Leetcode #1926: Nearest Exit from Entrance in Maze
In this guide, we solve Leetcode #1926 Nearest Exit from Entrance in Maze 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
You are given an m x n matrix maze (0-indexed) with empty cells (represented as '.') and walls (represented as '+'). You are also given the entrance of the maze, where entrance = [entrancerow, entrancecol] denotes the row and column of the cell you are initially standing at.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Breadth-First Search, Array, Matrix
Intuition
We need level-by-level exploration or shortest steps, which is ideal for BFS.
A queue naturally models the frontier of the search.
Approach
Push initial nodes into a queue and expand in layers.
Track visited nodes to prevent cycles.
Steps:
- Initialize queue with start nodes.
- Process level by level.
- Track visited nodes.
Example
Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2]
Output: 1
Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3].
Initially, you are at the entrance cell [1,2].
- You can reach [1,0] by moving 2 steps left.
- You can reach [0,2] by moving 1 step up.
It is impossible to reach [2,3] from the entrance.
Thus, the nearest exit is [0,2], which is 1 step away.
Python Solution
class Solution:
def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
m, n = len(maze), len(maze[0])
i, j = entrance
q = deque([(i, j)])
maze[i][j] = "+"
ans = 0
while q:
ans += 1
for _ in range(len(q)):
i, j = q.popleft()
for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and maze[x][y] == ".":
if x == 0 or x == m - 1 or y == 0 or y == n - 1:
return ans
q.append((x, y))
maze[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.