1730. Shortest Path to Get Food - Solutions and Explanation

Problem Statement

Imagine yourself in a matrix-like environment, represented by a m x n character grid, where you are desperately hungry and need to find the quickest way to food. This grid contains various elements:

'*' marks your current location. The layout guarantees that there is exactly one such location.
'#' symbolizes food cells. There can be one or more of these in the grid.
'O' stands for open paths which you are free to traverse.
'X' represents obstacles which block your path and cannot be traversed.

Your challenge is to determine the shortest path from your current position to any available food cell. Movement is possible to the north, south, east, or west, but only if the adjacent cell is not blocked by an obstacle. The task is to return the length of the shortest pathway to food. If no path exists (i.e., all potential paths are obstructed), the function should return -1.

Examples

Example 1

Input:

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

Output:

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:

It is not possible to reach the food.

Example 3

Input:

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

Output:

Explanation:

There can be multiple food cells. It only takes 6 steps to reach the bottom food.

Example 4

Input:

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

Output:

Constraints

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

Approach and Intuition

From the examples provided, we can flesh out the problem and our approach:

Initial Configuration Analysis:
- Identify the start point (your location, marked by '*') and any available food locations (marked by '#').
- Note the open paths ('O') and obstacles ('X').
Pathfinding Strategy:
- Use Breadth-First Search (BFS) which is ideal for finding the shortest path in an unweighted grid like this.
- Initialize the BFS with the starting position and explore all four possible directions.
Expand Node Conditionally:
- For each cell, move to adjacent cells only if they are not an obstacle ('X') and haven't been visited yet.
- Mark each visited cell appropriately to prevent revisiting and potential loops.
Handling Food Cells:
- On visiting any food cell, immediately return the current path length as the result since BFS guarantees this path is the shortest.
Completion and Return:
- If BFS completes without finding a food cell, return -1, indicating no available path to food.

This approach adheres to the constraints provided, including the boundaries on grid size and handles multiple scenarios whether direct paths are available or obstructed. Each movement in the grid is quantified as a single step, and movements are constraint to valid, unvisited cells. Thus, the exploration method (BFS) effectively ensures the minimal path length to the nearest food source is calculated.

Solutions

                            cpp
                            
                        
class Solution {
public:
    const vector<vector<int>> movements = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
    
    int findPathToFood(vector<vector<char>>& grid) {
        int totalRows = grid.size();
        int totalCols = grid[0].size();
        vector<int> startPosition = {-1, -1};
        vector<vector<int>> foodPositions;
    
        for (int row = 0; row < totalRows; ++row) {
            for (int col = 0; col < totalCols; ++col) {
                if (grid[row][col] == '*') {
                    startPosition = {row, col};
                } else if (grid[row][col] == '#') {
                    foodPositions.push_back({row, col});
                }
            }
        }
    
        if (foodPositions.empty()) {
            return -1;
        }
    
        vector<vector<bool>> visited(totalRows, vector<bool>(totalCols, false));
    
        auto comparator = [](const vector<int>& a, const vector<int>& b) {
            return a[0] > b[0];
        };
        priority_queue<vector<int>, vector<vector<int>>, decltype(comparator)> priorityQueue(
            comparator);
    
        int initialCost = manhattanDistance(startPosition[0], startPosition[1], foodPositions);
        priorityQueue.push({initialCost, 0, startPosition[0], startPosition[1]});
    
        while (!priorityQueue.empty()) {
            vector<int> current = priorityQueue.top();
            priorityQueue.pop();
            int estimatedCost = current[0];
            int stepsTaken = current[1];
            int row = current[2];
            int col = current[3];
    
            for (auto& move : movements) {
                int newRow = row + move[0];
                int newCol = col + move[1];
    
                if (!isPositionValid(grid, newRow, newCol) || visited[newRow][newCol]) {
                    continue;
                }
    
                if (grid[newRow][newCol] == '#') {
                    return stepsTaken + 1;
                } else {
                    visited[newRow][newCol] = true;
                    int newEstimatedCost = manhattanDistance(newRow, newCol, foodPositions);
                    priorityQueue.push({newEstimatedCost + stepsTaken + 1, stepsTaken + 1, newRow, newCol});
                }
            }
        }
    
        return -1;
    }
    
private:
    int manhattanDistance(int row, int col, vector<vector<int>>& foodPositions) {
        int minDistance = INT_MAX;
        for (const vector<int>& food : foodPositions) {
            int distance = abs(food[0] - row) + abs(food[1] - col);
            minDistance = min(minDistance, distance);
        }
        return minDistance;
    }
    
    bool isPositionValid(vector<vector<char>>& grid, int row, int col) {
        return row >= 0 && col >= 0 && row < grid.size() && col < grid[0].size() && grid[row][col] != 'X';
    }
};

This summary explains a C++ solution for finding the shortest path to food in a grid. The grid has cells marked as * for start, # for food, and X for obstacles. The solution uses breadth-first search (BFS) with a priority queue, where each state in the queue contains the estimated total cost and the number of steps taken so far. The estimation is based on the Manhattan distance to the closest food cell.

Ensure to follow these steps:

Initialize 2D vectors for tracking visited cells and representing movements in four directions (up, down, left, right).
Locate the starting position (*) and food positions (#) in the grid.
Use a priority queue to process positions by least estimated cost first, which combines steps taken and the Manhattan distance to the nearest food position.
In each BFS iteration, explore adjacent cells.
- If an adjacent cell is food (#), return the number of steps taken plus one.
- If an adjacent cell is within bounds and not visited, flag it as visited and push its details into the priority queue after recalculating the estimated cost based on distance from food positions.
If no food is accessible, return -1.

Functions used for utility:

manhattanDistance to compute the Manhattan distance between two points.
isPositionValid to check boundaries and obstacles.

This approach ensures an efficient traverse of the grid while always prioritizing the path that appears to lead to food quicker, incorporating both the actual path cost and the estimated remaining distance based on the heuristic.

Java

                            java
                            
                        
class Solution {
    private final int[][] moves = { { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };
    
    public int searchFood(char[][] grid) {
        int height = grid.length;
        int width = grid[0].length;
        int[] startPoint = { -1, -1 };
        List<int[]> foodLocations = new ArrayList<>();
    
        for (int i = 0; i < height; i++) {
            for (int j = 0; j < width; j++) {
                if (grid[i][j] == '*') {
                    startPoint = new int[] { i, j };
                } else if (grid[i][j] == '#') {
                    foodLocations.add(new int[] { i, j });
                }
            }
        }
    
        if (foodLocations.isEmpty()) return -1;
    
        boolean[][] visited = new boolean[height][width];
        PriorityQueue<int[]> pq = new PriorityQueue<>((x, y) -> x[0] - y[0]);
        int initialEstimate = calculateDistance(startPoint[0], startPoint[1], foodLocations);
        pq.add(new int[] { initialEstimate, 0, startPoint[0], startPoint[1] });
    
        while (!pq.isEmpty()) {
            int[] current = pq.poll();
            int estimatedCost = current[0];
            int stepCount = current[1];
            int row = current[2];
            int col = current[3];
    
            for (int[] move : moves) {
                int newRow = row + move[0];
                int newCol = col + move[1];
    
                if (!isWithinBounds(grid, newRow, newCol) || visited[newRow][newCol]) {
                    continue;
                }
    
                if (grid[newRow][newCol] == '#') {
                    return stepCount + 1;
                }
                visited[newRow][newCol] = true;
                int newDistanceEstimate = calculateDistance(newRow, newCol, foodLocations);
                pq.add(new int[] { newDistanceEstimate + stepCount + 1, stepCount + 1, newRow, newCol });
            }
        }
        return -1;
    }
    
    private int calculateDistance(int row, int col, List<int[]> foodLocations) {
        int minDistance = Integer.MAX_VALUE;
        for (int[] food : foodLocations) {
            int distance = Math.abs(food[0] - row) + Math.abs(food[1] - col);
            minDistance = Math.min(minDistance, distance);
        }
        return minDistance;
    }
    
    private boolean isWithinBounds(char[][] grid, int row, int col) {
        return row >= 0 && col >= 0 && row < grid.length && col < grid[0].length && grid[row][col] != 'X';
    }
}

The Java solution provided offers an efficient way to find the shortest path to food in a grid represented by a matrix of characters. The grid contains the following elements:

"*": start position
"#": food position
"X": obstacles
space (" "): walkable paths

To solve this, the solution adopts a best-first search strategy, leveraging a priority queue to always process the cell with the lowest estimated cost to a food cell first. This approach is akin to a heuristic-based search or A* search, where the heuristic is the Manhattan distance — the sum of absolute differences of their Cartesian coordinates. Here’s a breakdown of the execution flow:

Initialize the grid dimensions and structure to store food positions. Scan the grid to identify the starting point and all food cells. If no food is present, return -1 immediately indicating the food is not reachable.
Define a 'visited' boolean matrix of the same size as the grid to track visited cells, thus preventing cycles or redundant paths.
Begin the priority queue with the start position and an initial cost estimate based on the shortest Manhattan distance from the start to any of the food cells.
Proceed with the best-first search:
- Dequeue the cell with the lowest estimated cost.
- Check all four possible moves (up, down, left, right) to continue the path from the current cell:
  - Validate each move considering boundaries and whether the target cell is walkable and not previously visited.
  - If a move leads to a food cell, return the accumulated steps as the shortest path length.
  - For other valid moves, mark the cell as visited, calculate the new heuristic cost, and enqueue it with the new cost and step count.
If the queue depletes without reaching a food cell, return -1.

The helper functions calculateDistance to compute Manhattan distances and isWithinBounds to ensure move validity are crucial in maintaining efficiency and correctness. This method not only ensures optimal pathfinding through potentially complex landscapes but also efficiently handles sparse food distributions and larger grids through the use of heuristics and priority-based exploration.

Python

                            python
                            
                        
class Solution:
    # Possible movement directions vectors
    movements = [(0, 1), (0, -1), (-1, 0), (1, 0)]
    
    def findFood(self, land: list[list[str]]) -> int:
        rows, columns = len(land), len(land[0])
        start_point = None
        food_locations = []
    
        # Identify start point and food locations
        for row in range(rows):
            for column in range(columns):
                if land[row][column] == "*":
                    start_point = [row, column]
                elif land[row][column] == "#":
                    food_locations.append([row, column])
    
        if not food_locations:
            return -1
    
        # Set to track visited positions
        visited = set()
    
        # Priority queue: (estimated cost, steps, row, column)
        priority_queue = [
            (self._estimate_distance(start_point[0], start_point[1], food_locations), 0, start_point[0], start_point[1])
        ]
    
        while priority_queue:
            estimated_cost, steps_taken, row, col = heapq.heappop(priority_queue)
    
            # Navigate the four directions
            for dr, dc in self.movements:
                new_row, new_col = row + dr, col + dc
    
                if(
                    not self._check_validity(land, new_row, new_col)
                    or (new_row, new_col) in visited
                ):
                    continue
    
                # Return if food is found
                if land[new_row][new_col] == "#":
                    return steps_taken + 1
    
                visited.add((new_row, new_col))
                    
                # Update estimated distance
                new_distance_estimate = self._estimate_distance(new_row, new_col, food_locations)
                heapq.heappush(
                    priority_queue, (new_distance_estimate + steps_taken + 1, steps_taken + 1, new_row, new_col)
                )
    
        return -1
    
    # Calculate shortest Manhattan distance to any food source
    def _estimate_distance(self, row: int, col: int, food_sources: list[list[int]]) -> int:
        return min(abs(food_source[0] - row) + abs(food_source[1] - col) for food_source in food_sources)
    
    # Validate if the new position is within grid limits and not an obstacle
    def _check_validity(self, land: list[list[str]], row: int, col: int) -> bool:
        return (
            0 <= row < len(land) and 0 <= col < len(land[0]) and land[row][col] != "X"
        )

The provided Python code defines a Solution class to solve the problem of finding the shortest path to food in a grid. Here are the steps and logic used in the code:

Identify the Dimensions and Start Point:
- Determine the number of rows and columns in the grid land.
- Locate the start point (marked by *) and possible food locations (marked by #).
Edge Case Consideration:
- If no food locations are identified, return -1 to indicate that food is not reachable.
Initialize Priority Queue and Visited Set:
- A priority queue initialized with an estimated cost of reaching the nearest food location, starting steps (0), and the start position coordinates.
- A set visited gets used to keep track of visited grid cells to prevent revisiting.
Process Priority Queue with Breadth-First Search (BFS):
- Extract positions from the priority queue based on the estimated cost and actual steps taken so far.
- For each popped position, check four possible moves (up, down, left, right).
- Invalid moves are skipped based on boundary conditions and obstacles ('X').
- If a move reaches a food location (#), return the total steps taken to reach there.
Updates During Traversal:
- If the current position does not contain food, add its surrounding valid moves to the priority queue with updated distance estimates and increased step count.
- All explored positions are recorded in the visited set to avoid loops and redundant checks.
Support Methods:
- _estimate_distance: Calculates the shortest Manhattan distance from the current position to the nearest food location, used for making heuristic decisions in path exploration.
- _check_validity: Ensures a position is inside the grid and not blocked (no 'X').

In essence, the solution employs a BFS strategy augmented with a priority queue for efficient frontier exploration based on distance estimates to the nearest food location, ensuring the shortest path is favored.

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Shortest Path to Get Food

Problem Statement

Examples

Example 1

Example 2

Example 3

Example 4

Constraints

Approach and Intuition

Solutions

Comments

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