<p>Given a fixed-length integer array <code>arr</code>, duplicate each occurrence of zero, shifting the remaining elements to the right.</p>

<p><strong>Note</strong> that elements beyond the length of the original array are not written. Do the above modifications to the input array in place and do not return anything.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,0,2,3,0,4,5,0]
<strong>Output:</strong> [1,0,0,2,3,0,0,4]
<strong>Explanation:</strong> After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,2,3]
<strong>Output:</strong> [1,2,3]
<strong>Explanation:</strong> After calling your function, the input array is modified to: [1,2,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt;= 9</code></li>
</ul>

from typing import List

class Solution:
    def duplicateZeros(self, arr: List[int]) -> None:
    # First, count the number of zeros to be duplicated
        zeros = arr.count(0)
        n = len(arr)

    # We will iterate backwards, starting from the last element that will stay within bounds
        for i in range(n - 1, -1, -1):
            if i + zeros < n:
                arr[i + zeros] = arr[i]

            if arr[i] == 0:
                zeros -= 1
                if i + zeros < n:
                    arr[i + zeros] = 0
17300