#909

Medium

auto_awesomeArray plus Breadth-First Search

LeetCode Problem Workspace

Snakes and Ladders

Solve the Snakes and Ladders problem using a BFS strategy to efficiently navigate the board and reach the final square.

Problem Statement

You are given an n x n integer matrix representing a Snakes and Ladders game board. Each cell is numbered starting from 1, and cells are arranged in a Boustrophedon style. Movement can either involve a snake, which moves you backward, or a ladder, which moves you forward. A square labeled with a number other than -1 represents the destination of a snake or ladder.

Starting at square 1, the objective is to find the minimum number of moves required to reach the last square (n^2). Each move involves jumping from one square to another based on either a snake or a ladder. The challenge is to use a breadth-first search strategy to explore the board while taking the shortest route possible to the last square.

Examples

Example 1

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]]

Output: 4

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.

Example 2

Input: board = [[-1,-1],[-1,3]]

Output: 1

Example details omitted.

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 are not the starting points of any snake or ladder.

Solution Approach

Breadth-First Search (BFS) Approach

Use BFS to explore each possible move from the current square. This ensures that you always explore the shortest possible paths first, minimizing the number of moves to reach the final square. Each square is treated as a node, and each valid move is treated as an edge in the graph.

Tracking Visited Squares

To avoid cycles and redundant paths, maintain a 'visited' array that marks squares that have already been processed. This prevents revisiting the same square and ensures the algorithm doesn't get stuck in infinite loops.

Handling Snakes and Ladders

For each square, check if it contains a snake or ladder. If so, move directly to the destination indicated by the value of the square. These jumps represent direct connections in the BFS graph, allowing you to skip over some squares.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the number of squares, as we explore each one at most once. The space complexity is proportional to the number of squares, as we need space for tracking visited nodes and BFS queue elements.

What Interviewers Usually Probe

Understanding the BFS strategy is key to solving this problem efficiently.
Candidates should demonstrate the ability to handle dynamic graphs with directed edges (snakes and ladders).
Look for clear reasoning behind the BFS approach and the use of the visited set to avoid revisits.

Common Pitfalls or Variants

Common pitfalls

Failing to account for the Boustrophedon layout when processing the board.
Not properly managing the visited array, which can lead to infinite loops or redundant work.
Misunderstanding BFS and using a depth-first search approach, leading to suboptimal solutions.

Follow-up variants

You may be asked to solve the problem with a different board size or more complex variations of the snakes and ladders layout.
An extended version of the problem may require handling multiple boards in sequence, testing your approach's scalability.
The problem could introduce additional elements, such as obstacles on the board, that modify how you traverse the squares.

FAQ

How does BFS help in solving the Snakes and Ladders problem?

BFS ensures that you explore the board in layers, always processing the shortest possible move first, which guarantees the least number of moves to the last square.

What is the significance of the 'visited' array in this problem?

The 'visited' array helps prevent revisiting squares and ensures that each square is processed only once, avoiding infinite loops and redundant calculations.

Can the BFS approach be applied to larger boards?

Yes, BFS can be efficiently applied to larger boards, though the time and space complexity will increase with the board size.

What does the Boustrophedon layout mean in the Snakes and Ladders problem?

The Boustrophedon layout means that the board is filled row by row, alternating directions between left-to-right and right-to-left, creating a zig-zag pattern for square numbering.

How does GhostInterview help with this specific problem?

GhostInterview provides a step-by-step breakdown of BFS traversal and visual aids to enhance your understanding of the Snakes and Ladders problem.

terminal

Solution

Solution 1: BFS

We can use the Breadth-First Search (BFS) method, starting from the starting point, moving forward 1 to 6 steps each time, and then checking for snakes or ladders. If there are any, move to the destination of the snake or ladder; otherwise, move to the next square.

class Solution:
    def snakesAndLadders(self, board: List[List[int]]) -> int:
        n = len(board)
        q = deque([1])
        vis = {1}
        ans = 0
        m = n * n
        while q:
            for _ in range(len(q)):
                x = q.popleft()
                if x == m:
                    return ans
                for y in range(x + 1, min(x + 6, m) + 1):
                    i, j = divmod(y - 1, n)
                    if i & 1:
                        j = n - j - 1
                    i = n - i - 1
                    z = y if board[i][j] == -1 else board[i][j]
                    if z not in vis:
                        vis.add(z)
                        q.append(z)
            ans += 1
        return -1

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