Duplicate Zeros LeetCode Solution

Problem – Duplicate Zeros LeetCode Solution

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 <= 104
• 0 <= arr[i] <= 9

Duplicate Zeros LeetCode Solution in Python

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

Duplicate Zeros LeetCode Solution in C++

    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;
        }
    }

Duplicate Zeros LeetCode Solution in Java

    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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