# Snakes and Ladders

You are given an `n x n` integer matrix `board` where the cells are labeled from `1` to `n2` in a [**Boustrophedon style**](https://en.wikipedia.org/wiki/Boustrophedon) starting from the bottom left of the board (i.e. `board[n - 1][0]`) and alternating direction each row.

You start on square `1` of the board. In each move, starting from square `curr`, do the following:

* Choose a destination square `next` with a label in the range `[curr + 1, min(curr + 6, n2)]`.
  * This choice simulates the result of a standard **6-sided die roll**: i.e., there are always at most 6 destinations, regardless of the size of the board.
* If `next` has a snake or ladder, you **must** move to the destination of that snake or ladder. Otherwise, you move to `next`.
* The game ends when you reach the square `n2`.

A board square on row `r` and column `c` has a snake or ladder if `board[r][c] != -1`. The destination of that snake or ladder is `board[r][c]`. Squares `1` and `n2` do not have a snake or ladder.

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 **not** follow the subsequent snake or ladder.

* For example, suppose the board is `[[-1,4],[-1,3]]`, and on the first move, your destination square is `2`. You follow the ladder to square `3`, but do **not** follow the subsequent ladder to `4`.

Return _the least number of moves required to reach the square_ `n2`_. If it is not possible to reach the square, return_ `-1`.

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2018/09/23/snakes.png)

<pre><code><strong>Input: 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><strong>Output: 4
</strong><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.
</code></pre>

**Example 2:**

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

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**Constraints:**

* `n == board.length == board[i].length`
* `2 <= n <= 20`
* `board[i][j]` is either `-1` or in the range `[1, n2]`.
* The squares labeled `1` and `n2` do not have any ladders or snakes.



### Solutions

This problem can be solved using Breadth-First Search (BFS). The idea is to start from the first cell and visit all its reachable cells (from 1 to 6 steps ahead). If a cell has a snake or ladder, we jump to the destination of the snake or ladder. We continue this process until we reach the last cell. BFS ensures that we reach the last cell with the minimum number of steps because it visits cells in increasing order of their step count.

```csharp
public class Solution {
    public int SnakesAndLadders(int[][] board) {
        int n = board.Length;
        bool[] visited = new bool[n * n + 1];
        Queue<int> queue = new Queue<int>();
        queue.Enqueue(1);
        visited[1] = true;
        int steps = 0;

        while (queue.Count > 0) {
            int size = queue.Count;
            for (int i = 0; i < size; i++) {
                int num = queue.Dequeue();
                if (num == n * n) return steps;
                for (int j = 1; j <= 6; j++) {
                    int next = num + j;
                    if (next > n * n) break;
                    int value = GetBoardValue(board, next);
                    if (value != -1) next = value;
                    if (!visited[next]) {
                        visited[next] = true;
                        queue.Enqueue(next);
                    }
                }
            }
            steps++;
        }

        return -1;
    }

    private int GetBoardValue(int[][] board, int num) {
        int n = board.Length;
        int oldRow = (num - 1) / n;
        int row = n - 1 - oldRow;
        int oldCol = (num - 1) % n;
        int col = oldRow % 2 == 0 ? oldCol : n - 1 - oldCol;
        return board[row][col];
    }
}

```

The time complexity of this solution is O(n^2) and the space complexity is also O(n^2), where n is the size of the board.
