1778. Shortest Path in a Hidden Grid

MediumDepth-First SearchBreadth-First SearchGraphInteractive

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 returns true if it's possible, false otherwise.
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 – the robot remains in place.
isTarget(): Determines if the robot is currently on the target cell, returning true if it is, false otherwise.

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.

Flowchart Walkthrough

Let's analyze the problem using the Flowchart to determine the appropriate algorithm for LeetCode Problem 1778. "Shortest Path in a Hidden Grid". Here's the decision path:

Is it a graph?

Yes: The grid structure in the problem implicitly forms a graph where each cell is a node and edges exist between adjacent cells.

Is it a tree?

No: The graph represented by the grid is not hierarchical; it contains cycles and multiple possible paths.

Is the problem related to directed acyclic graphs (DAGs)?

No: The problem is set in a grid which does not have directed or acyclic constraints.

Is the problem related to shortest paths?

Yes: The problem explicitly asks for the shortest path, so we are dealing with a shortest path problem.

Is the graph weighted?

No: The grid or graph doesn't have different weights assigned to different edges; each move to an adjacent cell has uniform cost or weight.

The flowchart steers towards the use of BFS (Breadth-First Search), not DFS, for shortest path problems in unweighted graphs. Here's why:

BFS is optimal for unweighted graphs to find the shortest path because it explores all nodes at the present depth prior to moving on to nodes at the next depth level.

Conclusion: Based on the flowchart and the characteristics of the problem, BFS, not DFS, is the suitable algorithm for solving LeetCode Problem 1778, as it efficiently handles shortest paths in unweighted graphs.

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.

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 GridMaster reports that the robot is at the target (isTarget() returns true), we record the target's location in the variable target.
Exploration: For each of the four possible directions ('U', 'D', 'L', 'R'), the algorithm checks if the robot can move in that direction using canMove(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 set s to mark as visited.
Move and Recur: We move the robot with move(dir) and recursively call dfs(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) where ndir is 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 queue q.
Looping: BFS runs in a loop until the queue is empty, iterating through cells level by level.
Distance Counting: Increase ans by 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 ans value is the minimum distance, and we return it.
- Enqueue Neighbors: For valid neighboring cells that are in the set s, we remove them from s (to indicate they're visited) and enqueue them to be processed in the next 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.

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Example 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 set s.
The robot checks if it can move right with canMove('R'). It can, so it moves to (0, 1) with move('R').
On (0, 1), canMove('R') is true. However, after moving the robot, we find that isTarget() is false, and the cell is blocked. We backtrack with move('L') to (0, 0).
Next, we try moving down with canMove('D') and it's possible. We move to (1, 0) with move('D').
We continue trying to move right until the robot reaches (1, 2) with successive move('R'). Upon reaching (1, 2), isTarget() returns true. 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 q and add the starting cell (0, 0) to it, with an initial distance ans set 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 set s, so it is enqueued and removed from s.
Increase ans to 1 as we're moving to the next level.
Next, we dequeue (1, 0) and enqueue its unvisited neighbor (1, 1), and remove it from s.
The process repeats at each level. Increase ans to 2, dequeue (1, 1) and enqueue (1, 2), which is the target location.
Once at (1, 2), the target is reached. ans is 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.
55

1/**
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}
79

1#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 */
94

1// 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}
91

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 s stores all visited cells. In the worst case, this could be all N cells in the grid, which also contributes to O(N).

For BFS:

The queue q can store at most N cells 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.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

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How would you design a stack which has a function min that returns the minimum element in the stack, in addition to push and pop? All push, pop, min should have running time O(1).

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1778. Shortest Path in a Hidden Grid

Problem Description

Flowchart Walkthrough

Intuition

Solution Approach

Depth-First Search (DFS)

Breadth-First Search (BFS)

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Example Walkthrough

Depth-First Search (DFS)

Backtracking

Breadth-First Search (BFS)

Solution Implementation

Time and Space Complexity

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