Leetcode #505: The Maze II
In this guide, we solve Leetcode #505 The Maze II 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
There is a ball in a maze with empty spaces (represented as 0) and walls (represented as 1). The ball can go through the empty spaces by rolling up, down, left or right, but it won't stop rolling until hitting a wall.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Depth-First Search, Breadth-First Search, Graph, Array, Matrix, Shortest Path, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
Input: maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [4,4]
Output: 12
Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.
The length of the path is 1 + 1 + 3 + 1 + 2 + 2 + 2 = 12.
Python Solution
class Solution:
def shortestDistance(
self, maze: List[List[int]], start: List[int], destination: List[int]
) -> int:
m, n = len(maze), len(maze[0])
dirs = (-1, 0, 1, 0, -1)
si, sj = start
di, dj = destination
q = deque([(si, sj)])
dist = [[inf] * n for _ in range(m)]
dist[si][sj] = 0
while q:
i, j = q.popleft()
for a, b in pairwise(dirs):
x, y, k = i, j, dist[i][j]
while 0 <= x + a < m and 0 <= y + b < n and maze[x + a][y + b] == 0:
x, y, k = x + a, y + b, k + 1
if k < dist[x][y]:
dist[x][y] = k
q.append((x, y))
return -1 if dist[di][dj] == inf else dist[di][dj]
Complexity
The time complexity is O(n log n). The space complexity is O(n).
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.