HOw to make a loop in all the array untill finish using js [duplicate]

Question

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

That is not a classic permutation algorithm, btw. It's not permutation at all. — Sergio Tulentsev
– Sergio Tulentsev, Commented Jan 5, 2023 at 14:34
You can just sort [...arr].sort((a, b) => (a===0) - (b===0)) — pilchard
– pilchard, Commented Jan 5, 2023 at 14:36
@pilchard: ah, but that requires the sort algorithm to be stable. Is it guaranteed to be stable in javascript? — Sergio Tulentsev
– Sergio Tulentsev, Commented Jan 5, 2023 at 14:39
@pilchard: only in es2019+, according to stackoverflow.com/questions/3026281/… — Sergio Tulentsev
– Sergio Tulentsev, Commented Jan 5, 2023 at 14:40
also 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 — pilchard
– pilchard, Commented Jan 5, 2023 at 14:44

Ahmad ghoneim · Accepted Answer · 2023-01-05 14:42:15Z

3

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

filter the zeros
know how many zeros were there

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]

answered Jan 5, 2023 at 14:42

Ahmad ghoneim

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Comments

zer00ne · Accepted Answer · 2023-01-05 14:45:53Z

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])));

edited Jan 5, 2023 at 14:45

answered Jan 5, 2023 at 14:41

zer00ne

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4 Comments

Sergio Tulentsev Over a year ago

Not sure if this matters, but this approach looks like O(N^2).

pilchard Over a year ago

it's also mutating the array being iterated, and while forEach closes over the initial array it's still awkward at best

EugenSunic Over a year ago

splice is linear time thus making this n2

zer00ne Over a year ago

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.

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