Algorithms - Circular Array Rotation - c++

Question

Is this linear complexity implementation of circular array rotation correct?
n = number of elements
k = number of rotations
int write_to = 0;
int copy_current = 0;
int copy_final = a[0];
int rotation = k;
int position = 0;
for (int i = 0; i < n; i++) {
write_to = (position + rotation) % n;
copy_current = a[write_to];
a[write_to] = copy_final;
position = write_to;
copy_final = copy_current;
}

Answer 1

No.
Consider this example.
#include <iostream>
int main(void) {
int n = 6;
int k = 2;
int a[] = {1, 2, 3, 4, 5, 6};
int write_to = 0;
int copy_current = 0;
int copy_final = a[0];
int rotation = k;
int position = 0;
for (int i = 0; i < n; i++) {
write_to = (position + rotation) % n;
copy_current = a[write_to];
a[write_to] = copy_final;
position = write_to;
copy_final = copy_current;
}
for (int i = 0; i < n; i++) {
std::cout << a[i] << (i + 1 < n ? ' ' : '\n');
}
return 0;
}
Expected result:
5 6 1 2 3 4
Actual result:
3 2 1 4 1 6

Answer 2

Using stl::rotate on std::array, you can left rotate by, say 2, as:
std::array<int, 6> a{1, 2, 3, 4, 5, 6};
std::rotate(begin(a), begin(a) + 2, end(a)); // left rotate by 2
to yield: 3 4 5 6 1 2, or right-rotate by, say 2, as:
std::rotate(begin(a), end(a) - 2, end(a)); // right rotate by 2
to yield: 5 6 1 2 3 4, with linear complexity.

Answer 3

Rotate an Array of length n for k times in left or right directions.
The code is in Java
I define a Direction Enum:
public enum Direction {
L, R
};
Rotation with times and direction:
public static final void rotate(int[] arr, int times, Direction direction) {
if (arr == null || times < 0) {
throw new IllegalArgumentException("The array must be non-null and the order must be non-negative");
}
int offset = arr.length - times % arr.length;
if (offset > 0) {
int[] copy = arr.clone();
for (int i = 0; i < arr.length; ++i) {
int j = (i + offset) % arr.length;
if (Direction.R.equals(direction)) {
arr[i] = copy[j];
} else {
arr[j] = copy[i];
}
}
}
}
Complexity: O(n).
Example:
Input: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Rotate 3 times left
Output: [4, 5, 6, 7, 8, 9, 10, 1, 2, 3]
Input: [4, 5, 6, 7, 8, 9, 10, 1, 2, 3]
Rotate 3 times right
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]