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Given a fixed-length integer array `arr`

, duplicate each occurrence of zero, shifting the remaining elements to the right.

**Note** 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.

**Example 1:**

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

**Example 2:**

```
Input: arr = [1,2,3]
Output: [1,2,3]
Explanation: After calling your function, the input array is modified to: [1,2,3]
```

**Constraints:**

`1 <= arr.length <= 10`

^{4}`0 <= arr[i] <= 9`

```
def duplicateZeros(self, A):
A[:] = [x for a in A for x in ([a] if a else [0, 0])][:len(A)]
```

```
void duplicateZeros(vector<int>& A) {
int n = A.size(), j = n + count(A.begin(), A.end(), 0);
for (int i = n - 1; i >= 0; --i) {
if (--j < n)
A[j] = A[i];
if (A[i] == 0 && --j < n)
A[j] = 0;
}
}
```

```
public void duplicateZeros(int[] arr) {
int countZero = 0;
for (int i = 0; i < arr.length; i++) {
if (arr[i] == 0) countZero++;
}
int len = arr.length + countZero;
//We just need O(1) space if we scan from back
//i point to the original array, j point to the new location
for (int i = arr.length - 1, j = len - 1; i < j; i--, j--) {
if (arr[i] != 0) {
if (j < arr.length) arr[j] = arr[i];
} else {
if (j < arr.length) arr[j] = arr[i];
j--;
if (j < arr.length) arr[j] = arr[i]; //copy twice when hit '0'
}
}
}
```

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