LeetCode //C - 1089. Duplicate Zeros

1089. Duplicate Zeros

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 < = a r r . l e n g t h < = 10 4 1 <= arr.length <= 10^4 1<=arr.length<=104
• 0 <= arri <= 9

From: LeetCode

Link: 1089. Duplicate Zeros

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Solution:

Ideas:

• Treat the array as if it had extra space (arrSize + zeros)
• Only write values when index < arrSize
• Backward processing avoids overwriting unprocessed elements

Code:

void duplicateZeros(int* arr, int arrSize) {
    int zeros = 0;

    for (int i = 0; i < arrSize; i++) {
        if (arr[i] == 0) {
            zeros++;
        }
    }

    int i = arrSize - 1;
    int j = arrSize + zeros - 1;

    while (i >= 0) {
        if (j < arrSize) {
            arr[j] = arr[i];
        }

        if (arr[i] == 0) {
            j--;
            if (j >= 0 && j < arrSize) {
                arr[j] = 0;
            }
        }

        i--;
        j--;
    }
}