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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.
1var snakesAndLadders = function(board) {
2 const n = board.length;
3 function getBoardValue(number) {
4 const q = Math.floor((number - 1) / n), r = (number - 1) % n;
5 const row = n - 1 - q, col = (q % 2 === 0) ? r : (n - 1 - r);
6 return board[row][col];
7 }
8 const visited = new Array(n * n + 1).fill(false);
9 const queue = [[1, 0]];
10 visited[1] = true;
11
12 while (queue.length) {
13 const [pos, moves] = queue.shift();
14 for (let i = 1; i <= 6; i++) {
15 let nextPos = pos + i;
16 if (nextPos > n * n) continue;
17 const value = getBoardValue(nextPos);
18 if (value !== -1) nextPos = value;
19
20 if (nextPos === n * n) return moves + 1;
21 if (!visited[nextPos]) {
22 visited[nextPos] = true;
23 queue.push([nextPos, moves + 1]);
24 }
25 }
26 }
27 return -1;
28};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.
1
JavaScript's solution leverages a priority queue interfaced via a minimum heap, efficiently matching Dijkstra's logic style for shortest path discernment.