Solution Explanation for Cyclically Rotating a Grid

This problem involves cyclically rotating layers of a given matrix. The solution iterates through each layer, extracts its elements, performs a cyclic rotation on those elements, and then places them back into the matrix.

Approach

The core idea is to process the matrix layer by layer. Each layer can be considered a ring. We extract the elements of each ring, rotate them by k positions, and put them back. This rotation is done counter-clockwise. To handle large values of k, we use the modulo operator (%) to ensure that the rotation doesn't exceed the ring's length.

1. Layer Iteration: The solution uses nested loops to iterate through layers. The outer loop iterates from p = 0 to min(m, n) / 2, where m and n are the dimensions of the matrix. This represents the layer index, starting from the outermost layer.

2. Ring Extraction: For each layer p, the elements of the ring are extracted into a temporary list or array (nums). This is done by traversing the ring counter-clockwise.

3. Cyclic Rotation: The nums list is rotated by k positions using list slicing in Python or array manipulation in other languages. The modulo operator (k % len(nums)) handles cases where k is larger than the ring's length.

4. Ring Placement: The rotated elements from nums are placed back into the matrix in their new positions, again traversing the ring counter-clockwise.

Code Explanation (Python)

class Solution:
    def rotateGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
        def rotate(p: int, k: int): # Inner function to rotate a single layer
            nums = []
            # Extract elements from the layer (counter-clockwise)
            for j in range(p, n - p - 1):
                nums.append(grid[p][j])
            for i in range(p, m - p - 1):
                nums.append(grid[i][n - p - 1])
            for j in range(n - p - 1, p, -1):
                nums.append(grid[m - p - 1][j])
            for i in range(m - p - 1, p, -1):
                nums.append(grid[i][p])
 
            k %= len(nums) # Handle k larger than ring length
            nums = nums[k:] + nums[:k] # Perform cyclic rotation
 
            # Place rotated elements back into the grid
            idx = 0
            for j in range(p, n - p - 1):
                grid[p][j] = nums[idx]
                idx += 1
            for i in range(p, m - p - 1):
                grid[i][n - p - 1] = nums[idx]
                idx += 1
            for j in range(n - p - 1, p, -1):
                grid[m - p - 1][j] = nums[idx]
                idx += 1
            for i in range(m - p - 1, p, -1):
                grid[i][p] = nums[idx]
                idx += 1
 
        m, n = len(grid), len(grid[0])
        for p in range(min(m, n) >> 1): # Iterate through layers
            rotate(p, k)
        return grid

The other languages (Java, C++, Go, TypeScript) follow a very similar structure, adapting the syntax and data structures to their respective languages.

Time and Space Complexity

• Time Complexity: O(mn), where m and n are the dimensions of the matrix. This is because we iterate through each element of the matrix at least once during the extraction and placement phases. The rotation within a layer is O(l), where l is the layer's perimeter length, and the sum of all layer perimeters is proportional to mn.

• Space Complexity: O(min(m, n)), in the worst case. This is the space used by the temporary array (nums) to store the elements of a single layer. The space used is proportional to the length of the largest layer's perimeter, which is at most 2 * (m + n - 2) (approximately 2 * max(m,n)), hence O(min(m, n)). In-place rotation algorithms could reduce this to O(1), but they are more complex to implement.