Shortest Path in a Hidden Grid

#952 Most AskedMedium

Shortest Path in a Hidden Grid #952 Most Asked

Medium

Bloomberg LP

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Topics:

GraphsDepth-First SearchBreadth-First Search

This is an interactive problem.

There is a robot in a hidden grid, and you are trying to find the target. You are given the size of the grid as m x n, where m is the number of rows and n is the number of columns (0-indexed).

You don't know the location of the robot or the target. You are only allowed to ask the robot to move in one of the four directions: up, down, left, or right. The robot might not move if it is trying to move off the grid. You can also ask the robot for its current location.

You are given the API Grid:

interface Grid {
  public boolean move(char direction) : Returns true if the robot moves successfully, false otherwise.
  public int[] getPos() : Returns the current position of the robot as [row, col]
}

direction is 'U', 'D', 'L', or 'R' representing the directions up, down, left, and right, respectively.
getPos() returns the robot's position as an array of length 2, [row, col].

You are also given the following function:

int findShortestPath(Grid grid)

Return the minimum number of steps required to reach the target. If there is no solution, return -1.

Example 1:

Input: m = 5, n = 4
Output: 6
Explanation: One possible path to the target is UURRDD.

Example 2:

Input: m = 3, n = 4
Output: -1
Explanation: It is impossible to reach the target.

Constraints:

1 <= n, m <= 100

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for the grid cells and what does each value represent (e.g., traversable, blocked)?
If no path exists between the start and the target, what should the function return?
Are we given a starting point? If so, what data type is the starting point and is it guaranteed to be within the grid?
How do we interact with the 'hidden' grid? Is there an API or set of functions available to check neighboring cells and if so, how are the neighbors represented?
Is the hidden grid guaranteed to be fully connected (i.e., is there a path from every traversable cell to every other traversable cell)?

Brute Force Solution

Approach

The brute force strategy involves exploring every possible path from the starting point until we find the shortest one to the target. We essentially try every possible combination of moves, like up, down, left, and right, until we hit the destination. This ensures we don't miss any potential paths, however inefficient it might be.

Here's how the algorithm would work step-by-step:

Begin at the starting location.
Consider all possible moves from that location: up, down, left, and right.
For each of those possible moves, see where it leads you and mark that spot as visited.
From each of these new spots, again consider all possible moves (up, down, left, and right), and mark those new spots as visited.
Keep repeating this process, branching out in all possible directions from each spot we visit.
As we explore, keep track of the length of each path from the start to the current location.
If we reach the target location, record the length of that path.
Once we've explored all possible paths within a reasonable limit (to avoid infinite loops), compare all the recorded path lengths to the target.
The shortest of those paths to the target is the solution.

Code Implementation

def shortest_path_hidden_grid_brute_force(start_row, start_column, target_row, target_column, max_steps): 
    shortest_path_length = float('inf')
    visited = set()

    def explore(current_row, current_column, path_length): 
        nonlocal shortest_path_length

        # Check if we've reached the target
        if current_row == target_row and current_column == target_column:
            shortest_path_length = min(shortest_path_length, path_length)
            return

        # Avoid infinite loops, bound the search space.
        if path_length > max_steps:
            return

        current_position = (current_row, current_column)

        # Prevent revisiting the same cell
        if current_position in visited:
            return

        visited.add(current_position)

        # Explore possible directions: up, down, left, right
        explore(current_row - 1, current_column, path_length + 1)
        explore(current_row + 1, current_column, path_length + 1)
        explore(current_row, current_column - 1, path_length + 1)
        explore(current_row, current_column + 1, path_length + 1)

        # Backtrack: remove the current cell from visited
        visited.remove(current_position)

    explore(start_row, start_column, 0)

    if shortest_path_length == float('inf'):
        return -1
    else:
        return shortest_path_length

Big(O) Analysis

Time Complexity

O(4^n) – The provided brute force approach explores all possible paths from the starting point. In the worst-case scenario, where the grid is largely empty, from each cell, there are up to four possible moves (up, down, left, right). The depth of the search can potentially reach 'n', representing the number of cells in the grid. This creates a search tree where each node can have up to four children, making it an exponential search. Therefore, the time complexity is O(4^n), representing the exploration of all potential paths of length up to 'n'.

Space Complexity

O(N) – The brute force approach explores all possible paths and marks visited locations. This implies using a data structure, such as a set or matrix, to keep track of the visited cells in the grid. In the worst-case scenario, we might visit every cell in the grid once, so the size of this visited set/matrix could grow linearly with the number of cells in the grid, where N represents the total number of cells. Therefore, the auxiliary space used is proportional to N, which simplifies to O(N).

Optimal Solution

Approach

The challenge is to find the quickest route through an unknown maze. The best strategy is to explore the maze systematically, keeping track of where you've been and how far each spot is from the start. This ensures you find the shortest path without getting lost in circles or wasting time on longer routes.

Here's how the algorithm would work step-by-step:

Begin at the starting point and mark its distance as zero.
Explore the immediate surroundings (north, south, east, west).
For each new location found, mark its distance as one more than the current location's distance.
Add these new locations to a list of places to explore later.
Continue exploring locations from the list, always picking the location that's closest to the start.
Repeat the process of exploring surroundings, marking distances, and adding new locations to the list.
If you encounter a location you've already visited, only update its distance if the new path is shorter. This is crucial for finding the shortest path if there are loops.
Keep going until you find the destination or until there are no more places left to explore.
If you find the destination, the marked distance at that location is the length of the shortest path. Retrace your steps from the destination back to the start to find the actual path.

Code Implementation

from collections import deque

def find_shortest_path(grid):
    start_node = (0, 0)
    target_node = None
    distance_map = {start_node: 0}
    queue = deque([start_node])
    visited_nodes = {start_node}

    # Directions: (row change, col change)
    possible_directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    while queue:
        current_node = queue.popleft()

        if grid.is_target(current_node):
            target_node = current_node
            break

        for row_change, col_change in possible_directions:
            new_row = current_node[0] + row_change
            new_col = current_node[1] + col_change
            neighbor_node = (new_row, new_col)

            if grid.is_valid(neighbor_node) and neighbor_node not in visited_nodes:
                # This ensures we only explore valid and unvisited nodes.
                distance_map[neighbor_node] = distance_map[current_node] + 1
                visited_nodes.add(neighbor_node)
                queue.append(neighbor_node)
            elif grid.is_valid(neighbor_node) and neighbor_node in distance_map:
                # If a shorter path to an already visited node is found.
                new_distance = distance_map[current_node] + 1
                if new_distance < distance_map[neighbor_node]:
                    distance_map[neighbor_node] = new_distance

    if target_node is None:
        return -1  # No path found

    # If the target has been found return the path length.
    return distance_map[target_node]

Big(O) Analysis

Time Complexity

O(R*C) – The algorithm explores a grid of size R rows by C columns using a Breadth-First Search (BFS) approach. In the worst case, the algorithm visits every cell in the grid. For each cell, it examines up to four neighboring cells (north, south, east, west). While exploring neighbors technically adds a small constant factor, the dominant cost is visiting all cells in the grid, resulting in a time complexity proportional to the number of cells. Therefore, the time complexity is O(R*C), where R is the number of rows and C is the number of columns in the grid, as the algorithm may potentially visit all cells.

Space Complexity

O(N) – The algorithm uses a queue (the "list of places to explore later") to store locations to visit, and a data structure (implicitly described as marking distances at locations) to track visited locations and their distances. In the worst-case scenario, where N is the number of cells in the hidden grid, the algorithm might need to visit and store all cells in the queue if the destination is far away or unreachable. Thus, the space complexity is proportional to the number of cells in the grid, leading to O(N) auxiliary space.

Edge Cases

Case	How to Handle
Null grid or start node	Return -1 to indicate no path is found if the grid or start node is null.
Start node is an obstacle	Return -1 indicating no path found if the starting position itself is an obstacle.
Grid is a single cell and the start node is valid	Return 0 since the shortest path to itself is 0 if it's a valid cell.
No path exists from start to any reachable empty cell	The BFS should terminate and return -1 if no unvisited reachable cell is found.
Grid is very large, potentially causing memory issues	Use an iterative BFS approach with appropriate data structures (e.g., queue) to manage memory efficiently and avoid stack overflow from recursion.
The hidden grid has cycles.	BFS inherently avoids cycles by maintaining a 'visited' set, preventing re-exploration of already visited nodes.
The search algorithm explores a very long path before finding the shortest	BFS guarantees finding the shortest path in an unweighted graph, so it will eventually find it even if there are long paths.
Integer overflow when calculating distances in a very large grid	Use appropriate data types (e.g., long) to store distances to prevent overflow, or assume grid sizes are bounded to avoid it.

Null grid or start node

How to Handle:

Return -1 to indicate no path is found if the grid or start node is null.

Start node is an obstacle

How to Handle:

Return -1 indicating no path found if the starting position itself is an obstacle.

Grid is a single cell and the start node is valid

How to Handle:

Return 0 since the shortest path to itself is 0 if it's a valid cell.

No path exists from start to any reachable empty cell

How to Handle:

The BFS should terminate and return -1 if no unvisited reachable cell is found.

Grid is very large, potentially causing memory issues

How to Handle:

Use an iterative BFS approach with appropriate data structures (e.g., queue) to manage memory efficiently and avoid stack overflow from recursion.

The hidden grid has cycles.

How to Handle:

BFS inherently avoids cycles by maintaining a 'visited' set, preventing re-exploration of already visited nodes.

The search algorithm explores a very long path before finding the shortest

How to Handle:

BFS guarantees finding the shortest path in an unweighted graph, so it will eventually find it even if there are long paths.

Integer overflow when calculating distances in a very large grid

How to Handle:

Use appropriate data types (e.g., long) to store distances to prevent overflow, or assume grid sizes are bounded to avoid it.