499. The Maze III
Problem Description
In this problem, we are given a m x n
maze represented by a 2D grid with 0
indicating empty spaces and 1
indicating walls. A ball can roll in the maze in the cardinal directions (up, down, left, or right), but it only stops when it hits a wall. The ball stops moving further in the same direction once it encounters a wall. Additionally, there is a hole in the maze; if the ball goes over the hole, it will fall into it, and that is the target location the ball needs to reach.
The ball's start position and the hole's position are both given. The goal is to find a sequence of directions (as a string) that will lead the ball into the hole using the shortest path possible. If the shortest path is not unique, the lexicographically smallest sequence of directions should be returned. A sequence is considered lexicographically smaller if it has the smallest alphabetical order possible. If there is no possible way for the ball to fall into the hole, the function should return the string "impossible".
The valid directions the ball can roll are denoted by: 'u'
for up, 'd'
for down, 'l'
for left, and 'r'
for right. The distance is measured by the number of empty spaces the ball rolls over from its start to the hole, not counting the starting square but including the final square over the hole.
The maze is framed by walls, so the ball cannot roll beyond the limits of the grid.
Flowchart Walkthrough
First, let's analyze the problem using the Flowchart. Here's a step-by-step walkthrough:
Is it a graph?
- Yes: The maze can be treated as a graph where each cell is a node, and there are edges between adjacent cells.
Is it a tree?
- No: The maze does not form a hierarchy or a tree structure since it has potentially multiple paths and cycles.
Is the problem related to directed acyclic graphs (DAGs)?
- No: In this problem, although the maze has a directed nature (you can think of ball movements as directed edges), it is not acyclic as it is possible to come back to the starting position after making several moves.
Is the problem related to shortest paths?
- Yes: The goal is to find the shortest path to the hole.
Is the graph weighted?
- Yes: The problem does not use constant unit weights as the ball continues to roll until hitting a wall; covering potentially multiple cells, resulting in variable costs to move from one position to next in this way, making this a weighted problem.
We end up at the decision to use "Dijkstra's Algorithm", but when we consider the problem's unique properties like determining all possible paths and their lexicographical order evaluation, we need a modification.
Conclusion: Given the necessity to explore all possible paths, backtrack, and keep track of paths (for lexicographical order), a custom DFS approach or DFS enhanced with priority structures for handling path order and distances might be needed more than just a straightforward application of Dijkstra’s algorithm. Thus our flowchart and a deeper analysis hint at a DFS or a modified DFS being the most effective approach for solving LeetCode 499. The Maze III.
Intuition
The intuition behind the given solution is founded on Breadth-first Search (BFS), a common algorithm for exploring paths in a graph or grid-like structures like a maze.
The BFS works by exploring all potential moves from a given position before moving on to the next set of positions. In this context, the valid moves for the ball are to keep rolling in one of the four cardinal directions until it hits a wall.
We use BFS to explore each possible path the ball can take and keep track of the distance travelled and the path taken for each visited position in the maze:
-
Keep Distance and Path Information: Each position in the maze maintains two pieces of information. One is the minimum number of steps taken to reach that position from the initial ball position. The other is the lexicographically smallest sequence of moves that the ball takes to reach that position.
-
Exploring the Maze: From the current position, we consider all four possible directions the ball can roll. For each direction, we roll the ball until it hits a wall or falls into the hole.
-
Tracking Shortest Paths: When the ball stops, we check if the distance to this new stopped position is less than any previously recorded distance. If it is less, or if the distances are equal but the new path is lexicographically smaller, we update the distance and path for that position.
-
Queue for BFS: We add each new position that we can roll to (except the hole) to a queue to explore in the future. This ensures that all possible moves from each position are considered before moving further away.
-
Final Check: Once the entire maze has been explored, we check the information for the hole's position. If a path was found, it will contain the lexicographically smallest sequence of directions to get there. If no path was found, the original
None
assignment will remain, and we will return "impossible" instead.
The provided solution neatly consolidates this approach using a queue data structure to perform the BFS, a 2D array for storing the distance, and another 2D array for storing the path strings for each position the ball stops at.
Learn more about Depth-First Search, Breadth-First Search, Graph, Shortest Path and Heap (Priority Queue) patterns.
Solution Approach
The implementation uses a BFS algorithm to traverse the maze and search for the shortest path to the hole. Here's the detailed breakdown of the steps and the data structures used:
-
Initialization: We define the maze's dimensions
m
andn
. The starting positionr, c
(row, column) of the ball is extracted from theball
list. Similarly, the hole's positionrh, ch
is extracted from thehole
list. We initialize a queueq
usingdeque
from thecollections
module, which will hold the positions to explore. We also define two 2D lists:dist
to keep track of the distances andpath
to keep track of the paths. -
Distance and Path Setup: The
dist
list is filled withinf
(infinity) to indicate an unknown distance, whilepath
is filled withNone
, except for the starting position, which is set to0
indist
and the empty string''
inpath
. -
BFS Loop: A while loop runs as long as there are positions in the queue
q
. At each iteration, we dequeue a position and examine all four possible movements from that position. The movements are encoded as tuples(a, b, d)
, wherea
andb
are the directional increments, andd
is the corresponding direction character ('u', 'd', 'l', 'r'). -
Valid Movement Checking: Inside this loop, we check if moving in a particular direction will stop the ball at a different position. We simulate the ball's movement by incrementally adding the direction increments
a
andb
to the current positionx, y
until the ball hits a wall or the hole. -
Updating Distances and Paths: When the ball cannot move any further in a direction, we check if this new position is either closer than any previously recorded position or if it's the same distance but with a lexicographically smaller path. If either is true, we update
dist
andpath
. -
Enqueueing Next Positions: If the new stopped position is not the hole, it is added to the queue for further exploration.
-
Returning the Result: Once the BFS completes, the path at the hole's position
path[rh][ch]
represents the shortest path to the hole. If no path was found,path[rh][ch]
will beNone
, and we will return'impossible'
.
The algorithm efficiently finds the shortest path by only updating positions when a better path is found. It ensures that the path followed is both the shortest in terms of distance and lexicographically minimal due to the updates in dist
and path
.
The data structures used — the queue for BFS and 2D arrays for distances and paths — are well-suited for the task. These choices allow the algorithm to queue future positions to visit and maintain necessary state information for each position in the maze.
This breadth-first search ensures that we examine all moves from a starting point before moving outward, thus always finding the shortest path to any reachable position.
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Start EvaluatorExample Walkthrough
Let's consider a small maze, where 0
indicates an empty space, and 1
indicates a wall. We are given the ball's start position and the hole's target position. We will demonstrate the steps of the BFS algorithm to find the shortest path.
For the example, our maze (grid) will look like this:
1 1 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 1 1 1
Let the ball start position be at (1, 1)
and the hole's target position be at (3, 2)
. Note that we are using (row, column)
indexing.
Initial Setup
Start position (S): (1, 1) Hole position (H): (3, 2) Maze: 1 1 1 1 1 S 0 1 1 0 1 1 1 0 H 0 1 1 1 1
We initialize the dist
with inf
(indicating unknown distance) and path
with None
for each cell except the start (1, 1)
, which has a distance of 0
and a path of ''
. We enqueue the start position into our queue q
.
Step-by-Step BFS
-
We dequeue
(1, 1)
and start exploring all four directions. -
From
(1, 1)
, we can roll down(d)
to(3, 1)
, since it is the next wall or hole position vertically. We updatedist[3][1]
to2
andpath[3][1]
to'd'
, and enqueue(3, 1)
. -
Enqueue order:
(3, 1)
. -
Dequeue
(3, 1)
and explore. It can go right(r)
to the hole(3, 2)
. We updatedist[3][2]
to3
andpath[3][2]
to'dr'
. There are no more enqueues because we've reached the hole. -
No more positions to explore in the queue; the BFS ends.
Final Check
We consult the final dist
and path
values at the hole position (3, 2)
and find a distance of 3
and a path of 'dr'
.
Hence, the shortest path for the ball to fall into the hole is dr
(down, then right), with only 3
moves disregarding the starting square.
This example walk-through demonstrates the BFS approach, where we systematically explore all directions from the starting point, updating distances and paths, until we find the shortest lexicographically smallest path to the hole.
Solution Implementation
1from collections import deque
2from math import inf
3
4class Solution:
5 def findShortestWay(self, maze: List[List[int]], ball: List[int], hole: List[int]) -> str:
6 # Dimensions of the maze
7 num_rows, num_cols = len(maze), len(maze[0])
8
9 # Start position
10 start_row, start_col = ball
11
12 # Hole position
13 hole_row, hole_col = hole
14
15 # Queue for the BFS
16 queue = deque([(start_row, start_col)])
17
18 # Distance matrix initialized with infinity
19 distance = [[inf] * num_cols for _ in range(num_rows)]
20 distance[start_row][start_col] = 0
21
22 # Path matrix to store the path taken to reach each cell
23 path = [[""] * num_cols for _ in range(num_rows)]
24
25 # Perform a BFS
26 while queue:
27 current_row, current_col = queue.popleft()
28
29 # Directions: (row_change, col_change, direction_string)
30 for row_change, col_change, direction in [(-1, 0, 'u'), (1, 0, 'd'), (0, -1, 'l'), (0, 1, 'r')]:
31 x, y = current_row, current_col
32 steps = distance[current_row][current_col]
33
34 # Keep rolling in the direction until hitting a wall or the hole
35 while (0 <= x + row_change < num_rows
36 and 0 <= y + col_change < num_cols
37 and maze[x + row_change][y + col_change] == 0
38 and (x, y) != (hole_row, hole_col)):
39 x += row_change
40 y += col_change
41 steps += 1
42
43 # If the distance to the new position is less, or if the distance is the same,
44 # but the path is lexicographically smaller, update the distance and path.
45 if distance[x][y] > steps or \
46 (distance[x][y] == steps and path[current_row][current_col] + direction < path[x][y]):
47 distance[x][y] = steps
48 path[x][y] = path[current_row][current_col] + direction
49
50 # If the new position is not the hole, add it to the queue.
51 if (x, y) != (hole_row, hole_col):
52 queue.append((x, y))
53
54 # If no path was found to the hole, return "impossible".
55 return path[hole_row][hole_col] or 'impossible'
56
1class Solution {
2 public String findShortestWay(int[][] maze, int[] ball, int[] hole) {
3 int rows = maze.length; // Number of rows in the maze
4 int cols = maze[0].length; // Number of columns in the maze
5 int ballRow = ball[0], ballCol = ball[1]; // Starting point (ball)
6 int holeRow = hole[0], holeCol = hole[1]; // Ending point (hole)
7
8 // Queue for BFS search
9 Deque<int[]> queue = new LinkedList<>();
10 queue.offer(new int[]{ballRow, ballCol});
11
12 // Initialize distance array with max values
13 int[][] distance = new int[rows][cols];
14 for (int i = 0; i < rows; ++i) {
15 Arrays.fill(distance[i], Integer.MAX_VALUE);
16 }
17 distance[ballRow][ballCol] = 0; // Starting point distance is 0
18
19 // Path strings for each coordinate
20 String[][] path = new String[rows][cols];
21 path[ballRow][ballCol] = ""; // Empty path for the starting point
22
23 // Directions array with row and column movements (up, down, left, right)
24 int[][] directions = {
25 {-1, 0, 'u'},
26 {1, 0, 'd'},
27 {0, -1, 'l'},
28 {0, 1, 'r'}
29 };
30
31 // BFS search loop
32 while (!queue.isEmpty()) {
33 int[] point = queue.poll();
34 int i = point[0], j = point[1];
35
36 // Iterate through possible directions
37 for (int[] direction : directions) {
38 int vertical = direction[0];
39 int horizontal = direction[1];
40 String directionChar = String.valueOf((char) (direction[2]));
41 int x = i, y = j;
42 int steps = distance[i][j];
43
44 // Roll in the current direction until hitting a wall or the hole
45 while (x + vertical >= 0 && x + vertical < rows && y + horizontal >= 0 && y + horizontal < cols &&
46 maze[x + vertical][y + horizontal] == 0 && !(x == holeRow && y == holeCol)) {
47 x += vertical;
48 y += horizontal;
49 ++steps;
50 }
51
52 // Check if a shorter path to (x, y) is found or a lexicographically smaller path of the same distance
53 if (distance[x][y] > steps ||
54 (distance[x][y] == steps && (path[i][j] + directionChar).compareTo(path[x][y]) < 0)) {
55 distance[x][y] = steps;
56 path[x][y] = path[i][j] + directionChar;
57
58 // If we have not reached the hole, add the new point to the queue
59 if (x != holeRow || y != holeCol) {
60 queue.offer(new int[]{x, y});
61 }
62 }
63 }
64 }
65
66 // Return the path to the hole or "impossible" if no path is found
67 return path[holeRow][holeCol] == null ? "impossible" : path[holeRow][holeCol];
68 }
69}
70
1#include <vector>
2#include <queue>
3#include <string>
4#include <climits>
5
6using namespace std;
7
8class Solution {
9public:
10 // Method to find the shortest path for the ball to fall into the hole in the maze
11 string findShortestWay(vector<vector<int>>& maze, vector<int>& ball, vector<int>& hole) {
12 int rows = maze.size(); // Number of rows in the maze
13 int cols = maze[0].size(); // Number of columns in the maze
14 int ballRow = ball[0]; // Ball's starting row
15 int ballCol = ball[1]; // Ball's starting column
16 int holeRow = hole[0]; // Hole's row position
17 int holeCol = hole[1]; // Hole's column position
18
19 // Queue for BFS, storing pairs of (row, column)
20 queue<pair<int, int>> queue;
21 queue.push({ballRow, ballCol});
22
23 // Distance matrix, initialized to INT_MAX
24 vector<vector<int>> distance(rows, vector<int>(cols, INT_MAX));
25 distance[ballRow][ballCol] = 0;
26
27 // Path matrix, storing the path to reach each cell
28 vector<vector<string>> path(rows, vector<string>(cols, ""));
29
30 // Possible directions to move in the maze with corresponding direction characters
31 vector<vector<int>> directions = {{-1, 0, 'u'}, {1, 0, 'd'}, {0, -1, 'l'}, {0, 1, 'r'}};
32
33 // BFS to find the shortest path
34 while (!queue.empty()) {
35 auto current = queue.front();
36 queue.pop();
37 int currentRow = current.first;
38 int currentCol = current.second;
39
40 // Explore each direction
41 for (auto& dir : directions) {
42 int dRow = dir[0];
43 int dCol = dir[1];
44 char directionChar = (char)dir[2];
45 int nextRow = currentRow;
46 int nextCol = currentCol;
47 int steps = distance[currentRow][currentCol];
48
49 // Roll the ball until it hits a wall or the hole
50 while (nextRow + dRow >= 0 && nextRow + dRow < rows &&
51 nextCol + dCol >= 0 && nextCol + dCol < cols &&
52 maze[nextRow + dRow][nextCol + dCol] == 0 &&
53 (nextRow != holeRow || nextCol != holeCol)) {
54 nextRow += dRow;
55 nextCol += dCol;
56 ++steps;
57 }
58
59 // Update distance and path if a shorter path is found
60 if (distance[nextRow][nextCol] > steps ||
61 (distance[nextRow][nextCol] == steps && (path[currentRow][currentCol] + directionChar < path[nextRow][nextCol]))) {
62 distance[nextRow][nextCol] = steps;
63 path[nextRow][nextCol] = path[currentRow][currentCol] + directionChar;
64 if (nextRow != holeRow || nextCol != holeCol) {
65 queue.push({nextRow, nextCol});
66 }
67 }
68 }
69 }
70
71 return path[holeRow][holeCol].empty() ? "impossible" : path[holeRow][holeCol];
72 }
73};
74
1// Type definition for a point in the maze represented by row and column
2type Point = { row: number; col: number };
3
4// Function to find the shortest path for the ball to fall into the hole in the maze using BFS
5function findShortestWay(maze: number[][], ball: [number, number], hole: [number, number]): string {
6 const rows = maze.length; // Number of rows in the maze
7 const cols = maze[0].length; // Number of columns in the maze
8 const ballPoint: Point = { row: ball[0], col: ball[1] }; // Starting point of the ball
9 const holePoint: Point = { row: hole[0], col: hole[1] }; // Location of the hole
10
11 // Queue for BFS initialized with starting ball point
12 const queue: Point[] = [ballPoint];
13
14 // Distance matrix initialized to Infinity
15 const distance: number[][] = Array.from({ length: rows }, () => Array(cols).fill(Infinity));
16 distance[ballPoint.row][ballPoint.col] = 0;
17
18 // Path matrix, storing the path to reach each cell
19 const path: string[][] = Array.from({ length: rows }, () => Array(cols).fill(""));
20
21 // Possible directions to move in the maze with corresponding direction characters
22 const directions = [{ dr: -1, dc: 0, dcChar: 'u' }, { dr: 1, dc: 0, dcChar: 'd' }, { dr: 0, dc: -1, dcChar: 'l' }, { dr: 0, dc: 1, dcChar: 'r' }];
23
24 // BFS to find the shortest path
25 while (queue.length > 0) {
26 const current = queue.shift()!;
27 const currentRow = current.row;
28 const currentCol = current.col;
29
30 // Explore each direction
31 for (const { dr, dc, dcChar } of directions) {
32 let nextRow = currentRow;
33 let nextCol = currentCol;
34 let steps = distance[currentRow][currentCol];
35
36 // Roll the ball until it hits a wall or the hole
37 while (
38 nextRow + dr >= 0 &&
39 nextRow + dr < rows &&
40 nextCol + dc >= 0 &&
41 nextCol + dc < cols &&
42 maze[nextRow + dr][nextCol + dc] === 0 &&
43 (nextRow !== holePoint.row || nextCol !== holePoint.col)
44 ) {
45 nextRow += dr;
46 nextCol += dc;
47 steps++;
48 }
49
50 // Update distance and path if a shorter path is found
51 if (
52 distance[nextRow][nextCol] > steps ||
53 (
54 distance[nextRow][nextCol] === steps &&
55 (path[currentRow][currentCol] + dcChar) < path[nextRow][nextCol]
56 )
57 ) {
58 distance[nextRow][nextCol] = steps;
59 path[nextRow][nextCol] = path[currentRow][currentCol] + dcChar;
60 if (nextRow !== holePoint.row || nextCol !== holePoint.col) {
61 queue.push({ row: nextRow, col: nextCol });
62 }
63 }
64 }
65 }
66
67 return path[holePoint.row][holePoint.col] || "impossible";
68}
69
70// Variables and function can be called or used from here onwards as global variables
71
Time and Space Complexity
Time Complexity
The given code implements a breadth-first search (BFS) algorithm on a matrix to find the shortest path for a ball to roll from its starting position to the hole. Each move consists of the ball rolling in a straight line until it hits a wall or falls into the hole.
The time complexity of the BFS algorithm is generally O(V + E), where V is the number of vertices and E is the number of edges in the graph. However, for a two-dimensional matrix, we should consider the following:
- Each cell in the matrix can be visited multiple times because the ball can reach the same cell from different directions. There are
m * n
cells, wherem
is the number of rows andn
is the number of columns. - For each cell
(i, j)
, the ball can roll in four directions until it hits a wall or a boundary of the maze. Therefore, in the worst-case scenario, it might traverse the entire length or width of the maze. - On average, the ball rolls
m/2
orn/2
steps before hitting a wall or the boundary in either direction.
Taking these into consideration, the worst-case time complexity can be approximated as:
O((m * n) * (m + n))
This is because, for each of the m * n
possible starting points, the ball may have to traverse an entire row or column (m + n
steps in the worst case).
Space Complexity
The space complexity of the BFS algorithm depends on the size of the queue used, as well as the additional space for the dist
and path
arrays:
- The queue can grow up to
m * n
in size, as it contains positions representing the cells visited by the ball. - The
dist
andpath
arrays are both sizedm * n
.
Hence, the overall space complexity is:
O(m * n + m * n) = O(2 * m * n) = O(m * n)
Overall, both dist
and path
arrays contribute to the space complexity equally, but constants are dropped in Big O notation, resulting in a total space complexity of O(m * n).
Learn more about how to find time and space complexity quickly using problem constraints.
How many ways can you arrange the three letters A, B and C?
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