#2326

Medium

auto_awesomeLinked-list pointer manipulation

LeetCode Problem Workspace

Spiral Matrix IV

In this problem, you need to generate a matrix filled with values from a linked list in a spiral order.

Problem Statement

You are given two integers, m and n, representing the dimensions of a matrix, and the head of a linked list containing integers. Your task is to create an m x n matrix filled with the integers from the linked list in a spiral order, starting from the top-left corner.

If the linked list contains fewer elements than the matrix can hold, the remaining empty spaces in the matrix should be filled with -1.

Examples

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]]

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]]

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

Solution Approach

Matrix Initialization

Start by creating an m x n matrix initialized to -1. This step ensures that any unfilled spaces in the matrix are automatically set to -1, which is crucial when the linked list is shorter than the matrix.

Spiral Traversal

Use four boundaries (top, bottom, left, right) to traverse the matrix in a spiral order. Keep moving the boundaries inward while inserting linked list values into the matrix from left to right, top to bottom, right to left, and bottom to top.

Pointer Management

Iterate through the linked list by maintaining a pointer to the current node. Update the pointer as you place values in the matrix, ensuring that all nodes are used before reaching the end of the linked list.

Complexity Analysis

Metric	Value
Time	O(n \cdot m)
Space	O(1)

The time complexity of the solution is O(m * n) because we iterate through each cell of the matrix exactly once. The space complexity is O(1) since we are modifying the matrix in place and not using extra space proportional to the input size.

What Interviewers Usually Probe

Look for the candidate's ability to implement linked-list pointer manipulation effectively.
Evaluate how they manage matrix traversal boundaries and handle edge cases.
Check if they handle the matrix's unfilled spaces correctly when the list is smaller.

Common Pitfalls or Variants

Common pitfalls

Forgetting to initialize the matrix with -1s before filling it.
Not properly updating the boundaries while performing the spiral traversal.
Incorrectly handling cases where the linked list is shorter than the matrix.

Follow-up variants

Increase the matrix dimensions to test if the candidate can handle larger inputs.
Challenge the candidate with linked lists that have exactly m * n elements.
Test the candidate with a very small matrix (e.g., m = 1, n = 1).

FAQ

What is the primary pattern used in the Spiral Matrix IV problem?

The primary pattern involves linked-list pointer manipulation and matrix traversal in a spiral order.

How do I handle an incomplete linked list for the matrix?

Fill the remaining spaces with -1 once the linked list is exhausted, as described in the problem statement.

What are the time and space complexities of the solution?

The time complexity is O(m * n), and the space complexity is O(1).

How does GhostInterview help with linked-list pointer manipulation in this problem?

GhostInterview provides detailed guidance on how to manage the pointer to traverse the linked list while filling the matrix.

What should I do if the matrix is larger than the linked list?

Make sure to fill the remaining empty spaces in the matrix with -1 once the linked list is exhausted.

terminal

Solution

Solution 1: Simulation

We define a two-dimensional array $\textit{ans}$ to store the elements in the linked list, initially all filled with $-1$. We define three variables $i, j, k$, representing the current row, column, and direction respectively. We define an array $\textit{dirs}$ to represent the offsets of the four directions.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def spiralMatrix(self, m: int, n: int, head: Optional[ListNode]) -> List[List[int]]:
        ans = [[-1] * n for _ in range(m)]
        i = j = k = 0
        dirs = (0, 1, 0, -1, 0)
        while 1:
            ans[i][j] = head.val
            head = head.next
            if head is None:
                break
            while 1:
                x, y = i + dirs[k], j + dirs[k + 1]
                if 0 <= x < m and 0 <= y < n and ans[x][y] == -1:
                    i, j = x, y
                    break
                k = (k + 1) % 4
        return ans

Continue Practicing

#2352 equal row and column pairs #2257 count unguarded cells in the grid #2482 difference between ones and zeros in row and column