You are given an n x n integer matrix board where the cells are labeled from 1 to n2 in a Boustrophedon style 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:
next with a label in the range [curr + 1, min(curr + 6, n2)].
next has a snake or ladder, you must move to the destination of that snake or ladder. Otherwise, you move to next.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 are not the starting points of any snake or ladder.
Note that you only take a snake or ladder at most once per dice roll. 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.
[[-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 dice rolls required to reach the square n2. If it is not possible to reach the square, return -1.
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 Explanation: 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
Constraints:
n == board.length == board[i].length2 <= n <= 20board[i][j] is either -1 or in the range [1, n2].1 and n2 are not the starting points of any snake or ladder.The Snakes and Ladders board can be seen as a graph where each square is a node, and an edge exists between node i and node j if you can move from square i to square j with a dice roll. Use BFS to efficiently explore this graph, ensuring each move accounts for ladders and snakes.
The solution flattens the 2D board to a 1D array while considering the Boustrophedon order. Then, using BFS, it processes nodes (squares) ensuring each is visited based on dice moves accounting for any ladders or snakes directly. It effectively exits on reaching the end square.
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Time Complexity: O(n^2), where n is the board dimension, as each square is processed once.
Space Complexity: O(n^2), for storing the queue and visited status of squares.
Dijkstra's Algorithm typically finds shortest paths in weighted graphs. Adapting it for an unweighted snakes and ladders board can ensure the fewest moves to reach the final square by evaluating and using priority. Each move is computed implicitly based on its dice-distance weight.
The C implementation uses a modified Dijkstra-like approach with direct BFS priority management via a min-heap to identify the least cost number of rolls to navigate the board to its end.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n^2 log(n^2)), given heap operations for prioritized traversal.
Space Complexity: O(n^2) for heap management.
| Approach | Complexity |
|---|---|
| Breadth-First Search (BFS) | Time Complexity: O(n^2), where n is the board dimension, as each square is processed once. Space Complexity: O(n^2), for storing the queue and visited status of squares. |
| Dijkstra's Algorithm Adaptation | Time Complexity: O(n^2 log(n^2)), given heap operations for prioritized traversal. Space Complexity: O(n^2) for heap management. |
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