Shortest Path to Get Food

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You are starving and want to eat food as quickly as possible. You want to find the shortest path to arrive at any food cell.

You are given an m x n character matrix, grid, of these different types of cells:

'*' is your location. There is exactly one '*' cell.
'#' is a food cell. There may be multiple food cells.
'O' is free space, and you can travel through these cells.
'X' is an obstacle, and you cannot travel through these cells.

You can travel to any adjacent cell north, east, south, or west in one step.

Return the length of the shortest path to arrive at any food cell. If there is no food cell, return -1.

Example 1:

Input: grid = [["X","X","X","X","X","X","X","X"],["X","*","O","O","O","O","O","X"],["X","O","O","#","O","O","O","X"],["X","O","O","O","O","O","O","X"],["X","O","O","O","O","O","O","X"],["X","O","O","O","O","O","O","X"],["X","O","O","O","O","O","O","X"],["X","X","X","X","X","X","X","X"]]

Output: 3
Explanation: It takes 3 steps to reach the food.

Example 2:

Input: grid = [["X","X","X","X","X"],["X","*","X","O","X"],["X","O","X","#","X"],["X","X","X","X","X"]]

Output: -1
Explanation: There is no path to reach the food.

Example 3:

Input: grid = [["X","X","X","X","X","X","X","X","X","X"],["X","*","O","O","O","O","O","O","#","X"],["X","X","X","X","X","X","X","X","X","X"]]

Output: 7
Explanation: It takes 7 steps to reach the food.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 200
grid[row][col] is '*', 'X', 'O', or '#'.
The grid has exactly one '*' cell.

Try coding it in LeetCode

Shortest Path to Get Food

Medium

Google

8 views

Topics:

Graphs

You are starving and want to eat food as quickly as possible. You want to find the shortest path to arrive at any food cell.

You are given an m x n character matrix, grid, of these different types of cells:

'*' is your location. There is exactly one '*' cell.
'#' is a food cell. There may be multiple food cells.
'O' is free space, and you can travel through these cells.
'X' is an obstacle, and you cannot travel through these cells.

You can travel to any adjacent cell north, east, south, or west in one step.

Return the length of the shortest path to arrive at any food cell. If there is no food cell, return -1.

Example 1:

Input: grid = [["X","X","X","X","X","X","X","X"],["X","*","O","O","O","O","O","X"],["X","O","O","#","O","O","O","X"],["X","O","O","O","O","O","O","X"],["X","O","O","O","O","O","O","X"],["X","O","O","O","O","O","O","X"],["X","O","O","O","O","O","O","X"],["X","X","X","X","X","X","X","X"]]

Output: 3
Explanation: It takes 3 steps to reach the food.

Example 2:

Input: grid = [["X","X","X","X","X"],["X","*","X","O","X"],["X","O","X","#","X"],["X","X","X","X","X"]]

Output: -1
Explanation: There is no path to reach the food.

Example 3:

Input: grid = [["X","X","X","X","X","X","X","X","X","X"],["X","*","O","O","O","O","O","O","#","X"],["X","X","X","X","X","X","X","X","X","X"]]

Output: 7
Explanation: It takes 7 steps to reach the food.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 200
grid[row][col] is '*', 'X', 'O', or '#'.
The grid has exactly one '*' cell.

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 dimensions of the grid, and what are the maximum possible dimensions?
Besides 'X', 'O', and '*', are there any other possible characters in the grid, such as obstacles or walls?
If there is no path to food ('*'), what should I return?
Can the starting position ('X') be blocked or surrounded by obstacles, making it impossible to move?
Is it guaranteed that there will be at least one food location ('*') in the grid?

Brute Force Solution

Approach

Imagine exploring a maze where you need to find the shortest path to food. The brute force approach is like trying every single possible route, no matter how long or winding, until you find food. It’s like leaving breadcrumbs everywhere you go to remember where you've already been and avoid going in circles.

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

Start at your current location.
Consider all possible directions you can move: up, down, left, and right.
For each direction, take one step and mark that new spot as visited.
From each of those new spots, again consider all possible directions and take another step, marking those spots as visited too.
Keep repeating this process, exploring all possible paths, one step at a time.
Each time you reach a food location, record the number of steps it took to get there.
Once you've explored every possible path (or decided to stop after a very, very long time), compare the number of steps for each path that led to food.
Choose the path with the fewest steps. That's the shortest path to food.

Code Implementation

def shortest_path_to_food_brute_force(grid):
    start_row, start_col = -1, -1
    rows, cols = len(grid), len(grid[0])
    for row in range(rows):
        for col in range(cols):
            if grid[row][col] == '*':
                start_row, start_col = row, col
                break
        if start_row != -1:
            break

    shortest_path = float('inf')

    def explore(row, col, steps, visited):
        nonlocal shortest_path

        # Check for out-of-bounds or visited cells
        if row < 0 or row >= rows or col < 0 or col >= cols or (row, col) in visited or grid[row][col] == 'X':
            return

        # If we found food, update the shortest path
        if grid[row][col] == '#':
            shortest_path = min(shortest_path, steps)
            return

        visited.add((row, col))
        
        # Explore all 4 directions
        explore(row + 1, col, steps + 1, visited.copy())
        explore(row - 1, col, steps + 1, visited.copy())
        explore(row, col + 1, steps + 1, visited.copy())
        explore(row, col - 1, steps + 1, visited.copy())

    explore(start_row, start_col, 0, set())

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

Big(O) Analysis

Time Complexity

O(R * C) – The breadth-first search (BFS) explores each cell in the grid at most once. In the worst-case scenario, the entire grid needs to be explored to find the food or determine that it's unreachable. Assuming R is the number of rows and C is the number of columns, we might potentially visit every cell. Therefore, the time complexity is proportional to the number of cells in the grid, resulting in O(R * C).

Space Complexity

O(N) – The algorithm maintains a 'visited' set to track visited locations within the maze to avoid cycles. In the worst-case scenario, the algorithm might visit every cell in the maze, where N is the total number of cells. Therefore, the size of the 'visited' set can grow linearly with the input size N, leading to O(N) auxiliary space. Additionally, the breadth-first search implicitly uses a queue to store the next cells to explore which, in the worst case, may also contain all cells in the maze, leading to O(N) space complexity.

Optimal Solution

Approach

The quickest way to find food is to explore the map layer by layer, always moving to the nearest possible locations. This ensures that we find the food using the shortest path possible because we expand our search outwards in a systematic way.

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

Start at your current location.
Look at all the places you can reach in one step (up, down, left, right).
If food is there, you're done! That's the shortest path.
If not, mark these locations as 'visited' so you don't check them again.
Now, look at all the places you can reach in two steps from your starting location (again, up, down, left, right from the places you just checked, but don't go back to where you've been).
Check if food is in any of these new locations. If so, you're done! You've found the shortest path to food.
Keep repeating this process, expanding outwards layer by layer, checking for food at each new location. Keep track of the number of steps it takes to reach each spot.
The first time you find food, the number of steps you took is the length of the shortest path to food.

Code Implementation

def shortest_path_to_get_food(grid):
    queue = []
    visited = set()
    start_row, start_col = -1, -1
    rows = len(grid)
    cols = len(grid[0])

    for row_index in range(rows):
        for col_index in range(cols):
            if grid[row_index][col_index] == '*':
                start_row, start_col = row_index, col_index
                break
        if start_row != -1:
            break

    queue.append((start_row, start_col, 0))
    visited.add((start_row, start_col))
    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    while queue:
        current_row, current_col, path_length = queue.pop(0)

        # We found food, so return current path length
        if grid[current_row][current_col] == '#':
            return path_length

        for row_direction, col_direction in directions:
            new_row = current_row + row_direction
            new_col = current_col + col_direction

            # Ensure new coordinates are within grid bounds
            if 0 <= new_row < rows and 0 <= new_col < cols and \
               grid[new_row][new_col] != 'X' and (new_row, new_col) not in visited:

                # Mark as visited to prevent re-visits
                visited.add((new_row, new_col))
                queue.append((new_row, new_col, path_length + 1))

    return -1

Big(O) Analysis

Time Complexity

O(m*n) – The algorithm performs a Breadth-First Search (BFS) on the grid. In the worst-case scenario, the algorithm might need to visit every cell in the grid to find the food. If the grid has dimensions m x n, where m is the number of rows and n is the number of columns, then in the worst case, every cell is visited once. Thus, the time complexity is proportional to the total number of cells in the grid, which is m * n. Each cell is visited at most once due to the visited set.

Space Complexity

O(N) – The algorithm uses a queue to store locations to visit, and a set to keep track of visited locations. In the worst-case scenario, we might need to visit all cells in the grid before finding the food. Therefore, the queue and the set could potentially store all the cells of the grid. If N represents the number of cells in the grid, the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty grid	Return -1 immediately indicating no path exists as the grid is invalid.
Grid with only one cell, not food or start	Return -1 immediately as there's no food or starting point.
Grid with only starting point but no food	Return -1 after BFS completes without finding 'food'.
Grid with only food but no starting point	Return -1 immediately as no path can exist from the non-existent start point.
Starting point is surrounded by walls (no path possible)	BFS will enqueue the starting point, but no valid neighbor can be enqueued, eventually returning -1 when the queue is empty.
Food is unreachable due to walls or other obstacles	BFS will explore all reachable cells and return -1 if it never encounters the food cell.
Large grid size, potentially causing memory issues	BFS's queue size can grow linearly with the number of cells in the grid; therefore, if memory becomes a concern, consider using iterative deepening A* or other memory-efficient search algorithms.
Multiple food locations exist (find the shortest path to any food)	The BFS will find the shortest path to the first food encountered during the search, effectively solving the problem if multiple food sources were required.