Remove leading and trailing zeroes

Question

Given a non-empty list/array containing only non-negative integers like this:

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

Output the list with trailing and leading zeroes removed.

The output for this would be:

[8, 1, 4, 3, 5, 6, 4, 1, 2]

Some other test cases:

[0, 4, 1, 2, 0, 1, 2, 4, 0] > [4, 1, 2, 0, 1, 2, 4]
[0, 0, 0, 0, 0, 0] > nothing
[3, 4, 5, 0, 0] > [3, 4, 5]
[6] > [6]

Shortest code wins

Answer 1

Jelly, 2 bytes

Code:

t0

Explanation:

t   # Trim off...
 0  #  zero at both sides

Try it online!

Answer 2

JavaScript (ES6) 43

a=>(f=a=>a.reverse().filter(x=>a|=x))(f(a))

Less golfed

a=>{
  f=a=>a.reverse().filter(x=>a|=x) // reverse and remove leading 0
  // leverage js cast rules: operator | cast operands to integer
  // an array casted to integer is 0 unless the array is made of
  // a single integer value (that is ok for me in this case)
  return f(f(a)) // apply 2 times
}

Test

F=a=>(f=a=>a.reverse().filter(x=>a|=x))(f(a))

function test(){
  var l=(I.value.match(/\d+/g)||[]).map(x=>+x)
  O.textContent=F(l)
}

test()

#I { width:90%}

<input id=I oninput='test()' value='0 0 1 3 7 11 0 8 23 0 0 0'>
<pre id=O></pre>

Answer 3

Retina, 12 bytes

^0 ?

)` 0$

The trailing linefeed is significant.

Thanks to @Martin Büttner and @FryAmTheEggman for saving a few bytes.

Try it online

Answer 4

CJam, 13 bytes

l~{_{}#>W%}2*

With the array inputted.

Longer version:

l~             Puts input on the stack and parse as array
  {       }    Code block
   _           Duplicate the first thing on the stack
    {}#        Finds the index of the first non-0 value in the array, puts it on the stack
       >       Slices the array from that index
        W%     Reverses the array
           2*  Does the code block twice in total

Answer 5

Haskell, 29 bytes

t=f.f;f=reverse.dropWhile(<1)

Answer 6

Pyth, 4 bytes

.sQ0

Demo:

llama@llama:~$ pyth -c .sQ0
[0, 0, 0, 1, 2, 0, 3, 4, 0, 0, 5, 0, 0, 0, 0]
[1, 2, 0, 3, 4, 0, 0, 5]

From Pyth's rev-doc.txt:

.s <seq> <any>
    Strip from A maximal prefix and suffix of A consisting of copies of B.

Answer 7

05AB1E, 4 bytes

Code:

0Û0Ü

Try it online!

Explanation:

0Û    # Trim off leading zeroes
  0Ü  # Trim off trailing zeroes

Uses CP-1252 encoding.

Answer 8

Jelly, 10 bytes

Uo\U,o\PTị

This doesn't use the builtin.

Uo\U            Backward running logical OR
    ,           paired with
     o\         Forward running logical OR
       P        Product
        T       All indices of truthy elements
         ị      Index the input at those values.

Try it here.

Answer 9

R, 43 bytes

function(x)x[cummax(x)&rev(cummax(rev(x)))]

or as read/write STDIN/STDOUT

x=scan();cat(x[cummax(x)&rev(cummax(rev(x)))])

---

This finds the cumulative maximum from the beginning and the end (reversed) string. The & operator converts these two vectors to logical one of the same size as x, (zeroes will always converted to FALSE and everything else to TRUE), this way it makes it possible to subset from x according to the met conditions.

Answer 10

Mathematica 34 27 bytes

#//.{0,a___}|{a___,0}:>{a}&

This repeatedly applies replacement rules until such action fails to provide a new output. 7 bytes saved thanks to Alephalpha.

The first rule deletes a zero at the beginning; the second rule deletes a zero at the end of the array.

Answer 11

05AB1E, 4 bytes

0Û0Ü

Basically trimming leading then trailing zeroes of the input, given as an array.

Try it online !

Answer 12

Perl, 19 + 1 = 20 bytes

s/^(0 ?)+|( 0)+$//g

Requires -p flag:

$ perl -pE's/^(0 )+|( 0)+$//g' <<< '0 0 0 1 2 3 4 5 6 0 0 0'
1 2 3 4 5 6

Answer 13

Ruby, 49 44 bytes

->a{eval ?a+'.drop_while{|i|i<1}.reverse'*2}

Thanks to manatwork for chopping off 5 bytes with a completely different method!

This just drops the first element of the array while it's 0, reverses the array, repeats, and finally reverses the array to return it to the proper order.

Answer 14

Perl, 38 bytes

$\.=$_}{$\=~s/^(0\n)*|(0\n)*\n$//gs

Run with perl -p, (3 bytes added for -p).

Accepts numbers on STDIN, one per line; emits numbers on STDOUT, one per line, as a well-behaved unix utility should.

Only treats numbers represented exactly by '0' as zeroes; it would be possible to support other representations with a few more bytes in the regex.

Longer version, still to be run with -p:

    # append entire line to output record separator
    $\.=$_
}{
    # replace leading and trailng zeroes in output record separator
    $\ =~ s/^(0\n)*|(0\n)*\n$//gs
    # output record separator will be implicitly printed

Expanded version, showing interactions with -p flag:

# implicit while loop added by -p
while (<>) {
    # append line to output record separator
    $\.=$_
}{ # escape the implicit while loop
    # replace leading and traling 
    $\=~s/^(0\n)*|(0\n)*\n$//gs
    # print by default prints $_ followed by
    # the output record separator $\ which contains our answer
    ;print # implicit print added by -p
} # implicit closing brace added by -p

Answer 15

Elixir, 77 bytes

import Enum
z=fn x->x==0 end
reverse(drop_while(reverse(drop_while(l,z)),z))

l is the array.

Edit:wah! copy/pasta fail. of course one has to import Enum, which raises the byte count by 12 (or use Enum.function_name, which will make it even longer).

Answer 16

JavaScript (ES6), 47 bytes

a=>a.join(a="").replace(/(^0+|0+$)/g,a).split(a)

Where a is the array.

Answer 17

Vitsy, 13 bytes

Vitsy is slowly getting better... (I'm coming for you Jelly. ಠ_ಠ)

1mr1m
D)[X1m]

This exits with the array on the stack. For readability, the TryItOnline! link that I have provided below the explanation will output a formatted list.

Explanation:

1mr1m
1m      Do the second line of code.
  r     Reverse the stack.
   1m   I'ma let you figure this one out. ;)

D)[X1m]
D       Duplicate the top item of the stack.
 )[   ] If the top item of the stack is zero, do the stuff in brackets.
   X    Remove the top item of the stack.
    1m  Execute the second line of code.

Note that this will throw a StackOverflowException for unreasonably large inputs.

TryItOnline!

Answer 18

JavaScript (ES6), 34 bytes

a=>a.replace(/^(0 ?)*|( 0)*$/g,'')

Input and output are in the form of a space-delimited list, such as "0 4 1 2 0 1 2 4 0".

Answer 19

R, 39 bytes

function(x)x[min(i<-which(x>0)):max(i)]

Four bytes shorter than David Arenburg's R answer. This implementation finds the first and last index in the array which is greater than zero, and returns everything in the array between those two indices.

Answer 20

MATL, 9 bytes

2:"PtYsg)

Try it online!

Explanation

2:"     % For loop (do the following twice)
  P     %   Flip array. Implicitly asks for input the first time
  t     %   Duplicate
  Ys    %   Cumulative sum
  g     %   Convert to logical index
  )     %   Apply index
        % Implicitly end for
        % Implicitly display stack contents

Answer 21

Python 3, 48 bytes

lambda a:map(int,''.join(map(str,a)).strip('0'))

Answer 22

Brachylog, 12 bytes

{↔a₁.h>0∧}ⁱ²

Try it online!

The predicate fails in the case of "nothing".

  a₁.           Take the longest suffix of
 ↔              the input reversed
     h          the first element of which
      >0        is greater than 0
        ∧       (which is not the output).
{        }ⁱ²    Do it again.

If "nothing" has to be some actual value, like an empty list, it's only one more byte:

Brachylog, 13 bytes

{↔a₁.h>0∨Ė}ⁱ²

Try it online!

If the input can be assumed to contain only single digits, it's quite a bit shorter:

Brachylog, 6 bytes

c↔↔ℕ₁ẹ

Try it online!

Answer 23

Dyalog APL, 15 bytes

{⌽⍵↓⍨+/0=+\⍵}⍣2

               ⍣2     Apply this function twice:
{             }       Monadic function:
           +\⍵        Calculate the running sum.
       +/0=           Compare to zero and sum. Number of leading zeroes.
   ⍵↓⍨               Drop the first that many elements from the array.
 ⌽                   Reverse the result.

Try it here.

Answer 24

Vim 16 Keystrokes

i<input><esc>?[1-9]<enter>lD0d/<up><enter>

The input is to be typed by the user between i and esc, and does not count as a keystroke. This assumes that there will be at least one leading and one trailing zero. If that is not a valid assumption, we can use this slightly longer version: (18 Keystrokes)

i <input> <esc>?[1-9]<enter>lD0d/<up><enter>

Answer 25

ES6, 51 bytes

f=a=>a.map(x=>x?t=++i:f<i++||++f,f=i=0)&&a.slice(f,t)

t is set to the index after the last non-zero value, while f is incremented as long as only zeros have been seen so far.

Answer 26

Perl 6, 23 bytes

{.[.grep(?*):k.minmax]}
{.[minmax .grep(?*):k]}

Usage:

# replace the built-in trim subroutine
# with this one in the current lexical scope
my &trim = {.[.grep(?*):k.minmax]}

say trim [0, 0, 0, 8, 1, 4, 3, 5, 6, 4, 1, 2, 0, 0, 0, 0];
# (8 1 4 3 5 6 4 1 2)
say trim [0, 4, 1, 2, 0, 1, 2, 4, 0];
# (4 1 2 0 1 2 4)
say trim [0, 0, 0, 0, 0, 0];
# ()
say trim [3, 4, 5, 0, 0];
# (3 4 5)
say trim [6];
# (6)

Answer 27

Retina, 11 bytes

+`^0 ?| 0$
<empty>

Quite simple. Recursively replaces zeroes at beginning and end of line.

Try it online!

Answer 28

Python, 84 characters

def t(A):
 if set(A)<={0}:return[]
 for i in(0,-1):
  while A[i]==0:del A[i]
 return A

Answer 29

Javascript (ES6) 40 bytes

a=>/^(0,)*(.*?)(,0)*$/.exec(a.join())[2]

Answer 30

PHP, 56 54 52 bytes

Uses Windows-1252 encoding

String based solution

<?=preg_replace(~ÜÒ×ßÏÖÔƒ×ßÏÖÔÛÜ,"",join($argv,~ß));

Run like this:

echo '<?=preg_replace(~ÜÒ×ßÏÖÔƒ×ßÏÖÔÛÜ,"",join($argv,~ß));' | php -- 0 0 123 234 0 500 0 0 2>/dev/null;echo

If your terminal is set to UTF-8, this is the same:

echo '<?=preg_replace("#-( 0)+|( 0)+$#","",join($argv," "));' | php -- 0 0 123 234 0 500 0 0 2>/dev/null;echo

Tweaks

• Saved 2 bytes by negating strings and dropping string delimiters
• Saved 2 bytes by using short print tag