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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#include <stdio.h>
2#include <stdlib.h>
3#include <stdbool.h>
4#include <string.h>
5
6typedef struct {
7 int position;
8 int moves;
9} QueueNode;
10
11typedef struct {
12 QueueNode* nodes;
13 int front;
14 int rear;
15 int capacity;
16} Queue;
17
18Queue* createQueue(int capacity) {
19 Queue* queue = (Queue*)malloc(sizeof(Queue));
20 queue->capacity = capacity;
21 queue->front = queue->rear = 0;
22 queue->nodes = (QueueNode*)malloc(capacity * sizeof(QueueNode));
23 return queue;
24}
25
26bool isQueueEmpty(Queue* queue) {
27 return queue->front == queue->rear;
28}
29
30void enqueue(Queue* queue, int position, int moves) {
31 queue->nodes[queue->rear++] = (QueueNode){position, moves};
32}
33
34QueueNode dequeue(Queue* queue) {
35 return queue->nodes[queue->front++];
36}
37
38int snakesAndLadders(int** board, int boardSize, int* boardColSize) {
39 int n = boardSize;
40 int *flattenBoard = (int *)malloc((n * n + 1) * sizeof(int));
41 int index = 1;
42 bool flip = false;
43 for (int i = n - 1; i >= 0; i--) {
44 if (flip) {
45 for (int j = n - 1; j >= 0; j--) {
46 flattenBoard[index++] = board[i][j];
47 }
48 } else {
49 for (int j = 0; j < n; j++) {
50 flattenBoard[index++] = board[i][j];
51 }
52 }
53 flip = !flip;
54 }
55
56 bool *visited = (bool *)calloc(n * n + 1, sizeof(bool));
57 Queue* queue = createQueue(n * n);
58 enqueue(queue, 1, 0);
59 visited[1] = true;
60
61 while (!isQueueEmpty(queue)) {
62 QueueNode node = dequeue(queue);
63 int pos = node.position;
64 int moves = node.moves;
65
66 for (int i = 1; i <= 6; i++) {
67 int nextPos = pos + i;
68 if (nextPos <= n * n) {
69 if (flattenBoard[nextPos] != -1) {
70 nextPos = flattenBoard[nextPos];
71 }
72 if (!visited[nextPos]) {
73 if (nextPos == n * n) {
74 free(flattenBoard);
75 free(visited);
76 free(queue->nodes);
77 free(queue);
78 return moves + 1;
79 }
80 visited[nextPos] = true;
81 enqueue(queue, nextPos, moves + 1);
82 }
83 }
84 }
85 }
86 free(flattenBoard);
87 free(visited);
88 free(queue->nodes);
89 free(queue);
90 return -1;
91}
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.
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.
1var
JavaScript's solution leverages a priority queue interfaced via a minimum heap, efficiently matching Dijkstra's logic style for shortest path discernment.