LeetCode 909: Snakes and Ladders Solution

Master LeetCode problem 909 (Snakes and Ladders), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

909. Snakes and Ladders

Medium

Array Breadth-First Search Matrix

Problem Explanation

Explanation:

Algorithm:
1. We can treat this problem as a shortest path problem where each cell is a node and snakes/ladders represent edges between nodes.
2. We can flatten the 2D board into a 1D array for easier access.
3. We can perform a breadth-first search (BFS) starting from square 1 to find the minimum number of steps to reach square n^2.
4. During the BFS traversal, we need to consider the possible moves based on dice rolls and handle the cases where we encounter snakes or ladders.
5. We can keep track of visited nodes to avoid revisiting the same node.
Time Complexity: O(n^2) where n is the size of the board.
Space Complexity: O(n^2) for the visited array and queue.

Solution Code

class Solution {
    public int snakesAndLadders(int[][] board) {
        int n = board.length;
        int[] flatBoard = new int[n * n];
        boolean[] visited = new boolean[n * n];
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int idx = n * i + j;
                if (i % 2 == 0) {
                    flatBoard[idx] = board[n - 1 - i][j];
                } else {
                    flatBoard[idx] = board[n - 1 - i][n - 1 - j];
                }
            }
        }
        
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(0);
        visited[0] = true;
        int moves = 0;
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int curr = queue.poll();
                if (curr == n * n - 1) {
                    return moves;
                }
                for (int j = 1; j <= 6 && curr + j < n * n; j++) {
                    int next = flatBoard[curr + j] == -1 ? curr + j : flatBoard[curr + j] - 1;
                    if (!visited[next]) {
                        visited[next] = true;
                        queue.offer(next);
                    }
                }
            }
            moves++;
        }
        
        return -1;
    }
}

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