[codewars/JS] Sort the odd

Task

You will be given an array of numbers. You have to sort the odd numbers in ascending order while leaving the even numbers at their original positions.

Examples

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

<code>

function sortArray(array) {
  const result = array.filter((a)=>
     a%2 !== 0)
  result.sort(function(a,b){
    return a-b
  });
  console.log(result);
  let count = 0;
  const newarray = [];
  for(let i=0;i<array.length;i++){
    if(array[i]%2==0){
      newarray[i]=array[i];
    }
    if(array[i]%2!==0){
      newarray[i]=result[count];
      count++;
    }
    }console.log(newarray)
  return newarray
    
}