# Consolidate an Array

The task is simple: consolidate an array of ints. Consolidating this array consists of the following:

• All instances of 0 need to be moved to the end of the array.
• There should be no 0s between the non-zero integers.
• All non-zero indices should retain their order.

# Challenge

Consolidate an array in the least amount of bytes.

You are consolidating an array of random length with a size up to your language's max with random integers. Input may be any natural way for your language.

# Examples

Input

0 5 8 8 3 5 1 6 8 4 0 3 7 5 6 4 4 7 5 6 7 4 4 9 1 0 5 7 9 3 0 2 2 4 3 0 4 8 7 3 1 4 7 5 1 2 1 8 7 8 7 7 2 6 3 1 2 8 5 1 4 2 0 5 0 6 0 3

Output

5 8 8 3 5 1 6 8 4 3 7 5 6 4 4 7 5 6 7 4 4 9 1 5 7 9 3 2 2 4 3 4 8 7 3 1 4 7 5 1 2 1 8 7 8 7 7 2 6 3 1 2 8 5 1 4 2 5 6 3 0 0 0 0 0 0 0 0

Input

-1 -7 -6 5 1 -5 -2 7 -3 -8 0 8 9 1 -8 -1 6 -4 1 -2 1 -7 5 4 -6 7 -3 9 8 3 -1 0 -5 -7 3 8 1 1 3 -3 -2 -2 0 -7 0 -4 8 6 -3 6 0 5 3 2 2 2 -2 -7 -3 9 -1 6 0 6 -7 9 4 -2 8 -8 -4 1 -8 4 3 7 3 5 1 0 3 3 7 -1 -5 1 -3 4 -7 0 3 2 -2 7 -3 0 0 2 -5 8 -3 -2 -7 -5 7 -3 -9 -7 5 8 -3 9 6 7 -2 4 7

Output

-1 -7 -6 5 1 -5 -2 7 -3 -8 8 9 1 -8 -1 6 -4 1 -2 1 -7 5 4 -6 7 -3 9 8 3 -1 -5 -7 3 8 1 1 3 -3 -2 -2 -7 -4 8 6 -3 6 5 3 2 2 2 -2 -7 -3 9 -1 6 6 -7 9 4 -2 8 -8 -4 1 -8 4 3 7 3 5 1 3 3 7 -1 -5 1 -3 4 -7 3 2 -2 7 -3 2 -5 8 -3 -2 -7 -5 7 -3 -9 -7 5 8 -3 9 6 7 -2 4 7 0 0 0 0 0 0 0 0 0 0

# Example Code (Java)

public class Consolidate {
public static void main(String[] args) throws Exception {
int[] toConsolidate = new int[args.length];
for (int i=0; i<args.length; i++){
toConsolidate[i]=Integer.parseInt(args[i]);
}
for (int i=0; i<toConsolidate.length; i++) {
for (int k=0; k<toConsolidate.length-1; k++) {
if (toConsolidate[k] == 0){
toConsolidate[k] = toConsolidate[k+1];
toConsolidate[k+1] = 0;
}
}
}
for (int i:toConsolidate)
System.out.print(i+" ");
}
}
• Any integer or single digits like the examples? – edc65 Feb 2 '16 at 21:30
• @edc65 Any integer that your language supports. – Addison Crump Feb 2 '16 at 21:35
• How can the example be so complex when the shortest answers are 3 characters long? Is Java that much verbose? – A.L Feb 3 '16 at 0:48
• Isn't "There should be no 0s between the non-zero integers." redundant? – Martin Ender Feb 3 '16 at 9:48
• @immibis Might not be the right language for this challenge. :P – Addison Crump Feb 4 '16 at 7:48

## Pyth, 3 bytes

!DQ


Explanation:

  Q    Input
!D     Sort by logical NOT


Try it here.

• Q can be implicit at the end of any Pyth script (assuming it's outside of a lambda, which this is), making this 2 bytes. – hakr14 Mar 28 '18 at 2:33
• @hakr14 That feature didn't exist 2 years ago. – Dennis Mar 28 '18 at 16:26

# Jelly, 3 bytes

¬Ụị


Sorts the list by the logical NOT of its values. Try it online!

### How it works

¬Ụị    Main link. Input: A (list)

¬      Compute the logical NOT of each element of A.
Ụ     Grade up; sort the resulting list's indices by their corresponding values.
ị    Retrieve the elements of A at that indices.

• Oh hey, modern Jelly can do a 2-byte ¬Þ, even! – Lynn Sep 13 '16 at 0:42

# Octave, 18 bytes

@(A)[A(~~A) A(~A)]


sort() takes too many bytes. I'll just use logical indexing.

Examples on ideone.

• Nicely done. +1. – rayryeng - Reinstate Monica Mar 7 '16 at 17:36
• Wow, works in Matlab too! I didn't know that such indexing is possible – brainkz Sep 13 '16 at 9:13

# R, 2923 21 bytes

As noted by MarcoBreitig, we can shorten it to 21 bytes if we don't need to provide it as a function:

x=scan();x[order(!x)]


Previous versions:

function(x)x[order(!x)]


The function takes a vector as input and orders by the logical vector that results from negating the input.

function(x)c(x[x!=0],x[x==0])


The function takes a vector as input and the concatenates (c()) the non-zero values and then the zero-values.

• x=scan();x[order(!x)] is only 21 bytes long. – Marco Breitig Feb 3 '16 at 14:44
• @MarcoBreitig, that's right. I thought it should be a function (and initially, the requirement was a "full-fledged program"). Will update my answer – docendo discimus Feb 3 '16 at 15:13

# Retina, 15

Simple repeated regex substitution:

+\b0 (.*)
$1 0  Try it online. • I used the same method in my Java answer (if I'm reading this right). :D – Addison Crump Feb 2 '16 at 20:14 ## ES6, 23 bytes a=>a.sort((x,y)=>!x-!y)  It used to be the case that sort wasn't stable, in which case you needed 41 bytes: a=>a.filter(x=>x).concat(a.filter(x=>!x))  # Python byte code (2.7.9), 252 bytes, 33 opcodes, 0.0228 seconds This was build when the contest was still a contest Opens a file in the current directory called 'SourceArray' for use LOAD_CONST '' STORE_FAST no_zeroes# no_zeroes = '' LOAD_NAME open LOAD_CONST 'SourceArray' CALL_FUNCTION 0,1# open('SourceArray') LOAD_ATTR read CALL_FUNCTION 0,0# .read() LOAD_ATTR split CALL_FUNCTION 0,0# .split() DUP_TOP DUP_TOP #Start if BUILD_LIST 0 COMPARE_OP == POP_JUMP_IF_TRUE 35# if list == [], GOTO 35 LOAD_ATTR pop LOAD_CONST 0 CALL_FUNCTION 0,1# list.pop(0) DUP_TOP LOAD_CONST '0' COMPARE_OP == POP_JUMP_IF_TRUE 28# if list.pop(0) == '0', GOTO 28 PRINT_ITEM # print list.pop(0) JUMP_ABSOLUTE 13 POP_TOP LOAD_CONST '0%_'# '0 ' LOAD_FAST no_zeroes INPLACE_ADD STORE_FAST no_zeroes# no_zeroes = no_zeroes + '0 ' JUMP_ABSOLUTE 13 LOAD_FAST no_zeroes PRINT_ITEM # print no_zeroes LOAD_CONST None RETURN_VALUE  The co_code (The actual codey bit) 'd\x01\x00}\x00\x00\te\x00\x00\x83\x00\x00\tj\x01\x00\x83\x00\x00\t\x04\x04g\x00\x00k\x02\x00sG\x00j\x02\x00d\x02\x00\x83\x01\x00\x04d\x03\x00k\x02\x00s8\x00Gq\x15\x00\t\x01d\x04\x00|\x00\x007}\x00\x00q\x15\x00\t|\x00\x00G\td\x00\x00S'  Or a .pyc file version 03F3 03 F3 0D 0A 40 FD B0 56 63 00 00 00 00 01 00 00 00 03 00 00 00 00 00 00 00 73 59 00 00 00 64 01 00 7D 00 00 09 65 00 00 64 02 00 83 01 00 6A 01 00 83 00 00 09 6A 02 00 83 00 00 09 04 04 67 00 00 6B 02 00 73 50 00 6A 03 00 64 03 00 83 01 00 04 64 04 00 6B 02 00 73 41 00 47 71 1E 00 09 01 64 05 00 7C 00 00 37 7D 00 00 71 1E 00 09 7C 00 00 47 09 64 00 00 53 28 06 00 00 00 4E 74 00 00 00 00 74 0B 00 00 00 53 6F 75 72 63 65 41 72 72 61 79 69 00 00 00 00 74 01 00 00 00 30 73 02 00 00 00 30 20 28 04 00 00 00 74 04 00 00 00 6F 70 65 6E 74 04 00 00 00 72 65 61 64 74 05 00 00 00 73 70 6C 69 74 74 03 00 00 00 70 6F 70 28 01 00 00 00 74 09 00 00 00 6E 6F 5F 7A 65 72 6F 65 73 28 00 00 00 00 28 00 00 00 00 74 09 00 00 00 70 79 6B 65 5F 63 6F 64 65 52 08 00 00 00 01 00 00 00 52 00 00 00 00  You can try to compile my source code yourself using my library on github. I just posted a commit to it that allowed comments so I hope this is still competing as far as goes ;) Roughly equivalent to no_zeroes = '' unamed_variable = open('SourceArray').read().split() while unamed_variable != []: unamed_variable_2 = unamed_variable.pop() if unamed_variable_2 == '0': no_zeroes += '0 ' else: print unamed_variable_2, print no_zeroes,  • Wooow. You cut a ton of time off of that. – Addison Crump Feb 2 '16 at 19:42 • @VoteToClose that's about 1.5x the speed it ran on my laptop :O Who said Python was that slow? – Blue Feb 2 '16 at 19:46 ## Python, 32 bytes lambda x:sorted(x,key=0..__eq__)  Takes argument as any iterable (list, tuple, etc.). Thanks to @xnor for teaching me a new trick! • It's a bit shorter to use key=0..__eq__ (yes, two dots). – xnor Feb 3 '16 at 8:28 • @xnor That's neat... How does it work? – Mego Feb 3 '16 at 8:36 • Most Python objects have an equality method, so for example "abc".__eq__("abc")==True. It's what's called when you do "abc"==. For reasons, Python integers don't have it but floats do, and since 0. == 0, we can substitute its equality operator., which is 0..__eq__. – xnor Feb 3 '16 at 8:52 • @xnor ahh, I knew about the .__eq__ method, but the double dots were confusing me. I didn't catch that the first one was the decimal point in a float literal. – Mego Feb 3 '16 at 9:14 # Matlab: 21 bytes @(a)[a(a~=0),a(a==0)]  Prints nonzero elements first, then concatenates with zero elements @(a)____ create an anonymous function with one input argument a [___,___] concatenates horizontally vectors inside brackets, separated by commas a(a~=0) returns vector with all nonzero elements of vector a a(a==0) returns vector with all zero elements of vector a ## Haskell, 26 bytes f x=filter(/=0)x++[0|0<-x]  Take all non-zero numbers followed by all zeros. Filtering constants (here: 0) is quite short when using a list comprehension: [0|0<-x]. # Zsh, 22 bytes (input passed as arguments to the script/function ($@ aka $argv array), output on stdout as space separated list, newline terminated) <<<${@:#0}\ ${(M)@:#0}  • <<< string: here-string here passed as stdin to the $NULLCMD command (cat by default).
• ${@:#0} $@ except elements being 0.
• ${(M)@:#0} reverse of the above That assumes (like several other answers here) that zeroes in the input are all expressed as 0 (no 00 nor 0x0 nor 36#0). # Javascript, 5254 51 bytes s=>s.replace(/\b0 /g,x=>++i&&'',i=0)+' 0'.repeat(i)  • This doesn't work when the input does not contain any zeroes – rink.attendant.6 Feb 2 '16 at 22:07 • @rink.attendant.6. Thanks, I've updated and still looking for some bytes off :) – removed Feb 2 '16 at 22:46 # Mathematica, 14 bytes Sort[#,#!=0&]&  • Sort[#!=0&] should be enough. – Martin Ender Feb 3 '16 at 8:21 # APL: 8 bytes (⍴a)↑a~0  a~0 remove zeros from a (read "a without 0") (⍴a) original length of a (read "shape of a") ↑ pad a without zeros to a's original length Test data: a←1 0 1 2 3 4 0 1 0 0 0 0 1 2 3 4 5 • You should either write a full program and read the input from stdin, or write a function and read the input from its parameters. But you can use ⍴↑{⍵~0} and that's even shorter. – jimmy23013 Feb 3 '16 at 13:26 • Not so fast. ⍴↑{⍵~0} won't work everywhere, not in APL2000, nor in IBM APL2. – Lobachevsky Feb 7 '16 at 16:17 • ⍴↑{⍵~0} will return an empty vector. ⍴⍴↑{⍵~0} is a (one element vector) zero. – Lobachevsky Feb 8 '16 at 8:56 ## Java 7, 78 bytes void g(int[]a){int c=0;for(int o:a)a[o==0?c:c++]=o;for(;c<a.length;a[c++]=0);}  I'm not sure why the other Java entries are using strings. If you want to filter an integer array, it seems best to use an integer array. This modifies the input in place by keeping two indices, then just filling the remaining slots with zeros. • Heh, I used it because I felt like it. I think you should be able to declare o with int c=0,o;for(o:a).... You can also convert to Java 8 lambda syntax: a->{int c=0;for(int o:a)a[o==0?c:c++]=o;for(;c<a.length;a[c++]=0);} and state that it expects input as an int array. – Addison Crump Feb 4 '16 at 20:35 • Wait, scratch the o declaration thingy. But still, java 8 lambda. :D – Addison Crump Feb 4 '16 at 20:44 • @VoteToClose I thought it had to be self-contained. If I can declare types and stuff elsewhere without counting it, that doesn't seem right. – Marky Markov Feb 4 '16 at 22:28 • Since this is a function, input is passed to it by a previously executed statement anyways. The lambda can assume an input type, so its essentially the same. – Addison Crump Feb 4 '16 at 23:00 # Common Lisp, 46 bytes (lambda(a)(stable-sort a(lambda(_ b)(= 0 b))))  Sort the array so that for each couple (a,b), we have a < b if b is zero. When neither a < b or b < a, the sort is stable: the original order between elements is retained. I also tried with adjust-array and remove, but this was too long: (lambda(a)(adjust-array(remove 0 a)(length a):initial-element 0))  # PHP, 737170524948 46 bytes - BIG thanks to Ismael Miguel // Assuming$a = array(4,8,6,1,0,8,0,0,0,0,0,-4,'-5',-1,564,0);

// Produces a notice-level error
foreach($a as$v)$v?print"$v ":$b.="0 ";echo$b;

• $v==0 can be replaced with !$v, saving you 2 bytes. – Ismael Miguel Feb 2 '16 at 21:17
• @IsmaelMiguel thanks! – MonkeyZeus Feb 2 '16 at 21:25
• You're welcome. I see you did manage to cut a byte. Try this: foreach($argv as$v)$v?$f.=" $v":$b.=" $v";echo$f.$b;. It is.... some bytes, I don't know... – Ismael Miguel Feb 2 '16 at 21:41 • Or foreach($a as$v)$v?print("$v "):$b.="$v ";echo$b; for a more neat way, that looks exactly the same – Ismael Miguel Feb 2 '16 at 21:49
• @IsmaelMiguel Nice! I would cry if I ever had to pick up someone else's project and found this level of code-golfing lol – MonkeyZeus Feb 2 '16 at 21:54

# Bash + GNU utilities, 23

grep -v ^0 a
grep ^0 a


Assumes input is newline-separated entries in a file called a. Score includes +1 for this filename.

• @sch Yes, it should be bash - fixed. – Digital Trauma Feb 3 '16 at 16:24
• @TimmyD yes - thanks for the reminder. – Digital Trauma Feb 3 '16 at 16:24

# Perl 5, 26 bytes

23 plus three for -an (-E is free)

say for sort{!$a-!$b}@F


Thanks to Dennis for reminding me of -a, saving two bytes.

# CJam, 6 bytes

{{!}$}  An anonymous function. Sort using “whether or not an element is zero” as a key. # MATL, 7 bytes t~FT#S)  Try it online! t % input array. Duplicate ~ % logical negate: nonzero values become false, zeros become true FT#S % sort (false, then true) and output a vector with the indices of the sorting ) % apply that vector of indices to original array  ## Seriously, 12 bytes 4,nYM@░)░+  Try it online! Explanation: 4,nYM@░)░+ 4,n push 4 copies of input YM map logical negate @░) filter (take zeroes) and push to bottom of stack ░ filter (take non-zeroes) + append zeroes  # Swift, 13 bytes a.sort{$1==0}


# Perl6, 11 bytes

{.sort(!*)}


Produces a Block - which can be called on an array:

{.sort(!*)}.([1,2,0,3]).say


Although it would be more natural (and shorter) to write:

[1,2,0,3].sort(!*).say


How it works: if the perl6 sort routine is called with a block which accepts only one argument, the list elements are sorted according to by($a) cmp by($b). In this case, the block is !*, i.e. a negation of the whatever operator.

I notice that:

• The example in the question is a class which provides a method, not including boilerplate required to read in
• The description of the task does not require printing, and, except for the fact that the example prints, implies that an array might be returned

## TeX (Plain format), 160 bytes

Make the 0 character active (that is, make the interpreter process it as a command), then define that command to skip the character and increment a counter. At the end of the string, print as many zeros as were counted.

Save this as zero.tex and give the input through the command line with this command:

pdftex "\def\I{0 1 0 3 2 0 0 8 0 5 0 1 9 4}\input zero"

\def\I{}\newcount\Z\def\L{\loop\advance\Z by-1\ifnum\Z>00 \repeat}
\scantokens\expandafter{\I\empty}\endgroup\L\bye


# J, 4 bytes

/:0=


Explanation:

/:      NB. upward sort on
0=    NB. equality to zero


The sort function in J is guaranteed to be stable by the specification.

Alternative solution, 6 bytes:

#{.*#+


   *#+  NB. replicate each item by its sign (removing zeroes)
#{.     NB. take as many items from this as the original list had
NB.  (taking more items than there are in a list results in extra zeroes)


## Straw, 30 29 bytes

<:([^0 ])()/,0()/ +,+( +) /}>


Use the CP437 encoding

### Explanation

<:([^0 ])()/,0()/ +,+( +) /}>
<                             Take input
:                            Duplicate
([^0 ])()/                  Remove every character that is not a 0 or a space
,                 Swap the two items on the top of the stack
0()/             Remove every 0 on the top of the stack
+           Push a space and concatenate
,          Swap
+         Concatenate
( +) /   Remove duplicate spaces
}  Get the 'tail' of the string
> Output


Try it online! (The added code is to test all test cases)

# JavaScript ES6, 16 bytes

x=>x.sort(t=>!t)


Works on firefox

# Ruby, 25 bytes

->a{a-[0]+[0]*a.count(0)}


Try it online!

# 05AB1E, 15 14 bytes

Code:

ED0¢r0KRrFZ}|


Explanation:

E               # Evaluate input
D              # Duplicate top of the stack
0¢            # Count zeroes
r           # Reverse stack
0K         # Delete all zeroes
R        # Reverse top of the stack
# Flatten
r      # Reverse stack
FZ}   # For N in range(amount zeroes): push zero
|  # Print full stack


Uses CP-1252 encoding. Takes an array like this:

[0, 5, 8, 8, 3, 5, 1, 6, 8, 4, 0, 3, 7, 5, 6, 4, 4, 7, 5, 6, 7, 4, 4, 9, 1, 0, 5, 7, 9, 3, 0, 2, 2, 4, 3, 0, 4, 8, 7, 3, 1, 4, 7, 5, 1, 2, 1, 8, 7, 8, 7, 7, 2, 6, 3, 1, 2, 8, 5, 1, 4, 2, 0, 5, 0, 6, 0, 3]