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Challenges

Backspace an array

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Challenge

Given an array consisting of positive integers and 0s, return it with 0s acting like backspaces.

Leading backspaces do nothing, and more backspaces than elements also does nothing.

Credit: A comment on this video

Test Cases

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

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Japt, 9 bytes

ô rÈÔÅÔcY

Try it or run all test cases

ô rÈÔÅÔcY     :Implicit input of array
ô             :Split on falsey elements (i.e., 0)
  r           :Reduce by
   È          :Passing each Y through the following function, with the first element serving as the start value, X
    Ô         :  Reverse X
     Å        :  Slice off the first element
      Ô       :  Reverse
       cY     :  Concat Y
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+3
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Jelly, 6 bytes

ṣ0Ṗ;¥/

Try it online!

How it works

ṣ0Ṗ;¥/ - Main link. Takes a list on the left
ṣ0     - Split at zeroes
    ¥/ - Reduce by:
  Ṗ    -   Remove the last element
   ;   -   Concatenate
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Scala, 45 bytes

_./:(Seq[Int]())((a,x)=>a:+x dropRight-x*2+2)

Try it in Scastie!

I couldn't find a way to use underscores in the inner function :(. Explanation on its way.

_      //The input
./:(   //Fold over it,
 Seq[Int]() //starting with an empty list,
)(         //using this function
 (a,x) =>  //a is the accumulator, x is the current element
   a:+x    //Append x to a
    dropRight //then drop an element from the right if x is 0
    -x*2+2    //For 0, this is 2
              //(drop the 0 that we just appended and the previous element)
              //and for other numbers, it's negative, so it doesn't do anything
)
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+1
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Python 3.8 (pre-release), 62 bytes

f=lambda x,a=[]:f(x[1:],a+[x[0]]if x[0]else a[:-1])if x else a

Try it online!

-11 bytes thanks to user

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JavaScript (Node.js), 41 40 bytes

Saved 1 byte thanks to Shaggy

x=>x.map(e=>e?a.push(e):a.pop(),a=[])&&a

Try it online!

x =>      //x is the input
x.map(e=> //For every element e in x
 e?        //If e is not 0
 a.push(e) //Add it to the accumulator
 :a.pop(), //Otherwise, pop the last element
 a=[])     //Initial value of the accumulator
&&a       //Return the accumulator
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Husk, 6 bytes

Fo+hx0

Try it online!

there's got to be a smarter way to do this with grouping.

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