#1914

Medium

LeetCode Problem Workspace

Cyclically Rotating a Grid

This problem requires cyclically rotating a grid by rotating each layer of the matrix counter-clockwise for a specified number of times.

array matrix simulation

Problem Statement

You are given an m x n matrix grid, where m and n are even integers, and an integer k representing the number of rotations. The matrix is composed of several concentric layers, each treated as an array. For example, in a 4x4 matrix, the outermost layer consists of the elements at the boundary, and the next inner layer consists of the elements inside that boundary. Your task is to cyclically rotate the layers of the matrix counter-clockwise by k steps.

A cyclic rotation of each layer means that each element in the layer shifts to the adjacent element in a counter-clockwise direction. For example, for a given layer, the top-left element will move to the top-right position, and so on. The number of rotations k can be very large, so you need to handle the computation efficiently, possibly reducing redundant rotations.

Examples

Example 1

Input: grid = [[40,10],[30,20]], k = 1

Output: [[10,20],[40,30]]

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]]

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

Solution Approach

Extract Each Layer

The first step is to break the matrix into separate layers, where each layer represents an array of elements in a counter-clockwise fashion. This will allow us to treat each layer independently while rotating it.

Handle Large k Values

Since the number of rotations k can be very large, you should compute the effective rotations by using the modulo operation with the length of the layer. This will prevent unnecessary rotations and optimize the solution.

Rebuild the Matrix

After performing the rotations on each layer, rebuild the matrix by placing the rotated layers back in their respective positions. This will give the final matrix after all rotations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the number of elements in the matrix and the number of rotations. Extracting each layer and performing the rotation both take linear time in terms of the number of elements in the layer, while rebuilding the matrix is also linear in complexity. Therefore, the overall time complexity is O(m * n), where m and n are the matrix dimensions. Space complexity is O(m * n) due to the space required for storing the layers and rotated values.

What Interviewers Usually Probe

Check if the candidate recognizes the need to break the matrix into individual layers for independent rotation.
Look for efficient handling of large k values, ensuring that the number of rotations is minimized using modulo.
Evaluate how well the candidate handles matrix reconstruction after the rotations are completed.

Common Pitfalls or Variants

Common pitfalls

Failing to account for large values of k and performing redundant rotations.
Incorrectly reconstructing the matrix after rotating the layers, potentially missing positions.
Overcomplicating the problem by treating the matrix as a whole rather than focusing on individual layers.

Follow-up variants

Rotating the matrix in the opposite direction (clockwise).
Working with an odd-dimensional matrix, requiring adjustments in layer extraction.
Rotating a matrix that is not square but rectangular.

FAQ

How can I handle large k values in the Cyclically Rotating a Grid problem?

To handle large k values, reduce the number of rotations using the modulo operation with the length of the layer, ensuring you only perform necessary rotations.

What is the most important step in solving the Cyclically Rotating a Grid problem?

The most important step is to break the matrix into layers and treat each layer as an independent array for rotation, making the problem manageable.

How do I reconstruct the matrix after rotating the layers?

Once the layers are rotated, place them back into their respective positions in the matrix to form the final rotated grid.

Can the Cyclically Rotating a Grid problem be solved with a time complexity better than O(m * n)?

No, since every element in the matrix must be touched at least once to rotate the layers, the time complexity is O(m * n).

What are some common mistakes when solving the Cyclically Rotating a Grid problem?

Common mistakes include not reducing redundant rotations, incorrectly handling layer extraction, or failing to reconstruct the matrix properly after rotation.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def rotateGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
        def rotate(p: int, k: int):
            nums = []
            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)
            if k == 0:
                return
            nums = nums[k:] + nums[:k]
            k = 0
            for j in range(p, n - p - 1):
                grid[p][j] = nums[k]
                k += 1
            for i in range(p, m - p - 1):
                grid[i][n - p - 1] = nums[k]
                k += 1
            for j in range(n - p - 1, p, -1):
                grid[m - p - 1][j] = nums[k]
                k += 1
            for i in range(m - p - 1, p, -1):
                grid[i][p] = nums[k]
                k += 1

        m, n = len(grid), len(grid[0])
        for p in range(min(m, n) >> 1):
            rotate(p, k)
        return grid

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