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Formatted question description: https://leetcode.ca/all/1089.html

# 1089. Duplicate Zeros (Easy)

Given a fixed length array arr of integers, 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, do not return anything from your function.

Example 1:

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


Example 2:

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


Note:

1. 1 <= arr.length <= 10000
2. 0 <= arr[i] <= 9

Related Topics:
Array

## Solution 1.

// OJ: https://leetcode.com/problems/duplicate-zeros/
// Time: O(N)
// Space: O(1)
class Solution {
public:
void duplicateZeros(vector<int>& A) {
int cnt = 0;
for (int n : A) cnt += n == 0;
if (cnt == 0) return;
for (int i = A.size() - 1; i >= 0; --i) {
cnt -= A[i] == 0;
int j = i + cnt;
if (j < A.size()) A[j] = A[i];
if (j + 1 < A.size() && A[i] == 0) A[j + 1] = 0;
}
}
};

• class Solution {
public void duplicateZeros(int[] arr) {
int length = arr.length;
int slow = 0, fast = 0;
while (fast < length) {
int num = arr[slow];
slow++;
if (num == 0)
fast += 2;
else
fast++;
}
if (fast == slow)
return;
if (arr[slow] == 0)
fast++;
while (fast >= length) {
int num = arr[slow];
slow--;
if (num == 0) {
fast -= 2;
if (fast == length - 2)
arr[length - 1] = 0;
} else
fast--;
}
while (fast > slow) {
int num = arr[slow];
slow--;
arr[fast] = num;
fast--;
if (num == 0) {
arr[fast] = num;
fast--;
}
}
}
}

############

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

• // OJ: https://leetcode.com/problems/duplicate-zeros/
// Time: O(N)
// Space: O(1)
class Solution {
public:
void duplicateZeros(vector<int>& A) {
int cnt = 0;
for (int n : A) cnt += n == 0;
if (cnt == 0) return;
for (int i = A.size() - 1; i >= 0; --i) {
cnt -= A[i] == 0;
int j = i + cnt;
if (j < A.size()) A[j] = A[i];
if (j + 1 < A.size() && A[i] == 0) A[j + 1] = 0;
}
}
};

• 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

############

# 1089. Duplicate Zeros
# https://leetcode.com/problems/duplicate-zeros/

class Solution:
def duplicateZeros(self, arr: List[int]) -> None:
"""
Do not return anything, modify arr in-place instead.
"""

zeroes = arr.count(0)
n = len(arr)

for i in range(n - 1, -1, -1):

if i + zeroes < n:
arr[i + zeroes] = arr[i]

if arr[i] == 0:
zeroes -= 1

if i + zeroes < n:
arr[i + zeroes] = 0


• func duplicateZeros(arr []int) {
n := len(arr)
i, k := -1, 0
for k < n {
i, k = i+1, k+1
if arr[i] == 0 {
k++
}
}
j := n - 1
if k == n+1 {
arr[j] = 0
i, j = i-1, j-1
}
for j >= 0 {
arr[j] = arr[i]
if arr[i] == 0 {
j--
arr[j] = arr[i]
}
i, j = i-1, j-1
}
}

• impl Solution {
pub fn duplicate_zeros(arr: &mut Vec<i32>) {
let n = arr.len();
let mut i = 0;
let mut j = 0;
while j < n {
if arr[i] == 0 {
j += 1;
}
j += 1;
i += 1;
}
while i > 0 {
if arr[i - 1] == 0 {
if j <= n {
arr[j - 1] = arr[i - 1];
}
j -= 1;
}
arr[j - 1] = arr[i - 1];
i -= 1;
j -= 1;
}
}
}