<p>You are given two integers <code>m</code> and <code>n</code>, which represent the dimensions of a matrix.</p>

<p>You are also given the <code>head</code> of a linked list of integers.</p>

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

<p>Return <em>the generated matrix</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/09/ex1new.jpg" style="width: 240px; height: 150px;" />
<pre>
<strong>Input:</strong> m = 3, n = 5, head = [3,0,2,6,8,1,7,9,4,2,5,5,0]
<strong>Output:</strong> [[3,0,2,6,8],[5,0,-1,-1,1],[5,2,4,9,7]]
<strong>Explanation:</strong> The diagram above shows how the values are printed in the matrix.
Note that the remaining spaces in the matrix are filled with -1.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/11/ex2.jpg" style="width: 221px; height: 60px;" />
<pre>
<strong>Input:</strong> m = 1, n = 4, head = [0,1,2]
<strong>Output:</strong> [[0,1,2,-1]]
<strong>Explanation:</strong> 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.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li>The number of nodes in the list is in the range <code>[1, m * n]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
</ul>

auto speedUP = [](){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    return 'c';
}();

class Solution {
private:
    static constexpr array<pair<int, int>, 4> dirs = {make_pair(0, 1), make_pair(1, 0), make_pair(0, -1), make_pair(-1, 0)};

public:
    vector<vector<int>> spiralMatrix(int m, int n, ListNode* head) 
    {
        vector<vector<int>> result;
        result.reserve(m);

        for (int i = 0; i < m; i++)
        {
            vector<int> tmp(n, -1);
            result.push_back(move(tmp));
        }

        int row = 0;
        int col = 0;

        int dirIdx = 0;

        int initialStepsToTakeCol = n;
        int stepsToTakeCol = initialStepsToTakeCol;

        int initialStepsToTakeRow = m - 1;
        int stepsToTakeRow = initialStepsToTakeRow;

        int idx = 0;
        while (head != nullptr && idx < m * n)
        {
            result[row][col] = head->val;

            if (dirIdx % 2 == 0)
            {
                stepsToTakeCol--;
                if (stepsToTakeCol == 0)
                {
                    dirIdx = (dirIdx + 1) % 4;

                    initialStepsToTakeCol--;
                    stepsToTakeCol = initialStepsToTakeCol;
                }
            }
            else
            {
                stepsToTakeRow--;
                if (stepsToTakeRow == 0)
                {
                    dirIdx = (dirIdx + 1) % 4;

                    initialStepsToTakeRow--;
                    stepsToTakeRow = initialStepsToTakeRow;
                }
            }

            row += dirs[dirIdx].first;
            col += dirs[dirIdx].second;

            head = head->next;

            idx++;
        }

        return result;    
    }
};
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