1778. Shortest Path in a Hidden Grid
Problem Description
In this interactive problem, you are tasked with finding the minimum distance for a hidden robot to reach a target cell in an unknown grid. The grid is represented by m x n cells, where each cell can be empty or blocked. The exact positions of the robot's starting cell and the target cell are not known, nor are the dimensions of the grid. The only tool at your disposal is the GridMaster object, which offers three methods to interact with the robot:
canMove(char direction): Checks if the robot can move in a given direction ('U'for up,'D'for down,'L'for left,'R'for right) and returnstrueif it's possible,falseotherwise.move(char direction): Moves the robot in the specified direction. If the robot attempts to move into a blocked cell or off the grid, the move won't happen โ the robot remains in place.isTarget(): Determines if the robot is currently on the target cell, returningtrueif it is,falseotherwise.
Your goal is to return the minimum distance from the robot's start cell to the target cell or -1 if there isn't a valid path. This is conducted without knowledge of the grid layout or locations of the points of interest within it.
Intuition
To solve this problem, we must explore the grid to find the target cell while keeping track of where we've been. This process involves performing a Depth-First Search (DFS) first to explore the grid and locate the target cell. We employ the canMove and move methods of GridMaster to navigate the robot within the grid.
The DFS approach helps map out the accessible area within the grid by creating a set s to record explored cells as coordinates (i, j). For each direction the robot can move to, we recursively explore further until we hit a blocked cell or the grid's boundaries. Each valid move is tracked by adding the new coordinates to the s set. When the target is found during the DFS, we store its location for later use.
After completing the DFS and mapping all accessible areas (and hopefully finding the target cell), we use Breadth-First Search (BFS) to find the shortest path from the start to the target. BFS is initiated from the starting cell (0, 0), and we proceed level by level, adding neighboring cells to a queue if they have been marked as visited during the DFS phase (and thus, are part of the accessible area).
At each step of the BFS, we remove the current cell from the set s so that it is not revisited, and we increment a counter ans that keeps track of the distance we've covered. If we reach the target during the BFS, the current value of ans represents the minimum distance needed, and we return it. If BFS concludes without reaching the target, we return -1, indicating no valid path exists.
The combination of DFS for exploration and BFS for shortest path calculation is a powerful approach for this type of problem where the environment is initially unknown and needs to be explored systematically.
Learn more about Depth-First Search, Breadth-First Search and Graph patterns.
What is the best way of checking if an element exists in an unsorted array once in terms of time complexity? Select the best that applies.
Solution Approach
The implementation of the solution involves two main algorithms: Depth-First Search (DFS) and Breadth-First Search (BFS). Here's a walkthrough of how these algorithms are used in the code, and the data structures that support them:
Depth-First Search (DFS)
DFS is implemented through the recursive function dfs(i, j). It's initiated with the starting coordinates (0, 0). Here's the step-by-step process:
- Base Case: If the
GridMasterreports that the robot is at the target (isTarget()returnstrue), we record the target's location in the variabletarget. - Exploration: For each of the four possible directions (
'U','D','L','R'), the algorithm checks if the robot can move in that direction usingcanMove(dir). - Recording State: If the move is possible and the new coordinates
(x, y)have not been visited before, we add the coordinates to the setsto mark as visited. - Move and Recur: We move the robot with
move(dir)and recursively calldfs(x, y)to explore the new cell. - Backtrack: After the recursive call, we ensure the robot moves back to the previous position with
move(ndir)wherendiris the opposite of the direction we moved in. - Target Check: During DFS, once the target is found and recorded, we stop searching in that movement path since it's unnecessary to explore further.
The dirs variable is a list of lists, each containing a direction to move in, the opposite direction, and the corresponding change in the i and j coordinates.
Breadth-First Search (BFS)
Once the DFS completes, we use BFS to find the shortest path from the start (0, 0) to target. BFS is done iteratively, and we make use of a queue q and a distance counter ans initialized to -1.
- Initialization: Add the starting cell
(0, 0)to the queueq. - Looping: BFS runs in a loop until the queue is empty, iterating through cells level by level.
- Distance Counting: Increase
ansby one at the beginning of each level to represent the move to the next distance from the start. - Processing Cells:
- Dequeue: For each cell in the current level of BFS (
q.popleft()), we retrieve and remove the front of the queue. - Target Check: If we're at the target cell, the current
ansvalue is the minimum distance, and we return it. - Enqueue Neighbors: For valid neighboring cells that are in the set
s, we remove them froms(to indicate they're visited) and enqueue them to be processed in the next level of BFS.
- Dequeue: For each cell in the current level of BFS (
Using both DFS and BFS in unison allows us to effectively navigate and find the shortest path in an unseen grid. DFS helps us explore the maze and identifies which cells are reachable, including locating the target cell, while BFS then computes the shortest path to the target from the starting cell utilizing the information gathered by DFS.
What's the output of running the following function using input 56?
1KEYBOARD = {
2 '2': 'abc',
3 '3': 'def',
4 '4': 'ghi',
5 '5': 'jkl',
6 '6': 'mno',
7 '7': 'pqrs',
8 '8': 'tuv',
9 '9': 'wxyz',
10}
11
12def letter_combinations_of_phone_number(digits):
13 def dfs(path, res):
14 if len(path) == len(digits):
15 res.append(''.join(path))
16 return
17
18 next_number = digits[len(path)]
19 for letter in KEYBOARD[next_number]:
20 path.append(letter)
21 dfs(path, res)
22 path.pop()
23
24 res = []
25 dfs([], res)
26 return res
271private static final Map<Character, char[]> KEYBOARD = Map.of(
2 '2', "abc".toCharArray(),
3 '3', "def".toCharArray(),
4 '4', "ghi".toCharArray(),
5 '5', "jkl".toCharArray(),
6 '6', "mno".toCharArray(),
7 '7', "pqrs".toCharArray(),
8 '8', "tuv".toCharArray(),
9 '9', "wxyz".toCharArray()
10);
11
12public static List<String> letterCombinationsOfPhoneNumber(String digits) {
13 List<String> res = new ArrayList<>();
14 dfs(new StringBuilder(), res, digits.toCharArray());
15 return res;
16}
17
18private static void dfs(StringBuilder path, List<String> res, char[] digits) {
19 if (path.length() == digits.length) {
20 res.add(path.toString());
21 return;
22 }
23 char next_digit = digits[path.length()];
24 for (char letter : KEYBOARD.get(next_digit)) {
25 path.append(letter);
26 dfs(path, res, digits);
27 path.deleteCharAt(path.length() - 1);
28 }
29}
301const KEYBOARD = {
2 '2': 'abc',
3 '3': 'def',
4 '4': 'ghi',
5 '5': 'jkl',
6 '6': 'mno',
7 '7': 'pqrs',
8 '8': 'tuv',
9 '9': 'wxyz',
10}
11
12function letter_combinations_of_phone_number(digits) {
13 let res = [];
14 dfs(digits, [], res);
15 return res;
16}
17
18function dfs(digits, path, res) {
19 if (path.length === digits.length) {
20 res.push(path.join(''));
21 return;
22 }
23 let next_number = digits.charAt(path.length);
24 for (let letter of KEYBOARD[next_number]) {
25 path.push(letter);
26 dfs(digits, path, res);
27 path.pop();
28 }
29}
30Example Walkthrough
Imagine we have a 2 x 3 grid where the robot starts at position (0, 0) - the top left corner - and the target is located at position (1, 2) - the bottom right corner. The center cell (0, 1) is blocked. We do not know the layout beforehand, so we follow the solution approach.
Depth-First Search (DFS)
- The robot starts at
(0, 0). We mark it as visited(0, 0)in sets. - The robot checks if it can move right with
canMove('R'). It can, so it moves to(0, 1)withmove('R'). - On
(0, 1),canMove('R')is true. However, after moving the robot, we find thatisTarget()is false, and the cell is blocked. We backtrack withmove('L')to(0, 0). - Next, we try moving down with
canMove('D')and it's possible. We move to(1, 0)withmove('D'). - We continue trying to move right until the robot reaches
(1, 2)with successivemove('R'). Upon reaching(1, 2),isTarget()returnstrue. The target's coordinates are logged.
Backtracking
After each move, we ensure to backtrack to the previous cell to continue the DFS correctly.
Breadth-First Search (BFS)
After the DFS phase, we proceed with the BFS. Let's assume we haven't found the target during the DFS, and we are ignoring the found target to explain the BFS process.
- Initiate a queue
qand add the starting cell(0, 0)to it, with an initial distanceansset to 0. - We enter a loop. The starting cell
(0, 0)is dequeued and its neighbors are checked. - Cell
(1, 0)is a neighbor and is in sets, so it is enqueued and removed froms. - Increase
ansto 1 as we're moving to the next level. - Next, we dequeue
(1, 0)and enqueue its unvisited neighbor(1, 1), and remove it froms. - The process repeats at each level. Increase
ansto 2, dequeue(1, 1)and enqueue(1, 2), which is the target location. - Once at
(1, 2), the target is reached.ansis at 2, representing the minimum distance from the start to the target.
In practice, the DFS would have already located the target so the BFS would simply validate the shortest distance.
This example showcases how the strategy of DFS for exploration combined with BFS for finding the shortest distance works even when the grid layout is initially unknown.
Solution Implementation
1from collections import deque # deque is used for efficient queue operations
2
3class Solution:
4 def findShortestPath(self, master: 'GridMaster') -> int:
5 # Use depth-first search to explore the grid
6 def dfs(x, y):
7 nonlocal target # To modify the target variable outside of the nested function
8 if master.isTarget(): # Check if current position is the target
9 target = (x, y)
10 for direction, opposite_direction, delta_x, delta_y in DIRECTIONS:
11 next_x, next_y = x + delta_x, y + delta_y
12 if master.canMove(direction) and (next_x, next_y) not in visited:
13 visited.add((next_x, next_y)) # Mark the cell as visited
14 master.move(direction) # Move to the next cell
15 dfs(next_x, next_y) # Perform DFS from the next cell
16 master.move(opposite_direction) # Move back
17
18 # Define possible movement directions and their opposites
19 DIRECTIONS = [
20 ('U', 'D', -1, 0),
21 ('D', 'U', 1, 0),
22 ('L', 'R', 0, -1),
23 ('R', 'L', 0, 1),
24 ]
25
26 target = None # To keep track of the target's position
27 visited = set() # To keep track of visited cells
28 dfs(0, 0) # Start DFS from the origin (0, 0)
29
30 if target is None: # If target was not found during DFS
31 return -1 # Return -1 indicating no path exists to the target
32
33 # Use breadth-first search to find the shortest path to target
34 visited.remove((0, 0)) # Remove the starting cell from visited set
35 queue = deque([(0, 0)]) # Initialize queue with starting cell
36 steps = 0 # Counter for the number of steps taken
37
38 while queue:
39 for _ in range(len(queue)):
40 x, y = queue.popleft()
41 if (x, y) == target: # If current cell is the target
42 return steps # Return the number of steps taken
43
44 for _, _, delta_x, delta_y in DIRECTIONS:
45 next_x, next_y = x + delta_x, y + delta_y
46 if (next_x, next_y) in visited:
47 visited.remove((next_x, next_y)) # Mark cell as visited
48 queue.append((next_x, next_y)) # Enqueue the next cell
49
50 steps += 1 # Increment step count after exploring all cells at current level
51
52 return -1 # If target cannot be reached, return -1
53
54# No need to modify the GridMaster's interface as per instructions in the problem statement.
551/**
2 * This class provides a solution to find the shortest path in a grid.
3 * It uses DFS for path discovery and BFS to find the shortest path.
4 */
5class Solution {
6 // Delta arrays to allow movement in grid: up, right, down, left
7 private static final char[] DIRECTIONS = {'U', 'R', 'D', 'L'};
8 // Opposite directions for backtracking: down, left, up, right
9 private static final char[] OPPOSITE_DIRECTIONS = {'D', 'L', 'U', 'R'};
10 // Row and column increments for corresponding directions
11 private static final int[] DELTAS = {-1, 0, 1, 0, -1};
12 // Grid dimension for marking visited coordinates
13 private static final int GRID_SIZE = 1010;
14 // Set to keep track of visited coordinates
15 private Set<Integer> visited;
16 // Coordinate pair for the target location
17 private int[] targetPosition;
18
19 // Method to find the shortest path using BFS and DFS traversals in a grid
20 public int findShortestPath(GridMaster master) {
21 targetPosition = null;
22 visited = new HashSet<>();
23 // Initial position as origin (0, 0)
24 visited.add(0);
25 // Use DFS to traverse through the grid and mark visited paths
26 dfs(0, 0, master);
27 if (targetPosition == null) {
28 // Target not found, return -1
29 return -1;
30 }
31 // Remove origin from visited as we will use BFS from origin to find the shortest path
32 visited.remove(0);
33 Deque<int[]> queue = new ArrayDeque<>();
34 queue.offer(new int[] {0, 0});
35 // Distance counter
36 int steps = -1;
37 // BFS to find the shortest path
38 while (!queue.isEmpty()) {
39 ++steps;
40 for (int size = queue.size(); size > 0; --size) {
41 int[] position = queue.poll();
42 int row = position[0], col = position[1];
43 // Check if current position is the target
44 if (targetPosition[0] == row && targetPosition[1] == col) {
45 return steps;
46 }
47 // Explore 4-directionally adjacent cells
48 for (int k = 0; k < 4; ++k) {
49 int newRow = row + DELTAS[k], newCol = col + DELTAS[k + 1];
50 int newPosKey = newRow * GRID_SIZE + newCol;
51 if (visited.contains(newPosKey)) {
52 visited.remove(newPosKey);
53 queue.offer(new int[] {newRow, newCol});
54 }
55 }
56 }
57 }
58 return -1;
59 }
60
61 // DFS to explore the grid and mark paths that have been visited
62 private void dfs(int row, int col, GridMaster master) {
63 if (master.isTarget()) {
64 targetPosition = new int[] {row, col};
65 }
66 for (int k = 0; k < 4; ++k) {
67 char d = DIRECTIONS[k], oppositeD = OPPOSITE_DIRECTIONS[k];
68 int newRow = row + DELTAS[k], newCol = col + DELTAS[k + 1];
69 int newPosKey = newRow * GRID_SIZE + newCol;
70 if (master.canMove(d) && !visited.contains(newPosKey)) {
71 visited.add(newPosKey);
72 master.move(d);
73 dfs(newRow, newCol, master);
74 master.move(oppositeD);
75 }
76 }
77 }
78}
791#include <vector>
2#include <queue>
3#include <unordered_set>
4
5class Solution {
6private:
7 // Delta arrays to facilitate movement in grid: up, right, down, left
8 const std::vector<char> directions = {'U', 'R', 'D', 'L'};
9 // Opposite directions for backtracking: down, left, up, right
10 const std::vector<char> oppositeDirections = {'D', 'L', 'U', 'R'};
11 // Row and column increments for the corresponding directions
12 const std::vector<int> deltas = {-1, 0, 1, 0, -1};
13 // Grid size for marking visited coordinates (assuming a maximum grid size)
14 static const int kGridSize = 1010;
15 // Set to keep track of visited coordinates
16 std::unordered_set<int> visited;
17 // Coordinate pair for the target location
18 std::vector<int> targetPosition;
19
20 // Method to find the shortest path using BFS and DFS traversals in a grid
21public:
22 int findShortestPath(GridMaster& master) {
23 targetPosition.clear();
24 visited.clear();
25 // Initial position as origin (0, 0)
26 visited.insert(0);
27 // Use DFS to explore the grid and mark visited paths
28 dfs(0, 0, master);
29 if (targetPosition.empty()) {
30 // Target not found, return -1
31 return -1;
32 }
33 // Remove origin from visited as we will use BFS from origin to find the shortest path
34 visited.erase(0);
35 std::queue<std::vector<int>> queue;
36 queue.push({0, 0});
37 // Distance counter
38 int steps = -1;
39 // BFS to find the shortest path
40 while (!queue.empty()) {
41 ++steps;
42 int size = queue.size();
43 while (size > 0) {
44 --size;
45 std::vector<int> position = queue.front();
46 queue.pop();
47 int row = position[0], col = position[1];
48 // Check if current position is the target
49 if (targetPosition[0] == row && targetPosition[1] == col) {
50 return steps;
51 }
52 // Explore the four adjacent cells in each direction
53 for (int k = 0; k < 4; ++k) {
54 int newRow = row + deltas[k], newCol = col + deltas[k + 1];
55 int newPosKey = newRow * kGridSize + newCol;
56 if (visited.find(newPosKey) != visited.end()) {
57 visited.erase(newPosKey);
58 queue.push({newRow, newCol});
59 }
60 }
61 }
62 }
63 return -1;
64 }
65
66private:
67 // DFS to explore the grid and record paths that have been visited
68 void dfs(int row, int col, GridMaster& master) {
69 if (master.isTarget()) {
70 targetPosition = {row, col};
71 }
72 for (int k = 0; k < 4; ++k) {
73 char d = directions[k], oppositeD = oppositeDirections[k];
74 int newRow = row + deltas[k], newCol = col + deltas[k + 1];
75 int newPosKey = newRow * kGridSize + newCol;
76 // Check if it's possible to move in the current direction and if it's unvisited
77 if (master.canMove(d) && visited.find(newPosKey) == visited.end()) {
78 visited.insert(newPosKey);
79 master.move(d);
80 dfs(newRow, newCol, master);
81 master.move(oppositeD); // Move back to the previous cell (backtrack)
82 }
83 }
84 }
85};
86
87/**
88 * The GridMaster class is not defined in the problem statement.
89 * It is assumed to be a provided API for the grid system which has the following methods:
90 * - bool canMove(char direction); // Returns true if it's possible to move in the given direction
91 * - void move(char direction); // Moves in the specified direction
92 * - bool isTarget(); // Returns true if current position is the target
93 */
941// Delta arrays to allow movement in grid: up, right, down, left
2const DIRECTIONS: string[] = ['U', 'R', 'D', 'L'];
3// Opposite directions for backtracking: down, left, up, right
4const OPPOSITE_DIRECTIONS: string[] = ['D', 'L', 'U', 'R'];
5// Row and column increments for corresponding directions
6const DELTAS: number[] = [-1, 0, 1, 0, -1];
7// Grid dimension for marking visited coordinates
8const GRID_SIZE: number = 1010;
9// Set to keep track of visited coordinates
10let visited: Set<number>;
11// Coordinate pair for the target location
12let targetPosition: number[] | null;
13
14/**
15 * Finds the shortest path using BFS and DFS traversals in a grid.
16 * @param master - An instance of GridMaster.
17 * @returns The number of steps to the target or -1 if the target cannot be reached.
18 */
19function findShortestPath(master: GridMaster): number {
20 targetPosition = null;
21 visited = new Set<number>();
22 // Initial position as origin (0, 0)
23 visited.add(0);
24 // Use DFS to traverse the grid and mark visited paths
25 dfs(0, 0, master);
26 if (targetPosition === null) {
27 // Target not found, return -1
28 return -1;
29 }
30 // Remove origin from visited as we will use BFS from origin to find the shortest path
31 visited.delete(0);
32 const queue: number[][] = [];
33 queue.push([0, 0]);
34 // Distance counter
35 let steps: number = -1;
36 // BFS to find the shortest path
37 while (queue.length > 0) {
38 steps += 1;
39 for (let size = queue.length; size > 0; size -= 1) {
40 const position = queue.shift()!;
41 const [row, col] = position;
42 // Check if the current position is the target
43 if (targetPosition[0] === row && targetPosition[1] === col) {
44 return steps;
45 }
46 // Explore 4-directionally adjacent cells
47 for (let k = 0; k < 4; ++k) {
48 const newRow = row + DELTAS[k], newCol = col + DELTAS[k + 1];
49 const newPosKey = newRow * GRID_SIZE + newCol;
50 if (visited.has(newPosKey)) {
51 visited.delete(newPosKey);
52 queue.push([newRow, newCol]);
53 }
54 }
55 }
56 }
57 return -1;
58}
59
60/**
61 * DFS to explore the grid and mark paths that have been visited.
62 * @param row - The current row position in the grid.
63 * @param col - The current column position in the grid.
64 * @param master - An instance of GridMaster.
65 */
66function dfs(row: number, col: number, master: GridMaster) {
67 if (master.isTarget()) {
68 targetPosition = [row, col];
69 }
70 for (let k = 0; k < 4; ++k) {
71 const d = DIRECTIONS[k];
72 const oppositeD = OPPOSITE_DIRECTIONS[k];
73 const newRow = row + DELTAS[k], newCol = col + DELTAS[k + 1];
74 const newPosKey = newRow * GRID_SIZE + newCol;
75 if (master.canMove(d) && !visited.has(newPosKey)) {
76 visited.add(newPosKey);
77 master.move(d);
78 dfs(newRow, newCol, master);
79 master.move(oppositeD);
80 }
81 }
82}
83
84// The GridMaster interface stub, for TypeScript understanding purpose.
85// Implementer will provide the real interface.
86interface GridMaster {
87 canMove(direction: string): boolean;
88 move(direction: string): boolean;
89 isTarget(): boolean;
90}
91What is an advantages of top-down dynamic programming vs bottom-up dynamic programming?
Time and Space Complexity
The given Python code represents an algorithm to find the shortest path from a starting point to a target in a grid-like structure, using depth-first search (DFS) to explore the grid and Breadth-first search (BFS) to find the shortest path. Analyzing the code segment provided:
Time Complexity:
The DFS part of the code explores each cell at most once. For each cell, the DFS function performs a constant amount of work. Therefore, if there are N cells that can be visited, the time complexity for the DFS portion is O(N).
Subsequently, the BFS portion of the code also visits each cell at most once. Similar to DFS, since BFS performs a constant amount of work for each cell, the time complexity for the BFS portion is also O(N).
Hence, considering both DFS and BFS combined, the overall time complexity of the code is O(N), where N is the number of cells that can be visited.
Space Complexity:
The space complexity is influenced by both the recursive call stack for DFS and the storage of cells in the set s, and the queue q.
For DFS:
- The call stack can grow up to
O(N)in the worst case when the grid forms a deep structure with only one path and no branching. - The set
sstores all visited cells. In the worst case, this could be allNcells in the grid, which also contributes toO(N).
For BFS:
- The queue
qcan store at mostNcells during the layer-by-layer exploration.
Combining DFS and BFS, the space required is the maximum space used by either algorithm since they are not run simultaneously, which leads to a combined space complexity of O(N).
Thus, the overall space complexity of the code is O(N), where N is the number of cells that can be visited.
Learn more about how to find time and space complexity quickly using problem constraints.
What is an advantages of top-down dynamic programming vs bottom-up dynamic programming?
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