Problem Statement
In this challenge, you are given a grid, represented by an m x n matrix, where various symbols mark different entities in the grid:
'.'symbolizes an empty cell where movement is possible.'#'represents a wall that blocks movement.'@'denotes the starting point from where you begin your journey.- Lowercase letters (
atof) symbolize keys that can be collected. - Corresponding uppercase letters (
AtoF) represent locks that can only be traversed if you hold the matching key.
From the starting point '@', your task is to navigate through the grid. You can move in one of the four cardinal directions (up, down, left, or right), one space at a time. However, movement is constricted by walls and locks; the latter requires possessing the respective key to pass. The objective is to collect all available keys in the grid in the fewest possible number of moves. If collecting all keys is not feasible, return -1.
Examples
Example 1
Input:
grid = ["@.a..","###.#","b.A.B"]
Output:
8
Explanation:
Note that the goal is to obtain all the keys not to open all the locks.
Example 2
Input:
grid = ["@..aA","..B#.","....b"]
Output:
6
Example 3
Input:
grid = ["@Aa"]
Output:
-1
Constraints
m == grid.lengthn == grid[i].length1 <= m, n <= 30grid[i][j]is either an English letter,'.','#', or'@'.- There is exactly one
'@'in the grid. - The number of keys in the grid is in the range
[1, 6]. - Each key in the grid is unique.
- Each key in the grid has a matching lock.
Approach and Intuition
Based on the examples provided, we can derive the following approach and intuitions:
Initialization:
- Begin at the starting point marked by
'@'. - Use a queue to keep track of positions to explore, and set the initial position to the starting cell. Include the initial number of moves (which is 0) and the keys collected so far (which initially is an empty set).
- Begin at the starting point marked by
Handling Moves:
- For each position, consider all possible movements (up, down, left, right).
- If a movement leads to a wall
'#', ignore that path. - If it leads to a lock (uppercase letter), check if you possess the corresponding key. If not, ignore that path.
- If it leads to a key (lowercase letter), add it to your collection of keys.
Goal Check:
- Whenever you move to a new position, check if you've collected all keys. The keys needed are determined from the grid by finding all unique lowercase letters present.
Optimization with Breadth-First Search (BFS):
- Utilize BFS to ensure that the shortest path to each key is found. Since BFS explores all nodes at the present depth level before moving on to nodes at the next depth level, it guarantees finding the shortest path to the nearest key, and sequentially to all keys.
Tracking State:
- Each state in the BFS queue should keep track of the current cell's position, the number of moves taken to reach there, and the set of keys collected so far.
Cycle Prevention:
- To prevent cycles and revisiting the same state (same position and same set of keys), maintain a set of visited states.
By implementing these steps, you can efficiently determine the minimum number of moves required to collect all keys in the grid. If after exploring all possible paths and states you have not collected all keys, return -1 as it indicates that some keys are unreachable due to layout constraints (like being surrounded by walls).
Solutions
- Java
java
import java.awt.Point;
import java.util.*;
class Solution {
final int INFINITY = Integer.MAX_VALUE;
String[] maze;
int rows, cols;
Map<Character, Point> charLocations;
int[] rowMove = new int[]{-1, 0, 1, 0};
int[] colMove = new int[]{0, -1, 0, 1};
public int findShortestPathForAllKeys(String[] maze) {
this.maze = maze;
rows = maze.length;
cols = maze[0].length();
charLocations = new HashMap<>();
for (int r = 0; r < rows; ++r)
for (int c = 0; c < cols; ++c) {
char ch = maze[r].charAt(c);
if (ch != '.' && ch != '#')
charLocations.put(ch, new Point(r, c));
}
int allKeysState = (1 << (charLocations.size() / 2)) - 1;
Map<Character, Map<Character, Integer>> distances = new HashMap<>();
for (char place: charLocations.keySet())
distances.put(place, breadthFirstSearch(place));
PriorityQueue<PathNode> priorityQueue = new PriorityQueue<>((a, b) ->
Integer.compare(a.distance, b.distance));
priorityQueue.offer(new PathNode(new SearchNode('@', 0), 0));
Map<SearchNode, Integer> optimalDistance = new HashMap<>();
optimalDistance.put(new SearchNode('@', 0), 0);
while (!priorityQueue.isEmpty()) {
PathNode pathNode = priorityQueue.poll();
SearchNode searchNode = pathNode.searchNode;
int dist = pathNode.distance;
if (optimalDistance.getOrDefault(searchNode, INFINITY) < dist) continue;
if (searchNode.state == allKeysState) return dist;
for (char destination: distances.get(searchNode.place).keySet()) {
int distanceToNext = distances.get(searchNode.place).get(destination);
int nextState = searchNode.state;
if (Character.isLowerCase(destination)) // if it's a key
nextState |= (1 << (destination - 'a'));
if (Character.isUpperCase(destination)) // if it's a lock
if ((searchNode.state & (1 << (destination - 'A'))) == 0) // key isn't available
continue;
if (dist + distanceToNext < optimalDistance.getOrDefault(new SearchNode(destination, nextState), INFINITY)) {
optimalDistance.put(new SearchNode(destination, nextState), dist + distanceToNext);
priorityQueue.offer(new PathNode(new SearchNode(destination, nextState), dist + distanceToNext));
}
}
}
return -1;
}
public Map<Character, Integer> breadthFirstSearch(char source) {
int sourceRow = charLocations.get(source).x;
int sourceCol = charLocations.get(source).y;
boolean[][] visited = new boolean[rows][cols];
visited[sourceRow][sourceCol] = true;
int depth = 0;
Queue<Point> queue = new LinkedList<>();
queue.offer(new Point(sourceRow, sourceCol));
queue.offer(null);
Map<Character, Integer> discoveredDistance = new HashMap<>();
while (!queue.isEmpty()) {
Point point = queue.poll();
if (point == null) {
depth++;
if (!queue.isEmpty())
queue.offer(null);
continue;
}
int r = point.x, c = point.y;
if (maze[r].charAt(c) != source && maze[r].charAt(c) != '.') {
discoveredDistance.put(maze[r].charAt(c), depth);
continue;
}
for (int i = 0; i < 4; ++i) {
int newRow = r + rowMove[i];
int newCol = c + colMove[i];
if (0 <= newRow && newRow < rows && 0 <= newCol && newCol < cols && !visited[newRow][newCol]) {
if (maze[newRow].charAt(newCol) != '#') {
queue.offer(new Point(newRow, newCol));
visited[newRow][newCol] = true;
}
}
}
}
return discoveredDistance;
}
}
class PathNode {
SearchNode searchNode;
int distance;
PathNode(SearchNode node, int dist) {
searchNode = node;
distance = dist;
}
}
class SearchNode {
char place;
int state;
SearchNode(char loc, int st) {
place = loc;
state = st;
}
@Override
public boolean equals(Object obj) {
if (this == obj) return true;
if (!(obj instanceof SearchNode)) return false;
SearchNode other = (SearchNode) obj;
return place == other.place && state == other.state;
}
@Override
public int hashCode() {
return 256 * state + place;
}
}
The provided Java solution tackles the problem of finding the shortest path to collect all keys in a maze represented by a string array. The maze contains obstacles, keys (denoted by lowercase letters), and doors (denoted by uppercase letters), with the starting point represented by '@'. Here's an overview of how the code approaches this complex pathfinding problem:
Initialization and Data Extraction:
- The maze dimensions and character locations are stored.
- All keys and significant points of interest (like the starting point '@') are mapped with their respective coordinates.
Breadth-First Search (BFS) Functionality:
- A dedicated BFS method,
breadthFirstSearch(char source), calculates and returns distances from the source to all reachable targets (other significant points) within the maze. It efficiently navigates through the maze, avoiding obstacles and counting steps to reach various points.
Pathfinding with Priority Queue:
- Utilizes a priority queue to explore paths through the maze from the start position '@'. For each node (or state) in the priority queue:
- If it represents a state where all keys have been collected, the accumulated distance (cost) is returned as the result.
- The code explores transitions to neighboring nodes (reachable points), accounting for newly collected keys or bypassable doors based on the keys already collected.
- The distance to each neighboring node is updated in the
optimalDistancemap if a shorter path is found.
Handling Key-Door Mechanics:
- Keys are requisite for passing through doors. The algorithm bitmasks states to manage the keys collected so far, allowing it to efficiently check and transition through openable doors.
State and Transition Management:
SearchNodeencapsulates each unique state comprising a location and bitmask of obtained keys.PathNodeassists in associating aSearchNodewith its respective cost in terms of path distance.- The solution dynamically updates distances and checks for optimal paths, ensuring that the shortest path will be determined by the time all keys are collected or all possible paths are exhausted.
Efficiency and Completeness:
- The methodology ensures a comprehensive search of all potentially viable paths considering the presence of maze walls, multiple keys, and doors, integrating both spatial and key-collection aspects to compute an optimal solution. Comprehensive use of structures like maps, queues, and priority queues aids in managing the spatial and state complexities of the problem.
This solution works effectively to solve the "Shortest Path to Get All Keys" problem by combining pathfinding algorithms with state management to navigate through mazes with various constraints.
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