LeetCode 1914: Cyclically Rotating a Grid Solution

Master LeetCode problem 1914 (Cyclically Rotating a Grid), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1914. Cyclically Rotating a Grid

Medium

Array Matrix Simulation

Problem Explanation

Explanation

To solve this problem, we need to perform cyclic rotations on each layer of the matrix k times. We can achieve this by iteratively rotating each layer of the matrix. We can determine the number of layers in the matrix and then rotate each layer one by one. For each rotation, we can shift the elements of the layer in a cyclic manner.

Determine the number of layers in the matrix.
For each layer, calculate the number of elements in that layer.
Perform k rotations on each layer by shifting the elements in a cyclic manner.
Reconstruct the final matrix after all rotations.

Time Complexity: O(m * n) where m and n are the dimensions of the matrix. Space Complexity: O(1) as we are rotating the matrix in place.

Solution Code

class Solution {
    public int[][] rotateGrid(int[][] grid, int k) {
        int m = grid.length, n = grid[0].length;
        for (int layer = 0; layer < Math.min(m, n) / 2; layer++) {
            int row = m - 2 * layer, col = n - 2 * layer;
            int totalElements = 2 * (row + col) - 4;
            int rotation = k % totalElements;

            while (rotation-- > 0) {
                int temp = grid[layer][layer];
                for (int i = layer; i < layer + row - 1; i++) grid[i][layer] = grid[i + 1][layer];
                for (int i = layer; i < layer + col - 1; i++) grid[layer + row - 1][i] = grid[layer + row - 1][i + 1];
                for (int i = layer + row - 1; i > layer; i--) grid[i][layer + col - 1] = grid[i - 1][layer + col - 1];
                for (int i = layer + col - 1; i > layer + 1; i--) grid[layer][i] = grid[layer][i - 1];
                grid[layer][layer + 1] = temp;
            }
        }
        return grid;
    }
}

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