How do you solve Spiral Matrix II on LeetCode?

Spiral Matrix II (LeetCode #59) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: Understanding how to manage boundaries is crucial for spiral filling.

What is the brute force solution for Spiral Matrix II?

The brute force approach for Spiral Matrix II is Brute Force, with O(n²) time complexity and O(1) space complexity. We can fill the matrix layer by layer, starting from the outermost layer and moving inward. This involves keeping track of the boundaries of the current layer and filling it in a clockwise direction. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Spiral Matrix II?

Spiral Matrix II uses the following concepts: Array, Matrix, Simulation. The recommended patterns to study are: Layered filling, Matrix traversal.

What are the interview tips for Spiral Matrix II?

Explain your thought process clearly while coding. Use comments to clarify your logic if needed. Practice similar problems to get comfortable with matrix manipulations.

What are common mistakes when solving Spiral Matrix II?

Forgetting to adjust boundaries correctly after filling each side. Not handling the case when n is odd properly.

#59

Spiral Matrix II

Medium

ArrayMatrixSimulationLayered fillingMatrix traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(1)

The brute-force method is already quite efficient for this problem, as it directly fills the matrix in a spiral order. However, we can optimize the way we manage boundaries and the filling process to make the code cleaner and potentially faster.

⚙️

Algorithm

3 steps

1Step 1: Initialize an n x n matrix filled with zeros.
2Step 2: Use a single loop to fill the matrix in a spiral order by adjusting the boundaries dynamically.
3Step 3: Continue filling until all elements are placed.

solution.py18 lines

1def generateMatrix(n):
2    matrix = [[0] * n for _ in range(n)]
3    num = 1
4    layers = (n + 1) // 2
5    for layer in range(layers):
6        for i in range(layer, n - layer):
7            matrix[layer][i] = num
8            num += 1
9        for i in range(layer + 1, n - layer):
10            matrix[i][n - layer - 1] = num
11            num += 1
12        for i in range(n - layer - 1, layer - 1, -1):
13            matrix[n - layer - 1][i] = num
14            num += 1
15        for i in range(n - layer - 2, layer, -1):
16            matrix[i][layer] = num
17            num += 1
18    return matrix

ℹ

Complexity note: The time complexity remains O(n²) as we still fill n² elements. The space complexity is O(1) since we are using the input matrix for storage.

1Understanding how to manage boundaries is crucial for spiral filling.
2Recognizing that filling in layers can simplify the logic.

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