End If Next Loop For j = 0 To UBound(nums) largestint = largestint & nums(j) Next End Function
WScript.StdOut.Write largestint(WScript.Arguments(0)) WScript.StdOut.WriteLine </lang>
- Output:
F:\>cscript /nologo largestint.vbs 1,34,3,98,9,76,45,4 998764543431 F:\>cscript /nologo largestint.vbs 54,546,548,60 6054854654
Vim Script
This solution is intended to be run as an Ex command within a buffer containing the integers to be processed, one per line. <lang Vim>%s/\(.\+\)/\1\1/ | sort! | %s/\(.\+\)\1\n/\1/</lang>
- Demonstration
<lang Bash>$ paste -s nums 1 34 3 98 9 76 45 4 $ vim -S icsort.vim nums 998764543431</lang>
Wren
<lang ecmascript>import "/sort" for Sort
var cmp = Fn.new { |x, y|
var xy = Num.fromString(x.toString + y.toString) var yx = Num.fromString(y.toString + x.toString) return (xy - yx).sign
}
var findLargestSequence = Fn.new { |a|
var b = Sort.merge(a, cmp) return b[-1..0].join()
}
var arrays = [
[1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]
] for (a in arrays) {
System.print("%(a) -> %(findLargestSequence.call(a))")
}</lang>
- Output:
[1, 34, 3, 98, 9, 76, 45, 4] -> 998764543431 [54, 546, 548, 60] -> 6054854654
zkl
<lang zkl>fcn bigCI(ns){
ns.apply("toString").sort(fcn(a,b){ (a+b)>(b+a) }).concat();
}</lang> <lang zkl>bigCI(T(1, 34, 3, 98, 9, 76, 45, 4)).println(); bigCI(T(54, 546, 548, 60)).println();</lang>
- Output:
998764543431 6054854654
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