Problem Description

The problem presents a grid, which is a 2-dimensional matrix of integers of m rows and n columns, where both m and n are even numbers. This grid has several concentric rectangular layers, like an onion. The task is to perform cyclic rotations on each of these layers. A single rotation moves each element to the position of the next element in the counter-clockwise direction within the layer it belongs to.

Imagine holding the outermost ring (layer) of the grid and rotating it left so that each item shifts one place along the ring. A full rotation would bring every element back to its original position. To achieve a cyclic rotation, you'd keep rotating until the specified number of rotations k is reached. However, to reduce unnecessary rotations, the real number of rotations needed is k modulo the total number of elements in the ring, as repeating the total number of elements in the ring's rotations would result in the original ring configuration.

The problem asks us to perform this operation k times on each layer and return the new grid configuration.

Intuition

To solve this problem, an intuitive approach is to treat each layer of the grid as a linear list of numbers. By flattening each layer into a list, we can perform rotations on this list directly, which simplifies the process of shifting elements. To flatten a layer, we traverse the perimeter of the rectangular layer, adding each element to the list. Once the list is assembled, we perform the rotations.

Since a full rotation of the list would result in the original configuration of the layer, we only need to rotate k times modulo the size of the list. This is done to eliminate the extra full rotations, leaving us with the remainder of k rotations that actually change the element positions.

The rotation itself is simple—we split the list into two parts: the first k elements and the remaining elements. We then concatenate these two parts by placing the second part at the front and the first part at the end.

The final step is to put the rotated list back into the correct positions on the grid. We must take care to place each element back into its original layer, going counter-clockwise around the rectangle. This process is reversed from the flattening step and requiring careful index calculation to ensure each element goes back to the correct position in the grid.

Overall, the main algorithm can be divided into 3 steps for each layer:

1. Flatten the layer into a list.
2. Rotate the list.
3. Put the rotated elements back into their positions on the grid.

We apply this process starting from the outermost layer and moving inward until all layers have been rotated.

Solution Approach

The reference solution defines an inner function rotate that performs the rotation for a specific layer p of the grid, with p starting from 0 for the outermost layer and increasing to reach the inner layers. The layers are processed from outermost to innermost by a loop that calls the rotate function for each layer.

Here is the step-by-step breakdown of the solution:

1. Initialization and Outer Loop:

   • First, the dimensions of the grid m and n are stored.
   • An outer loop iterates over each layer of the grid, starting with p = 0 for the outermost layer and incrementing p until it reaches the innermost layer. The stopping condition of the loop is until p reaches min(m, n) >> 1, i.e., half of the minimum of m and n, because there are that many layers in an m x n grid where both m and n are even.

2. Layer Flattening:

   • Inside the rotate function, an empty list nums is created to represent the flattened layer.
   • Four for loops are used to traverse the rectangle's perimeter and fill nums with the elements in the layer, going counter-clockwise starting from the top-left corner of the layer.

3. Calculating Effective Rotations:

   • The number of effective rotations k is calculated using the modulo operator (%) with respect to the length of nums. This is because rotating a list by its own length returns it to its original order, so any multiples of the list length can be ignored in the count of rotations.

4. Rotating the Flattened Layer:

   • Rotations are performed by reassigning the values in nums such that it becomes nums[k:] + nums[:k]. This effectively moves the first k elements to the end of the list while moving the rest to the front.

5. Putting Elements Back:

   • The rotated list is then put back into the grid. This involves reversing the process of flattening the layer: it uses four for loops that assign the rotated values from nums back into their respective positions in the grid. The variable k used here is distinct from the number of rotations and is instead an index into the nums list. With each assignment, k is incremented to move to the next element in nums.

6. Return the Result:

   • After all layers have been rotated, the modified grid is returned as the final result.

This solution makes use of space to simplify the problem—flattening each layer into a list allows for straightforward rotation. The loops for putting the elements back ensure that the new values are assigned into proper positions according to the original layers' topology. This layer-by-layer approach is an effective way to modularize the problem and handle complex, multi-dimensional manipulations in a controllable manner.

Example Walkthrough

Let's take a grid of size 4x4 as an example to walk you through the solution process for cyclic rotation. Assume our grid looks as follows and we want to perform k = 2 rotations:

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16

With m = 4 and n = 4, we have two layers to rotate. We will go through the steps of the solution approach for the outermost layer (p = 0), as the same steps would apply to the inner layers:

1. Initialization and Outer Loop:

   • We start with p = 0, and since min(m, n) >> 1 equals 2, our loop will process two layers.

2. Layer Flattening:

   • We flatten the outermost layer (when p = 0) by adding the elements in counter-clockwise order to a list nums:

     nums = [1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5]

3. Calculating Effective Rotations:

   • There are 12 elements in nums, and we want to rotate k = 2 times. 2 % 12 is 2, so we will rotate the list by 2 elements.

4. Rotating the Flattened Layer:

   • We rotate nums by 2 places to get the new rotated list:

     nums = [3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 1, 2]

5. Putting Elements Back:

   • We put the elements back into the grid starting from where we began unrolling the layer, in the counter-clockwise direction:

     1 becomes 3, 2 becomes 4, 3 becomes 8, and so on, resulting in the grid:
     
     3 4 8 12
     2 6 7 16
     1 10 11 15
     5 9  14 13

6. Repeat for Next Layers:

   • Now, we'd repeat the process for the next layer (when p = 1, the inner 2x2 grid). Since k = 2, the inner grid would simply swap positions of elements.
   • The final grid after applying the process for the inner layer would be:

     3 4  8 12
     2 7 11 16
     1 6 10 15
     5 9 14 13

After performing the above steps for all layers, we have completed the rotations for each concentric rectangle in the grid, and the problem is solved. The grid we return represents the state of the original grid after 2 cyclic rotations have been applied to each layer.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the function primarily comes from the rotations performed layer by layer (from the outermost layer to the innermost). For each layer, we traverse all the elements in that layer to collect them into a list, and then write them back after the k rotations are applied.

Assuming the grid is m x n:

• There are min(m, n) / 2 layers to rotate because the deepest layer is limited by the shorter dimension of the grid.
• Each layer contains approximately 2*(m - 2p) + 2*(n - 2p) - 4 elements, where p is the current layer index (starting at 0). The -4 accounts for the corner elements being counted twice.
• Summing this over all layers gives a series which is somewhat less than 2*m*n in total since the inner layers have fewer elements.
• Therefore, the time complexity is essentially linear with respect to the total number of elements, leading to a total time complexity of O(m*n).

Space Complexity

The extra space complexity comes from two sources: the nums list that stores the border elements for each layer, and the recursion stack due to the rotate helper function. However, the helper function does not call itself recursively, so there's only ever one rotate function call active at a time.

The space required for the nums list is the longest when handling the outermost layer. Maximum space used would be for the next-to-largest possible layer, which is roughly 2*m + 2*n - 4. However, since the layers are processed one by one and the content of nums is overwritten for each layer rotation, this does not scale with the number of layers.

As a result, the space complexity of the algorithm is O(m + n) due to the nums list needed for rotating each layer.

Learn more about how to find time and space complexity quickly using problem constraints.