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 <= 1040 <= arr[i] <= 9program main
implicit none
integer, parameter :: n = 8
integer :: arr(n)
arr = [1, 0, 2, 3, 0, 4, 5, 0]
call duplicate_zeros(arr)
write (*, '(*(I1, 1X))') arr
arr = [1, 2, 3]
call duplicate_zeros(arr)
write (*, '(*(I1, 1X))') arr
contains
subroutine duplicate_zeros(arr)
implicit none
integer, intent(inout) :: arr(:)
integer :: i, j
do i = 1, size(arr) - 1
if (arr(i) == 0) then
do j = size(arr), i + 1, -1
arr(j) = arr(j - 1)
end do
arr(i) = 0
end if
end do
end subroutine duplicate_zeros
end program maintemp.f95:11:4:
11 | arr = [1, 2, 3]
| 1
Error: Different shape for array assignment at (1) on dimension 1 (8 and 3)
! Declare the input array and its length
integer, intent(inout) :: arr(100000)
integer :: n
! Initialize the length of the input array
n = size(arr)
! Loop through the input array and duplicate each occurrence of zero
do i = 1, n
if (arr(i) == 0) then
! Shift the remaining elements to the right
do j = i, n - 1
arr(j) = arr(j + 1)
end do
! Increment the length of the input array
n = n + 1
end if
end do
! Print the modified input array to stdout
do i = 1, n
write (*,*) arr(i)
end do
end
temp.f95:2:37:
2 | integer, intent(inout) :: arr(100000)
| 1
Error: Symbol at (1) is not a DUMMY variable
def duplicateZeros(arr):
n = len(arr)
i = 0
while i < n - 1:
if arr[i] == 0:
for j in range(n - 1, i + 1, -1):
arr[j] = arr[j - 1]
arr[i + 1] = 0
i += 1
i += 1
The algorithm iterates through the input array arr. When a zero is encountered at position i, a second for loop or equivalent control structure (for, while, or range depending on the language) is used to shift all elements at positions greater than i to the right by one. The element at position i + 1 is then set to zero, and the iterator i is incremented to skip the newly added zero. This process is repeated until the end of the array is reached, ensuring that consecutive zeros are processed correctly.
void duplicateZeros(vector<int>& arr) {
int n = arr.size();
for (int i = 0; i < n - 1; ++i) {
if (arr[i] == 0) {
for (int j = n - 1; j > i + 1; --j) {
arr[j] = arr[j - 1];
}
arr[++i] = 0;
}
}
}
The algorithm iterates through the input array arr. When a zero is encountered at position i, a second for loop or equivalent control structure (for, while, or range depending on the language) is used to shift all elements at positions greater than i to the right by one. The element at position i + 1 is then set to zero, and the iterator i is incremented to skip the newly added zero. This process is repeated until the end of the array is reached, ensuring that consecutive zeros are processed correctly.