Leetcode #1778: Shortest Path in a Hidden Grid
In this guide, we solve Leetcode #1778 Shortest Path in a Hidden Grid in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Depth-First Search, Breadth-First Search, Array, Interactive, Matrix
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
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.
Python Solution
# """
# 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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.