Introduction

In this blog post, we will explore how to solve the LeetCode problem "1914" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an m x n integer matrix grid​​​, where m and n are both even integers, and an integer k. The matrix is composed of several layers, which is shown in the below image, where each color is its own layer: A cyclic rotation of the matrix is done by cyclically rotating each layer in the matrix. To cyclically rotate a layer once, each element in the layer will take the place of the adjacent element in the counter-clockwise direction. An example rotation is shown below: Return the matrix after applying k cyclic rotations to it. Example 1:
Input: grid = [[40,10],[30,20]], k = 1 Output: [[10,20],[40,30]] Explanation: The figures above represent the grid at every state. Example 2:
Input: grid = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]], k = 2 Output: [[3,4,8,12],[2,11,10,16],[1,7,6,15],[5,9,13,14]] Explanation: The figures above represent the grid at every state. Constraints: m == grid.length n == grid[i].length 2 <= m, n <= 50 Both m and n are even integers. 1 <= grid[i][j] <= 5000 1 <= k <= 109

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Layer Extraction and Rotation: The code iterates through each layer of the matrix, extracts the elements of the layer into a 1D array, performs the cyclic rotation on this array, and then updates the original matrix with the rotated layer.

• Efficient Rotation: To avoid repeated element swapping during rotation, the code uses the modulo operator (%) to efficiently determine the new position of each element after k rotations. This dramatically reduces the number of operations, especially when k is large.

• In-place Update: The code updates the matrix in place, minimizing additional memory usage.

• Complexity:

  • Runtime Complexity: O(m * n), where m and n are the dimensions of the grid.
  • Storage Complexity: O(L), where L is the maximum length of a layer. In the worst case, L can be O(m+n).

Code

    def rotateGrid(grid, k):
    m = len(grid)
    n = len(grid[0])
    layers = min(m, n) // 2

    for layer in range(layers):
        # Extract layer elements
        layer_elements = []
        # Top row
        for j in range(layer, n - layer):
            layer_elements.append(grid[layer][j])
        # Right column
        for i in range(layer + 1, m - layer - 1):
            layer_elements.append(grid[i][n - layer - 1])
        # Bottom row
        for j in range(n - layer - 1, layer - 1, -1):
            layer_elements.append(grid[m - layer - 1][j])
        # Left column
        for i in range(m - layer - 2, layer, -1):
            layer_elements.append(grid[i][layer])

        layer_len = len(layer_elements)
        rotated_elements = [0] * layer_len
        for i in range(layer_len):
            rotated_elements[(i + k) % layer_len] = layer_elements[i]

        # Update the grid with rotated elements
        idx = 0
        # Top row
        for j in range(layer, n - layer):
            grid[layer][j] = rotated_elements[idx]
            idx += 1
        # Right column
        for i in range(layer + 1, m - layer - 1):
            grid[i][n - layer - 1] = rotated_elements[idx]
            idx += 1
        # Bottom row
        for j in range(n - layer - 1, layer - 1, -1):
            grid[m - layer - 1][j] = rotated_elements[idx]
            idx += 1
        # Left column
        for i in range(m - layer - 2, layer, -1):
            grid[i][layer] = rotated_elements[idx]
            idx += 1

    return grid