-1

i need to put all items that equals zero att the end of the array, i used a classic permutation code to do that, it works but it does not continu the comparison untill the end.

function moveZeros(arr) {
  var permut = 0;
  var i=0;
 
    while( i <= arr.length) {
      if(arr[i] === 0) {
      permut = arr[i];
      arr[i] = arr[i+1]
       arr[i+1] = "0";
    }
      i++
  }
  return arr.join()
}
console.log(moveZeros([1,2,0,1,0,1,0,3,0,1]))
// i have this : 1,2,1,0,1,0,3,0,1,0
// But Need to have this result : 1, 2, 1, 1, 3, 1, 0, 0, 0, 0
Meryem ACHEMLAL
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  • That is not a classic permutation algorithm, btw. It's not permutation at all. – Sergio Tulentsev Jan 05 '23 at 14:34
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    You can just sort `[...arr].sort((a, b) => (a===0) - (b===0))` – pilchard Jan 05 '23 at 14:36
  • @pilchard: ah, but that requires the sort algorithm to be stable. Is it guaranteed to be stable in javascript? – Sergio Tulentsev Jan 05 '23 at 14:39
  • @pilchard: only in es2019+, according to https://stackoverflow.com/questions/3026281/what-is-the-stability-of-the-array-sort-method-in-different-browsers – Sergio Tulentsev Jan 05 '23 at 14:40
  • also [How to move all elements that are zero to the end of an array?](https://stackoverflow.com/questions/56330270/how-to-move-all-elements-that-are-zero-to-the-end-of-an-array) and [shifting zeros to end of the array and keeping the non-zero elements without changing order](https://stackoverflow.com/questions/58067831/shifting-zeros-to-end-of-the-array-and-keeping-the-non-zero-elements-without-cha) – pilchard Jan 05 '23 at 14:44

2 Answers2

3

maybe there's another neat way to do it but that's what came into my mind

  1. filter the zeros

  2. know how many zeros were there

  3. add the filtered array to an array with the number of zeros that were there

     let myarr = [1, 2, 0, 1, 0, 1, 0, 3, 0, 1];
 const filtered = myarr.filter(function (ele) {
   return ele !== 0;
 });
 let diff = myarr.length - filtered.length;
 let arrOfZeros = Array(diff).fill(0);
 let reqArr = [...filtered, ...arrOfZeros];
 console.log(reqArr); // (10) [1, 2, 1, 1, 3, 1, 0, 0, 0, 0]
Ahmad ghoneim
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0

Simply remove each with .splice() 0 and push() a 0 to the end of the array. Note, .forEach(), .splice(), and .push() are array methods that mutate the array original array.

function moveZeros(array) {
  array.forEach((number, index) => {
    if (number === 0) {
      array.splice(index, 1);
      array.push(0);
    }
  });
  return array;
}
console.log(JSON.stringify(moveZeros([1,2,0,1,0,1,0,3,0,1])));
zer00ne
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    Not sure if this matters, but this approach looks like O(N^2). – Sergio Tulentsev Jan 05 '23 at 14:44
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    it's also mutating the array being iterated, and while forEach closes over the initial array it's still awkward at best – pilchard Jan 05 '23 at 14:48
  • splice is linear time thus making this n2 – EugenSunic Jan 05 '23 at 14:53
  • I answered for the end result, OP didn't mention that that the array would be huge. `splice()` is probably O(n) since it is linearly deleting at each position. If the input array was like 1000+ degrading performance should be noticeable, but not at a length of 10. – zer00ne Jan 05 '23 at 15:21