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 {
public:
    void duplicateZeros(vector<int>& arr) {
        int n = arr.size();
        int i = -1, k = 0;
        while (k < n) {
            ++i;
            k += arr[i] ? 1 : 2;
        }
        int j = n - 1;
        if (k == n + 1) {
            arr[j--] = 0;
            --i;
        }
        while (~j) {
            arr[j] = arr[i];
            if (arr[i] == 0) arr[--j] = arr[i];
            --i;
            --j;
        }
    }
};