How to Solve Spiral Matrix on LeetCode
The Spiral Matrix problem is a favorite array traversal question on LeetCode, designed to test your boundary management and multi-dimensional array skills.
Problem Statement
Given an m x n matrix, return all elements of the matrix in spiral order.
Example:
Input: matrix = [
[1,2,3],
[4,5,6],
[7,8,9]
]
Output: [1,2,3,6,9,8,7,4,5]
Why This Problem Matters for Interviews
This problem:
- Evaluates your 2D array traversal ability
- Checks boundary condition logic
- Encourages writing clear, bug-free code
Approach to Spiral Matrix
Layer-by-Layer Traversal (Optimal)
Maintain four boundaries: top, bottom, left, right. Traverse in spiral order, updating boundaries after each layer.
Time Complexity: O(mn)
Space Complexity: O(1) extra (output not counted)
def spiralOrder(matrix: list[list[int]]) -> list[int]:
res = []
if not matrix: return res
top, bottom = 0, len(matrix) - 1
left, right = 0, len(matrix[0]) - 1
while left <= right and top <= bottom:
for j in range(left, right + 1):
res.append(matrix[top][j])
top += 1
for i in range(top, bottom + 1):
res.append(matrix[i][right])
right -= 1
if top <= bottom:
for j in range(right, left - 1, -1):
res.append(matrix[bottom][j])
bottom -= 1
if left <= right:
for i in range(bottom, top - 1, -1):
res.append(matrix[i][left])
left += 1
return res
Key Interview Talking Points
- How boundaries are updated each loop
- Edge cases: empty matrix, single row/column
- Traversal order (right, down, left, up)
- Avoiding duplicates on tight corners
Big O Complexity Recap
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Layer-by-Layer Spiral | O(mn) | O(1) |
Pro Interview Tips
- Draw spiral layers visually in interviews.
- Clarify updates of all four pointers/boundaries.
- Practice with test cases