<p>You are given an <code>n x n</code> integer matrix <code>board</code> where the cells are labeled from <code>1</code> to <code>n<sup>2</sup></code> in a <a href="https://en.wikipedia.org/wiki/Boustrophedon" target="_blank"><strong>Boustrophedon style</strong></a> starting from the bottom left of the board (i.e. <code>board[n - 1][0]</code>) and alternating direction each row.</p>

<p>You start on square <code>1</code> of the board. In each move, starting from square <code>curr</code>, do the following:</p>

<ul>
	<li>Choose a destination square <code>next</code> with a label in the range <code>[curr + 1, min(curr + 6, n<sup>2</sup>)]</code>.

	<ul>
		<li>This choice simulates the result of a standard <strong>6-sided die roll</strong>: i.e., there are always at most 6 destinations, regardless of the size of the board.</li>
	</ul>
	</li>
	<li>If <code>next</code> has a snake or ladder, you <strong>must</strong> move to the destination of that snake or ladder. Otherwise, you move to <code>next</code>.</li>
	<li>The game ends when you reach the square <code>n<sup>2</sup></code>.</li>
</ul>

<p>A board square on row <code>r</code> and column <code>c</code> has a snake or ladder if <code>board[r][c] != -1</code>. The destination of that snake or ladder is <code>board[r][c]</code>. Squares <code>1</code> and <code>n<sup>2</sup></code> are not the starting points of any snake or ladder.</p>

<p>Note that you only take a snake or ladder at most once per move. If the destination to a snake or ladder is the start of another snake or ladder, you do <strong>not</strong> follow the subsequent&nbsp;snake or ladder.</p>

<ul>
	<li>For example, suppose the board is <code>[[-1,4],[-1,3]]</code>, and on the first move, your destination square is <code>2</code>. You follow the ladder to square <code>3</code>, but do <strong>not</strong> follow the subsequent ladder to <code>4</code>.</li>
</ul>

<p>Return <em>the least number of moves required to reach the square </em><code>n<sup>2</sup></code><em>. If it is not possible to reach the square, return </em><code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/09/23/snakes.png" style="width: 500px; height: 394px;" />
<pre>
<strong>Input:</strong> board = [[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,35,-1,-1,13,-1],[-1,-1,-1,-1,-1,-1],[-1,15,-1,-1,-1,-1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 
In the beginning, you start at square 1 (at row 5, column 0).
You decide to move to square 2 and must take the ladder to square 15.
You then decide to move to square 17 and must take the snake to square 13.
You then decide to move to square 14 and must take the ladder to square 35.
You then decide to move to square 36, ending the game.
This is the lowest possible number of moves to reach the last square, so return 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> board = [[-1,-1],[-1,3]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>n == board.length == board[i].length</code></li>
	<li><code>2 &lt;= n &lt;= 20</code></li>
	<li><code>board[i][j]</code> is either <code>-1</code> or in the range <code>[1, n<sup>2</sup>]</code>.</li>
	<li>The squares labeled <code>1</code> and <code>n<sup>2</sup></code> are not the starting points of any snake or ladder.</li>
</ul>

class Solution {
public:
    // Helper function to convert 1D position to 2D board position
    pair<int, int> getPosition(int curr, int n) {
        int r = (curr - 1) / n;
        int c = (curr - 1) % n;
        if (r % 2 == 1) {
            c = n - 1 - c;  // If it's an odd row, reverse the column
        }
        return {n - 1 - r, c};  // We start from the bottom, so reverse the row
    }

    int snakesAndLadders(vector<vector<int>>& board) {
        int n = board.size();
        
        // BFS setup
        queue<pair<int, int>> q;  // {square number, moves}
        vector<bool> visited(n * n + 1, false);  // Track visited squares
        q.push({1, 0});  // Start at square 1 with 0 moves
        visited[1] = true;
        
        while (!q.empty()) {
            auto [curr, moves] = q.front();
            q.pop();
            
            // If we reach the last square, return the number of moves
            if (curr == n * n) return moves;
            
            // Try all possible moves (roll the die: 1 to 6 steps)
            for (int i = 1; i <= 6; ++i) {
                int next = curr + i;
                if (next > n * n) break;  // Skip if beyond the last square
                
                // Get the 2D position of the next square
                auto [row, col] = getPosition(next, n);
                
                // If there's a ladder or snake, move to the destination
                if (board[row][col] != -1) {
                    next = board[row][col];
                }
                
                // If the square has not been visited yet, visit it
                if (!visited[next]) {
                    visited[next] = true;
                    q.push({next, moves + 1});
                }
            }
        }
        
        // If we cannot reach the last square, return -1
        return -1;
    }
};
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