This is really a conceptual (language-agnostic) question, but for the sake of explanation, I'll use C++. I'd prefer an answer that can be ported to other languages (no pointer arithmetic or memory tricks).

---

Let's say we have:

* `arr`, our **rectangular** 2D array of some arbitrary type `T`
* `void shift(int dx, int dy)`, the function that performs the "rotation"
* `numRows`, the number of rows
* `numCols`, the number of columns

`shift()` shifts the array such that all rows are moved down `dx` places, and the rows pushed out of bounds will wrap-around to the beginning. (Likewise for the columns and `dy`.) Let's say this is what our array looks like initially:

```
{{a1, a2, a3, a4},
 {b1, b2, b3, b4},
 {c1, c2, c3, c4},
 {d1, d2, d3, d4}};

```

After we call our function: `shift(2,1)`, `arr` should look like this:

```
{{c4, c1, c2, c3},
 {d4, d1, d2, d3},
 {a4, a1, a2, a3},
 {b4, b1, b2, b3}};

```

In this case, `dx` was **2**, so everything moved **down** two places, and `dy` was **1**, so everything also moved to the **right** one place.

Here was my approach to solving this problem:

```
void shift(int dx, int dy)
{
   T newArr[numRows][numCols];
   for(int r = 0; r < numRows; r++)
   {
      for(int c = 0; c < numCols; c++)
         newArr[(r + dx) % numRows][(c + dy) % numCols] = arr[r][c];
   }
   for(int r = 0; r < numRows; r++)
   {
      for(int c = 0; c < numCols; c++)
         arr[r][c] = newArr[r][c];
   }
}

```

I'm not satisfied with this code because it is neither time-efficient nor space-efficient. I'm looking for a more elegant solution that does more with fewer loops *and* uses less memory.