Is Spiral Matrix IV easy or hard?

Spiral Matrix IV is rated Medium because the spiral traversal requires careful boundary updates and direction changes. The linked list integration adds another layer since you must stop when the list runs out and fill remaining cells with -1.

Spiral Matrix IV Python/Java solution

A typical implementation uses four boundary variables and a pointer to the current linked list node. Iterate across the top row, right column, bottom row, and left column while updating boundaries after each pass. The same logic works across Python, Java, C++, and C with O(mn) time complexity.

How to solve Spiral Matrix IV in O(n)?

Treat the matrix fill as a spiral simulation and iterate through each cell once. Use boundary pointers (top, bottom, left, right) to control traversal direction while reading values from the linked list. Since each matrix position is written exactly once, the algorithm runs in O(mn) time with constant extra space.

What is the best approach for Spiral Matrix IV?

The boundary-based spiral traversal is the most practical approach. Maintain four boundaries (top, bottom, left, right) and traverse the matrix in four directions while consuming nodes from the linked list. Each cell is filled exactly once, resulting in O(mn) time and O(1) extra space.

Is Spiral Matrix IV asked at Google/Amazon/Meta?

Spiral traversal and matrix simulation problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variations such as Spiral Matrix I, II, and linked-list-driven fills test boundary control and careful iteration logic.

What data structure is used in Spiral Matrix IV?

The solution combines a singly linked list with a 2D matrix. The linked list provides the sequence of values, while the matrix is filled using spiral traversal controlled by boundary pointers.

What is the time complexity of Spiral Matrix IV?

The time complexity is O(mn), where m and n are the matrix dimensions. Every cell in the matrix is visited once during the spiral traversal. The linked list pointer simply advances as values are written into the matrix.

Description Hints 3 Solution Video 21:38

2326. Spiral Matrix IV

Medium82.3% acceptance4 companies

You are given two integers m and n, which represent the dimensions of a matrix.

You are also given the head of a linked list of integers.

Generate an m x n matrix that contains the integers in the linked list presented in spiral order (clockwise), starting from the top-left of the matrix. If there are remaining empty spaces, fill them with -1.

Return the generated matrix.

Example 1:

Input: m = 3, n = 5, head = [3,0,2,6,8,1,7,9,4,2,5,5,0]
Output: [[3,0,2,6,8],[5,0,-1,-1,1],[5,2,4,9,7]]
Explanation: The diagram above shows how the values are printed in the matrix.
Note that the remaining spaces in the matrix are filled with -1.

Example 2:

Input: m = 1, n = 4, head = [0,1,2]
Output: [[0,1,2,-1]]
Explanation: The diagram above shows how the values are printed from left to right in the matrix.
The last space in the matrix is set to -1.

Constraints:

1 <= m, n <= 10⁵
1 <= m * n <= 10⁵
The number of nodes in the list is in the range [1, m * n].
0 <= Node.val <= 1000

Asked at 4 companies

DocuSign

Google Microsoft Meta

Topics

Hints open one at a time

Hint 1

First, generate an m x n matrix filled with -1s.

Hint 2

Navigate within the matrix at (i, j) with the help of a direction vector ⟨di, dj⟩. At (i, j), you need to decide if you can keep going in the current direction.

Hint 3

If you cannot keep going, rotate the direction vector clockwise by 90 degrees.

21:38

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