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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.
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.
1
The JavaScript solution applies BFS using a queue to navigate the board squares while considering snakes and ladders using helper functions for board position calculations.
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.
Time Complexity: O(n^2 log(n^2)), given heap operations for prioritized traversal.
Space Complexity: O(n^2) for heap management.
1import heapq
2
3class Solution:
4 def snakesAndLadders(self, board):
5 n = len(board)
6 def get_board_value(num):
7 r, c = divmod(num-1, n)
8 if r % 2 == 0:
9 return board[n - 1 - r][c]
10 else:
11 return board[n - 1 - r][n - 1 - c]
12
13 dist = {i: float('inf') for i in range(1, n*n + 1)}
14 dist[1] = 0
15 pq = [(0, 1)]
16
17 while pq:
18 current_dist, u = heapq.heappop(pq)
19 if u == n * n:
20 return current_dist
21 for dice in range(1, 7):
22 v = u + dice
23 if v <= n * n:
24 board_value = get_board_value(v)
25 if board_value != -1:
26 v = board_value
27 if current_dist + 1 < dist[v]:
28 dist[v] = current_dist + 1
29 heapq.heappush(pq, (dist[v], v))
30
31 return -1Employing Python's native min-heap via heapq, the solution adapts shortest path logic with move cost checks and prioritization of paths.