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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.
1import java.util.*;
2
3class Solution {
4 public int snakesAndLadders(int[][] board) {
5 int n = board.length;
6 int[] flattenBoard = new int[n * n + 1];
7 int index = 1;
8 boolean leftToRight = true;
9 for (int i = n - 1; i >= 0; i--) {
10 if (leftToRight) {
11 for (int j = 0; j < n; j++) {
12 flattenBoard[index++] = board[i][j];
13 }
14 } else {
15 for (int j = n - 1; j >= 0; j--) {
16 flattenBoard[index++] = board[i][j];
17 }
18 }
19 leftToRight = !leftToRight;
20 }
21
22 Queue<int[]> queue = new LinkedList<>();
23 boolean[] visited = new boolean[n * n + 1];
24 queue.offer(new int[]{1, 0});
25 visited[1] = true;
26
27 while (!queue.isEmpty()) {
28 int[] node = queue.poll();
29 int pos = node[0];
30 int moves = node[1];
31 for (int i = 1; i <= 6; i++) {
32 int nextPos = pos + i;
33 if (nextPos <= n * n) {
34 if (flattenBoard[nextPos] != -1) {
35 nextPos = flattenBoard[nextPos];
36 }
37 if (!visited[nextPos]) {
38 if (nextPos == n * n) return moves + 1;
39 visited[nextPos] = true;
40 queue.offer(new int[]{nextPos, moves + 1});
41 }
42 }
43 }
44 }
45
46 return -1;
47 }
48}The Java solution uses BFS to explore each reachable position from the current one while flattening the board. The queue holds each square's position and move count.
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.