Leetcode #2258: Escape the Spreading Fire
In this guide, we solve Leetcode #2258 Escape the Spreading Fire 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 a 0-indexed 2D integer array grid of size m x n which represents a field. Each cell has one of three values: 0 represents grass, 1 represents fire, 2 represents a wall that you and fire cannot pass through.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Breadth-First Search, Array, Binary Search, Matrix
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: grid = [[0,2,0,0,0,0,0],[0,0,0,2,2,1,0],[0,2,0,0,1,2,0],[0,0,2,2,2,0,2],[0,0,0,0,0,0,0]]
Output: 3
Explanation: The figure above shows the scenario where you stay in the initial position for 3 minutes.
You will still be able to safely reach the safehouse.
Staying for more than 3 minutes will not allow you to safely reach the safehouse.
Python Solution
class Solution:
def maximumMinutes(self, grid: List[List[int]]) -> int:
def spread(q: Deque[int]) -> Deque[int]:
nq = deque()
while q:
i, j = q.popleft()
for a, b in pairwise(dirs):
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and not fire[x][y] and grid[x][y] == 0:
fire[x][y] = True
nq.append((x, y))
return nq
def check(t: int) -> bool:
for i in range(m):
for j in range(n):
fire[i][j] = False
q1 = deque()
for i, row in enumerate(grid):
for j, x in enumerate(row):
if x == 1:
fire[i][j] = True
q1.append((i, j))
while t and q1:
q1 = spread(q1)
t -= 1
if fire[0][0]:
return False
q2 = deque([(0, 0)])
vis = [[False] * n for _ in range(m)]
vis[0][0] = True
while q2:
for _ in range(len(q2)):
i, j = q2.popleft()
if fire[i][j]:
continue
for a, b in pairwise(dirs):
x, y = i + a, j + b
if (
0 <= x < m
and 0 <= y < n
and not vis[x][y]
and not fire[x][y]
and grid[x][y] == 0
):
if x == m - 1 and y == n - 1:
return True
vis[x][y] = True
q2.append((x, y))
q1 = spread(q1)
return False
m, n = len(grid), len(grid[0])
l, r = -1, m * n
dirs = (-1, 0, 1, 0, -1)
fire = [[False] * n for _ in range(m)]
while l < r:
mid = (l + r + 1) >> 1
if check(mid):
l = mid
else:
r = mid - 1
return int(1e9) if l == m * n else l
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.