LeetCode 1778: Shortest Path in a Hidden Grid Solution

Master LeetCode problem 1778 (Shortest Path in a Hidden Grid), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1778. Shortest Path in a Hidden Grid

Medium

Array Depth-First Search Breadth-First Search Matrix Interactive

Problem Explanation

Explanation:

The problem requires finding the shortest path in a hidden grid. Each cell in the grid can be either empty or blocked. We can move in four directions - up, down, left, or right. The hidden grid is represented by an interface GridMaster with the following methods:

boolean canMove(char direction): Returns true if we can move in the specified direction.
void move(char direction): Moves in the specified direction.
boolean isTarget(): Returns true if the current cell is the target cell.

To solve this problem, we can use backtracking along with DFS to explore all possible paths from the starting cell to the target cell. We can keep track of visited cells to avoid revisiting the same cell and ensure that we find the shortest path.

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Solution Code

interface GridMaster {
    boolean canMove(char direction);
    void move(char direction);
    boolean isTarget();
}

class Solution {
    public int findShortestPath(GridMaster master) {
        int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        Map<Integer, Integer> visited = new HashMap<>();
        visited.put(0, 0);
        int[] start = {0, 0};
        dfs(master, start, dirs, visited);
        return visited.getOrDefault(1, -1);
    }

    private void dfs(GridMaster master, int[] curr, int[][] dirs, Map<Integer, Integer> visited) {
        if (master.isTarget()) {
            visited.put(1, visited.get(curr[0] * 1000 + curr[1]) + 1);
            return;
        }
        for (int i = 0; i < 4; i++) {
            int newX = curr[0] + dirs[i][0];
            int newY = curr[1] + dirs[i][1];
            if (!visited.containsKey(newX * 1000 + newY) && master.canMove(getDirection(i))) {
                visited.put(newX * 1000 + newY, visited.get(curr[0] * 1000 + curr[1]) + 1);
                master.move(getDirection(i));
                dfs(master, new int[]{newX, newY}, dirs, visited);
                master.move(getDirection((i + 2) % 4)); // Move back
            }
        }
    }

    private char getDirection(int i) {
        return i == 0 ? 'R' : i == 1 ? 'L' : i == 2 ? 'D' : 'U';
    }
}

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