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1778. Shortest Path in a Hidden Grid

Description

This is an interactive problem.

There is a robot in a hidden grid, and you are trying to get it from its starting cell to the target cell in this grid. The grid is of size m x n, and each cell in the grid is either empty or blocked. It is guaranteed that the starting cell and the target cell are different, and neither of them is blocked.

You want to find the minimum distance to the target cell. However, you do not know the grid's dimensions, the starting cell, nor the target cell. You are only allowed to ask queries to the GridMaster object.

Thr GridMaster class has the following functions:

  • boolean canMove(char direction) Returns true if the robot can move in that direction. Otherwise, it returns false.
  • void move(char direction) Moves the robot in that direction. If this move would move the robot to a blocked cell or off the grid, the move will be ignored, and the robot will remain in the same position.
  • boolean isTarget() Returns true if the robot is currently on the target cell. Otherwise, it returns false.

Note that direction in the above functions should be a character from {'U','D','L','R'}, representing the directions up, down, left, and right, respectively.

Return the minimum distance between the robot's initial starting cell and the target cell. If there is no valid path between the cells, return -1.

Custom testing:

The test input is read as a 2D matrix grid of size m x n where:

  • grid[i][j] == -1 indicates that the robot is in cell (i, j) (the starting cell).
  • grid[i][j] == 0 indicates that the cell (i, j) is blocked.
  • grid[i][j] == 1 indicates that the cell (i, j) is empty.
  • grid[i][j] == 2 indicates that the cell (i, j) is the target cell.

There is exactly one -1 and 2 in grid. Remember that you will not have this information in your code.

 

Example 1:

Input: grid = [[1,2],[-1,0]]
Output: 2
Explanation: One possible interaction is described below:
The robot is initially standing on cell (1, 0), denoted by the -1.
- master.canMove('U') returns true.
- master.canMove('D') returns false.
- master.canMove('L') returns false.
- master.canMove('R') returns false.
- master.move('U') moves the robot to the cell (0, 0).
- master.isTarget() returns false.
- master.canMove('U') returns false.
- master.canMove('D') returns true.
- master.canMove('L') returns false.
- master.canMove('R') returns true.
- master.move('R') moves the robot to the cell (0, 1).
- master.isTarget() returns true. 
We now know that the target is the cell (0, 1), and the shortest path to the target cell is 2.

Example 2:

Input: grid = [[0,0,-1],[1,1,1],[2,0,0]]
Output: 4
Explanation: The minimum distance between the robot and the target cell is 4.

Example 3:

Input: grid = [[-1,0],[0,2]]
Output: -1
Explanation: There is no path from the robot to the target cell.

 

Constraints:

  • 1 <= n, m <= 500
  • m == grid.length
  • n == grid[i].length
  • grid[i][j] is either -1, 0, 1, or 2.
  • There is exactly one -1 in grid.
  • There is exactly one 2 in grid.

Solutions

Solution 1: DFS for Graph Construction + BFS for Shortest Path

We can assume that the robot starts from the coordinate $(0, 0)$. Then, we can use DFS to find all reachable coordinates and record them in the hash table $vis$. In addition, we also need to record the coordinates of the endpoint $target$.

If the endpoint cannot be found, we directly return $-1$. Otherwise, we can use BFS to find the shortest path.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the grid, respectively.

  • /**
 * // This is the GridMaster's API interface.
 * // You should not implement it, or speculate about its implementation
 * class GridMaster {
 *     boolean canMove(char direction);
 *     void move(char direction);
 *     boolean isTarget();
 * }
 */

class Solution {
    private int[] target;
    private GridMaster master;
    private final int n = 2010;
    private final String s = "URDL";
    private final int[] dirs = {-1, 0, 1, 0, -1};
    private final Set<Integer> vis = new HashSet<>();

    public int findShortestPath(GridMaster master) {
        this.master = master;
        dfs(0, 0);
        if (target == null) {
            return -1;
        }
        vis.remove(0);
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {0, 0});
        for (int ans = 0; !q.isEmpty(); ++ans) {
            for (int m = q.size(); m > 0; --m) {
                var p = q.poll();
                if (p[0] == target[0] && p[1] == target[1]) {
                    return ans;
                }
                for (int k = 0; k < 4; ++k) {
                    int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
                    if (vis.remove(x * n + y)) {
                        q.offer(new int[] {x, y});
                    }
                }
            }
        }
        return -1;
    }

    private void dfs(int i, int j) {
        if (master.isTarget()) {
            target = new int[] {i, j};
            return;
        }
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (master.canMove(s.charAt(k)) && vis.add(x * n + y)) {
                master.move(s.charAt(k));
                dfs(x, y);
                master.move(s.charAt((k + 2) % 4));
            }
        }
    }
}

  • /**
 * // This is the GridMaster's API interface.
 * // You should not implement it, or speculate about its implementation
 * class GridMaster {
 *   public:
 *     bool canMove(char direction);
 *     void move(char direction);
 *     boolean isTarget();
 * };
 */

class Solution {
private:
    const int n = 2010;
    int dirs[5] = {-1, 0, 1, 0, -1};
    string s = "URDL";
    int target;
    unordered_set<int> vis;

public:
    int findShortestPath(GridMaster& master) {
        target = n * n;
        vis.insert(0);
        dfs(0, 0, master);
        if (target == n * n) {
            return -1;
        }
        vis.erase(0);
        queue<pair<int, int>> q;
        q.emplace(0, 0);
        for (int ans = 0; q.size(); ++ans) {
            for (int m = q.size(); m; --m) {
                auto [i, j] = q.front();
                q.pop();
                if (i * n + j == target) {
                    return ans;
                }
                for (int k = 0; k < 4; ++k) {
                    int x = i + dirs[k], y = j + dirs[k + 1];
                    if (vis.count(x * n + y)) {
                        vis.erase(x * n + y);
                        q.emplace(x, y);
                    }
                }
            }
        }
        return -1;
    }

    void dfs(int i, int j, GridMaster& master) {
        if (master.isTarget()) {
            target = i * n + j;
        }
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (master.canMove(s[k]) && !vis.count(x * n + y)) {
                vis.insert(x * n + y);
                master.move(s[k]);
                dfs(x, y, master);
                master.move(s[(k + 2) % 4]);
            }
        }
    }
};

  • # """
# This is GridMaster's API interface.
# You should not implement it, or speculate about its implementation
# """
# class GridMaster(object):
#    def canMove(self, direction: str) -> bool:
#
#
#    def move(self, direction: str) -> bool:
#
#
#    def isTarget(self) -> None:
#
#


class Solution(object):
    def findShortestPath(self, master: "GridMaster") -> int:
        def dfs(i: int, j: int):
            if master.isTarget():
                nonlocal target
                target = (i, j)
                return
            for k, c in enumerate(s):
                x, y = i + dirs[k], j + dirs[k + 1]
                if master.canMove(c) and (x, y) not in vis:
                    vis.add((x, y))
                    master.move(c)
                    dfs(x, y)
                    master.move(s[(k + 2) % 4])

        s = "URDL"
        dirs = (-1, 0, 1, 0, -1)
        target = None
        vis = set()
        dfs(0, 0)
        if target is None:
            return -1
        vis.discard((0, 0))
        q = deque([(0, 0)])
        ans = -1
        while q:
            ans += 1
            for _ in range(len(q)):
                i, j = q.popleft()
                if (i, j) == target:
                    return ans
                for a, b in pairwise(dirs):
                    x, y = i + a, j + b
                    if (x, y) in vis:
                        vis.remove((x, y))
                        q.append((x, y))
        return -1

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