How do you solve Spiral Matrix on LeetCode?

Spiral Matrix (LeetCode #54) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Spiral traversal can be visualized as layers of an onion being peeled away.

What is the brute force solution for Spiral Matrix?

The brute force approach for Spiral Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates the spiral traversal by explicitly iterating through the boundaries of the matrix. We collect elements layer by layer, starting from the outermost layer and moving inward. The optimal Optimal Solution approach improves this to O(n²).

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

Spiral Matrix uses the following concepts: Array, Matrix, Simulation. The recommended patterns to study are: Two Pointers, Simulation.

What are the interview tips for Spiral Matrix?

Clearly explain your thought process and how you manage boundaries. Consider edge cases like empty matrices or single-row/column matrices. Practice writing code on a whiteboard or in a shared document to simulate interview conditions.

What are common mistakes when solving Spiral Matrix?

Forgetting to check boundary conditions, leading to index errors. Not handling empty matrices correctly.

#54

Spiral Matrix

Medium

ArrayMatrixSimulationTwo PointersSimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

The optimal solution is essentially the same as the brute force but can be optimized in terms of clarity and efficiency by using a single loop to manage the boundaries. This reduces the number of checks and makes the code cleaner.

⚙️

Algorithm

3 steps

1Step 1: Initialize boundaries: top, bottom, left, right.
2Step 2: Use a while loop to iterate until all boundaries converge.
3Step 3: In each iteration, traverse the top row, right column, bottom row, and left column, adjusting boundaries accordingly.

solution.py22 lines

1# Full working Python code
2
3def spiralOrder(matrix):
4    if not matrix:
5        return []
6    result = []
7    top, bottom, left, right = 0, len(matrix) - 1, 0, len(matrix[0]) - 1
8    while top <= bottom and left <= right:
9        result.extend(matrix[top][left:right + 1])
10        top += 1
11        for i in range(top, bottom + 1):
12            result.append(matrix[i][right])
13        right -= 1
14        if top <= bottom:
15            result.extend(matrix[bottom][left:right + 1][::-1])
16            bottom -= 1
17        if left <= right:
18            for i in range(bottom, top - 1, -1):
19                result.append(matrix[i][left])
20            left += 1
21    return result
22

ℹ

Complexity note: The time complexity remains O(n²) as we still visit each element once. The space complexity is O(n) because we store the result in a separate list.

1Spiral traversal can be visualized as layers of an onion being peeled away.
2Managing boundaries is crucial to avoid revisiting elements.

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