Move zeros to end of array in place

MoveZerosInPlace.swift

/*
Move zeros to end of array in place.

Brute force
Loop through elements n times
  When 0 found
  Bubble up 0 to end, perform swaps until swap performed with last element
O(n^2) time | O(1) space

Loop through elements n times
  When 0 found
  Shift all alements to right by index -= 1
  Place zero at last index
O(n^2) time | O(1) space

Can optimize this by having a pointer segmenting the 0's from the other numbers
Would only need to shift, or bubble up to the index at the pointer

Optimized approach
Loop through elements, while indexOfFirstZero or indexOfNextNonZero < arr.count
  indexOfFirstZero, pretty self explanitory
  indexOfNextNonZero, pretty self explanitory
  Increment indexOfFirstZero until zero found
    Increment indexOfNextNonZero until non zero found
      Perform a swap
O(n) time | O(1) space

 n 
 z
[1, 10, 2, 8, 3, 6, 4, 5, 7, 0, 0, 0, 0, 0]
*/

func moveZerosToEnd(arr: [Int]) -> [Int] {
  var arr = arr
  var indexOfFirstZero = 0
  var indexOfNextNonZero = 0
  
  while indexOfFirstZero < arr.count && indexOfNextNonZero < arr.count {
      
    (arr[indexOfFirstZero], arr[indexOfNextNonZero]) = (arr[indexOfNextNonZero], arr[indexOfFirstZero])

    while indexOfFirstZero < arr.count, arr[indexOfFirstZero] != 0 {
      indexOfFirstZero += 1
    }
    
    indexOfNextNonZero = indexOfFirstZero
    
    while indexOfNextNonZero < arr.count, arr[indexOfNextNonZero] == 0 {
      indexOfNextNonZero += 1
    }
  }
  
  return arr
}