Duplicate Zeros
<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>
Array; Two Pointers
class Solution:
    def duplicateZeros(self, arr: List[int]) -> None:
        """
        Do not return anything, modify arr in-place instead.
        """
        n = len(arr)
        i, k = -1, 0
        while k < n:
            i += 1
            k += 1 if arr[i] else 2
        j = n - 1
        if k == n + 1:
            arr[j] = 0
            i, j = i - 1, j - 1
        while ~j:
            if arr[i] == 0:
                arr[j] = arr[j - 1] = arr[i]
                j -= 1
            else:
                arr[j] = arr[i]
            i, j = i - 1, j - 1
