The Maze III is a pathfinding problem where you are given a 2D grid representing a maze, a ball starting position, and a hole (target) position. The maze consists of empty spaces (represented by 0) and walls (represented by 1). The ball can roll in four directions: up, down, left, or right. However, once the ball starts rolling in a direction, it continues until it hits a wall or falls into the hole. The ball cannot stop before hitting a wall or the hole.
The goal is to find the shortest path for the ball to fall into the hole, and if there are multiple paths of the same shortest length, return the lexicographically smallest path in terms of a string made of 'u', 'd', 'l', and 'r' for up, down, left, and right, respectively. If the ball cannot reach the hole, return "impossible".
maze (2D list), ball (start position), hole (target position).Initially, it might seem like a simple shortest-path problem, similar to Breadth-First Search (BFS) in a grid. However, the twist is that the ball keeps rolling until it hits a wall or the hole, and cannot stop arbitrarily. This means we must simulate the rolling motion for each direction, not just move one cell at a time.
A brute-force approach would be to try every possible path, but this quickly becomes infeasible due to the exponential number of possible paths and the rolling constraint. Instead, we need an optimized way to find the shortest path while also keeping track of the lexicographically smallest path in case of ties.
The problem is well-suited for a variant of Dijkstra's algorithm, where we track both the minimum distance and the path string for each position. We use a priority queue to always expand the path with the smallest distance and, for ties, the smallest lexicographical path.
To solve this problem efficiently, we use Dijkstra's algorithm with a priority queue. This ensures we always process the most promising state first and find the shortest and lexicographically smallest path.
We use a hash map (or 2D array) to efficiently look up the best (distance, path) for each cell. This keeps the algorithm efficient and avoids redundant processing.
Let's consider the following maze:
maze = [
[0, 0, 1, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 1, 0],
[1, 1, 0, 1, 1],
[0, 0, 0, 0, 0]
]
ball = [0, 4]
hole = [4, 4]
The process ensures we always find the shortest and lexicographically smallest path.
Brute-force Approach: The brute-force approach would try all possible paths, leading to exponential time complexity: O(4n*m), where n and m are the maze dimensions.
Optimized (Dijkstra's) Approach:
The Maze III problem requires simulating the rolling behavior of a ball in a maze and finding the shortest path to a hole, with a tie-breaker for lexicographically smallest paths. By modeling the problem as a graph and using Dijkstra's algorithm with a priority queue, we efficiently find the optimal solution. The key insight is to always roll until hitting a wall or the hole, and to track both distance and path string to ensure correctness and optimality.
import heapq
class Solution:
def findShortestWay(self, maze, ball, hole):
m, n = len(maze), len(maze[0])
dirs = [(-1,0,'u'), (1,0,'d'), (0,-1,'l'), (0,1,'r')]
heap = [(0, "", ball[0], ball[1])]
visited = dict()
while heap:
dist, path, x, y = heapq.heappop(heap)
if (x, y) == (hole[0], hole[1]):
return path
if (x, y) in visited and (visited[(x, y)][0] < dist or (visited[(x, y)][0]==dist and visited[(x, y)][1] <= path)):
continue
visited[(x, y)] = (dist, path)
for dx, dy, dchar in dirs:
nx, ny, steps = x, y, 0
# Roll until hit wall or hole
while 0 <= nx+dx < m and 0 <= ny+dy < n and maze[nx+dx][ny+dy] == 0:
nx += dx
ny += dy
steps += 1
if [nx, ny] == hole:
break
if (nx, ny) not in visited or dist+steps < visited[(nx, ny)][0] or (dist+steps == visited[(nx, ny)][0] and path+dchar < visited[(nx, ny)][1]):
heapq.heappush(heap, (dist+steps, path+dchar, nx, ny))
return "impossible"
#include <vector>
#include <string>
#include <queue>
#include <tuple>
#include <unordered_map>
using namespace std;
class Solution {
public:
string findShortestWay(vector<vector<int>>& maze, vector<int>& ball, vector<int>& hole) {
int m = maze.size(), n = maze[0].size();
vector<vector<pair<int, string>>> visited(m, vector<pair<int, string>>(n, {INT_MAX, ""}));
vector<tuple<int, int, int, string>> dirs = {{-1,0, 'u'}, {1,0, 'd'}, {0,-1, 'l'}, {0,1, 'r'}};
priority_queue<tuple<int, string, int, int>, vector<tuple<int, string, int, int>>, greater<>> pq;
pq.emplace(0, "", ball[0], ball[1]);
while (!pq.empty()) {
auto [dist, path, x, y] = pq.top(); pq.pop();
if (x == hole[0] && y == hole[1]) return path;
if (visited[x][y].first < dist || (visited[x][y].first == dist && visited[x][y].second <= path)) continue;
visited[x][y] = {dist, path};
for (auto& [dx, dy, dchar] : dirs) {
int nx = x, ny = y, steps = 0;
while (nx+dx >= 0 && nx+dx < m && ny+dy >= 0 && ny+dy < n && maze[nx+dx][ny+dy] == 0) {
nx += dx; ny += dy; steps++;
if (nx == hole[0] && ny == hole[1]) break;
}
if (visited[nx][ny].first > dist+steps || (visited[nx][ny].first == dist+steps && path + (char)dchar < visited[nx][ny].second)) {
pq.emplace(dist+steps, path + (char)dchar, nx, ny);
}
}
}
return "impossible";
}
};
import java.util.*;
class Solution {
public String findShortestWay(int[][] maze, int[] ball, int[] hole) {
int m = maze.length, n = maze[0].length;
String[][] visited = new String[m][n];
int[][] dist = new int[m][n];
for (int[] d : dist) Arrays.fill(d, Integer.MAX_VALUE);
PriorityQueue<State> pq = new PriorityQueue<>();
pq.offer(new State(ball[0], ball[1], 0, ""));
int[][] dirs = {{-1,0},{1,0},{0,-1},{0,1}};
char[] dchar = {'u','d','l','r'};
while (!pq.isEmpty()) {
State cur = pq.poll();
int x = cur.x, y = cur.y;
if (x == hole[0] && y == hole[1]) return cur.path;
if (dist[x][y] < cur.dist || (dist[x][y] == cur.dist && visited[x][y] != null && visited[x][y].compareTo(cur.path) <= 0)) continue;
dist[x][y] = cur.dist;
visited[x][y] = cur.path;
for (int d = 0; d < 4; d++) {
int nx = x, ny = y, steps = 0;
while (nx+dirs[d][0] >= 0 && nx+dirs[d][0] < m && ny+dirs[d][1] >= 0 && ny+dirs[d][1] < n && maze[nx+dirs[d][0]][ny+dirs[d][1]] == 0) {
nx += dirs[d][0];
ny += dirs[d][1];
steps++;
if (nx == hole[0] && ny == hole[1]) break;
}
if (dist[nx][ny] > cur.dist+steps || (dist[nx][ny] == cur.dist+steps && (visited[nx][ny]==null || cur.path+dchar[d].compareTo(visited[nx][ny]) < 0))) {
pq.offer(new State(nx, ny, cur.dist+steps, cur.path + dchar[d]));
}
}
}
return "impossible";
}
class State implements Comparable<State> {
int x, y, dist;
String path;
State(int x, int y, int dist, String path) {
this.x = x; this.y = y; this.dist = dist; this.path = path;
}
public int compareTo(State other) {
if (this.dist != other.dist) return this.dist - other.dist;
return this.path.compareTo(other.path);
}
}
}
class PriorityQueue {
constructor() { this.data = []; }
enqueue(item) {
this.data.push(item);
this.data.sort((a, b) => a[0] - b[0] || a[1].localeCompare(b[1]));
}
dequeue() { return this.data.shift(); }
isEmpty() { return this.data.length === 0; }
}
var findShortestWay = function(maze, ball, hole) {
let m = maze.length, n = maze[0].length;
let dirs = [[-1,0,'u'], [1,0,'d'], [0,-1,'l'], [0,1,'r']];
let visited = Array.from({length: m}, () => Array(n).fill(null));
let dist = Array.from({length: m}, () => Array(n).fill(Infinity));
let pq = new PriorityQueue();
pq.enqueue([0, "", ball[0], ball[1]]);
while (!pq.isEmpty()) {
let [d, path, x, y] = pq.dequeue();
if (x === hole[0] && y === hole[1]) return path;
if (dist[x][y] < d || (dist[x][y] === d && visited[x][y] !== null && visited[x][y] <= path)) continue;
dist[x][y] = d;
visited[x][y] = path;
for (let [dx, dy, dchar] of dirs) {
let nx = x, ny = y, steps = 0;
while (nx+dx >= 0 && nx+dx < m && ny+dy >= 0 && ny+dy < n && maze[nx+dx][ny+dy] === 0) {
nx += dx;
ny += dy;
steps++;
if (nx === hole[0] && ny === hole[1]) break;
}
if (dist[nx][ny] > d+steps || (dist[nx][ny] === d+steps && (visited[nx][ny] === null || path+dchar < visited[nx][ny]))) {
pq.enqueue([d+steps, path+dchar, nx, ny]);
}
}
}
return "impossible";
};