Sponsored
Sponsored
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.
1from collections import deque
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 visited = set()
14 queue = deque([(1, 0)])
15 visited.add(1)
16
17 while queue:
18 position, moves = queue.popleft()
19 for i in range(1, 7):
20 next_position = position + i
21 if next_position > n * n:
22 continue
23
24 board_value = get_board_value(next_position)
25 if board_value != -1:
26 next_position = board_value
27
28 if next_position == n * n:
29 return moves + 1
30
31 if next_position not in visited:
32 visited.add(next_position)
33 queue.append((next_position, moves + 1))
34
35 return -1The Python solution uses BFS for exploring possible moves, ensuring all reachable squares account for snakes and ladders. A helper function maps board squares to positions in a flattened manner.
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.
1
Employing Python's native min-heap via heapq, the solution adapts shortest path logic with move cost checks and prioritization of paths.