How do you solve Cyclically Rotating a Grid on LeetCode?

Cyclically Rotating a Grid (LeetCode #1914) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding layers is crucial for matrix manipulation.

What is the brute force solution for Cyclically Rotating a Grid?

The brute force approach for Cyclically Rotating a Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves extracting each layer of the matrix, rotating it by one position for k times, and then placing it back into the matrix. This is straightforward but inefficient for large k values. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Cyclically Rotating a Grid?

Cyclically Rotating a Grid uses the following concepts: Array, Matrix, Simulation. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Cyclically Rotating a Grid?

Explain your thought process clearly. Discuss edge cases, like k being 0 or larger than the layer size. Practice layer extraction and reinsertion separately.

What are common mistakes when solving Cyclically Rotating a Grid?

Not accounting for k being larger than the layer size. Failing to properly extract and reinsert layers.

#1914

Cyclically Rotating a Grid

Medium

ArrayMatrixSimulationArrayMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach focuses on reducing the number of rotations needed by calculating the effective rotations using modulo. This way, we avoid unnecessary full rotations and directly manipulate the layers.

⚙️

Algorithm

4 steps

1Step 1: Extract each layer from the grid and store it in a list.
2Step 2: Calculate effective rotations using k % length of the layer.
3Step 3: Rotate the layer using slicing instead of multiple shifts.
4Step 4: Place the rotated layers back into their respective positions in the grid.

solution.py37 lines

1def rotateGrid(grid, k):
2    m, n = len(grid), len(grid[0])
3    layers = []
4    for layer in range(min(m, n) // 2):
5        current_layer = []
6        # Extract layer
7        for i in range(layer, n - layer):
8            current_layer.append(grid[layer][i])
9        for i in range(layer + 1, m - layer):
10            current_layer.append(grid[i][n - layer - 1])
11        for i in range(n - layer - 1, layer - 1, -1):
12            current_layer.append(grid[m - layer - 1][i])
13        for i in range(m - layer - 2, layer, -1):
14            current_layer.append(grid[i][layer])
15        layers.append(current_layer)
16
17    # Rotate each layer
18    for i in range(len(layers)):
19        k_mod = k % len(layers[i])
20        layers[i] = layers[i][-k_mod:] + layers[i][:-k_mod]
21
22    # Place layers back
23    for layer in range(len(layers)):
24        idx = 0
25        for i in range(layer, n - layer):
26            grid[layer][i] = layers[layer][idx]
27            idx += 1
28        for i in range(layer + 1, m - layer):
29            grid[i][n - layer - 1] = layers[layer][idx]
30            idx += 1
31        for i in range(n - layer - 1, layer - 1, -1):
32            grid[m - layer - 1][i] = layers[layer][idx]
33            idx += 1
34        for i in range(m - layer - 2, layer, -1):
35            grid[i][layer] = layers[layer][idx]
36            idx += 1
37    return grid

ℹ

Complexity note: The time complexity is O(n) because we only traverse each layer once for extraction and placement, and the space complexity is O(n) due to storing the layers.

1Understanding layers is crucial for matrix manipulation.
2Effective rotation reduces unnecessary computations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.