In this guide, you will get 2258. Escape the Spreading Fire LeetCode Solution with the best time and space complexity. The solution to Escape the Spreading Fire problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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.
You are situated in the top-left cell, (0, 0), and you want to travel to the safehouse at the bottom-right cell, (m – 1, n – 1). Every minute, you may move to an adjacent grass cell. After your move, every fire cell will spread to all adjacent cells that are not walls.
Return the maximum number of minutes that you can stay in your initial position before moving while still safely reaching the safehouse. If this is impossible, return -1. If you can always reach the safehouse regardless of the minutes stayed, return 109.
Note that even if the fire spreads to the safehouse immediately after you have reached it, it will be counted as safely reaching the safehouse.
A cell is adjacent to another cell if the former is directly north, east, south, or west of the latter (i.e., their sides are touching).
Example 1:
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.
Example 2:
Input: grid = [[0,0,0,0],[0,1,2,0],[0,2,0,0]]
Output: -1
Explanation: The figure above shows the scenario where you immediately move towards the safehouse.
Fire will spread to any cell you move towards and it is impossible to safely reach the safehouse.
Thus, -1 is returned.
Input: grid = [[0,0,0],[2,2,0],[1,2,0]]
Output: 1000000000
Explanation: The figure above shows the initial grid.
Notice that the fire is contained by walls and you will always be able to safely reach the safehouse.
Thus, 109 is returned.
Constraints:
m == grid.length
n == grid[i].length
2 <= m, n <= 300
4 <= m * n <= 2 * 104
grid[i][j] is either 0, 1, or 2.
grid[0][0] == grid[m – 1][n – 1] == 0
Complexity Analysis
Time Complexity: O(mn\log mn)
Space Complexity: O(mn)
2258. Escape the Spreading Fire LeetCode Solution in C++
class Solution {
public:
int maximumMinutes(vector<vector<int>>& grid) {
const int kMax = grid.size() * grid[0].size();
vector<vector<int>> fireMinute(grid.size(),
vector<int>(grid[0].size(), -1));
buildFireGrid(grid, fireMinute);
int ans = -1;
int l = 0;
int r = kMax;
while (l <= r) {
const int m = (l + r) / 2;
if (canStayFor(grid, fireMinute, m)) {
ans = m;
l = m + 1;
} else {
r = m - 1;
}
}
return ans == kMax ? 1'000'000'000 : ans;
}
private:
static constexpr int dirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
void buildFireGrid(const vector<vector<int>>& grid,
vector<vector<int>>& fireMinute) {
queue<pair<int, int>> q;
for (int i = 0; i < grid.size(); ++i)
for (int j = 0; j < grid[0].size(); ++j)
if (grid[i][j] == 1) { // the fire
q.emplace(i, j);
fireMinute[i][j] = 0;
}
for (int minuteFromFire = 1; !q.empty(); ++minuteFromFire)
for (int sz = q.size(); sz > 0; --sz) {
const auto [i, j] = q.front();
q.pop();
for (const auto& [dx, dy] : dirs) {
const int x = i + dx;
const int y = j + dy;
if (x < 0 || x == grid.size() || y < 0 || y == grid[0].size())
continue;
if (grid[x][y] == 2) // the wall
continue;
if (fireMinute[x][y] != -1)
continue;
fireMinute[x][y] = minuteFromFire;
q.emplace(x, y);
}
}
}
bool canStayFor(const vector<vector<int>>& grid,
const vector<vector<int>>& fireMinute, int minute) {
queue<pair<int, int>> q{{{0, 0}}}; // the start position
vector<vector<bool>> seen(grid.size(), vector<bool>(grid[0].size()));
seen[0][0] = true;
while (!q.empty()) {
++minute;
for (int sz = q.size(); sz > 0; --sz) {
const auto [i, j] = q.front();
q.pop();
for (const auto& [dx, dy] : dirs) {
const int x = i + dx;
const int y = j + dy;
if (x < 0 || x == grid.size() || y < 0 || y == grid[0].size())
continue;
if (grid[x][y] == 2) // the wall
continue;
if (x == grid.size() - 1 && y == grid[0].size() - 1) {
if (fireMinute[x][y] != -1 && fireMinute[x][y] < minute)
continue;
return true;
}
if (fireMinute[x][y] != -1 && fireMinute[x][y] <= minute)
continue;
if (seen[x][y])
continue;
q.emplace(x, y);
seen[x][y] = true;
}
}
}
return false;
}
}; /* code provided by PROGIEZ */
2258. Escape the Spreading Fire LeetCode Solution in Java
class Solution {
public int maximumMinutes(int[][] grid) {
final int kMax = grid.length * grid[0].length;
int[][] fireMinute = new int[grid.length][grid[0].length];
Arrays.stream(fireMinute).forEach(A -> Arrays.fill(A, -1));
buildFireGrid(grid, fireMinute);
int ans = -1;
int l = 0;
int r = kMax;
while (l <= r) {
final int m = (l + r) / 2;
if (canStayFor(grid, fireMinute, m)) {
ans = m;
l = m + 1;
} else {
r = m - 1;
}
}
return ans == kMax ? 1_000_000_000 : ans;
}
private static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
private void buildFireGrid(int[][] grid, int[][] fireMinute) {
Queue<Pair<Integer, Integer>> q = new ArrayDeque<>();
for (int i = 0; i < grid.length; ++i)
for (int j = 0; j < grid[0].length; ++j)
if (grid[i][j] == 1) { // the fire
q.offer(new Pair<>(i, j));
fireMinute[i][j] = 0;
}
for (int minuteFromFire = 1; !q.isEmpty(); ++minuteFromFire)
for (int sz = q.size(); sz > 0; --sz) {
final int i = q.peek().getKey();
final int j = q.poll().getValue();
for (int[] dir : dirs) {
final int x = i + dir[0];
final int y = j + dir[1];
if (x < 0 || x == grid.length || y < 0 || y == grid[0].length)
continue;
if (grid[x][y] == 2) // the wall
continue;
if (fireMinute[x][y] != -1)
continue;
fireMinute[x][y] = minuteFromFire;
q.offer(new Pair<>(x, y));
}
}
}
boolean canStayFor(int[][] grid, int[][] fireMinute, int minute) {
Queue<Pair<Integer, Integer>> q =
new ArrayDeque<>(List.of(new Pair<>(0, 0))); // the start position
boolean[][] seen = new boolean[grid.length][grid[0].length];
seen[0][0] = true;
while (!q.isEmpty()) {
++minute;
for (int sz = q.size(); sz > 0; --sz) {
final int i = q.peek().getKey();
final int j = q.poll().getValue();
for (int[] dir : dirs) {
final int x = i + dir[0];
final int y = j + dir[1];
if (x < 0 || x == grid.length || y < 0 || y == grid[0].length)
continue;
if (grid[x][y] == 2) // the wall
continue;
if (x == grid.length - 1 && y == grid[0].length - 1) {
if (fireMinute[x][y] != -1 && fireMinute[x][y] < minute)
continue;
return true;
}
if (fireMinute[x][y] != -1 && fireMinute[x][y] <= minute)
continue;
if (seen[x][y])
continue;
q.offer(new Pair<>(x, y));
seen[x][y] = true;
}
}
}
return false;
}
} // code provided by PROGIEZ
2258. Escape the Spreading Fire LeetCode Solution in Python
class Solution:
def maximumMinutes(self, grid: list[list[int]]) -> int:
dirs = ((0, 1), (1, 0), (0, -1), (-1, 0))
kMax = len(grid) * len(grid[0])
fireGrid = [[-1] * len(grid[0]) for _ in range(len(grid[0]))]
self._buildFireGrid(grid, fireGrid, dirs)
ans = -1
l = 0
r = kMax
while l <= r:
m = (l + r) // 2
if self._canStayFor(grid, fireGrid, m, dirs):
ans = m
l = m + 1
else:
r = m - 1
return 1e9 if ans == kMax else ans
def _buildFireGrid(
self,
grid: list[list[int]],
fireMinute: list[list[int]],
dirs: list[int],
) -> None:
minuteFromFire = 0
q = collections.deque()
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1: # the fire
q.append((i, j))
fireMinute[i][j] = 0
while q:
minuteFromFire += 1
for _ in range(len(q)):
i, j = q.popleft()
for dx, dy in dirs:
x = i + dx
y = j + dy
if x < 0 or x == len(grid) or y < 0 or y == len(grid[0]):
continue
if grid[x][y] == 2: # the wall
continue
if fireMinute[x][y] != -1:
continue
fireMinute[x][y] = minuteFromFire
q.append((x, y))
def _canStayFor(
self,
grid: list[list[int]],
fireMinute: list[list[int]],
minute: int, dirs: list[int],
) -> bool:
q = collections.deque([(0, 0)]) # the start position
seen = {(0, 0)}
while q:
minute += 1
for _ in range(len(q)):
i, j = q.popleft()
for dx, dy in dirs:
x = i + dx
y = j + dy
if x < 0 or x == len(grid) or y < 0 or y == len(grid[0]):
continue
if grid[x][y] == 2: # the wall
continue
if x == len(grid) - 1 and y == len(grid[0]) - 1:
if fireMinute[x][y] != -1 and fireMinute[x][y] < minute:
continue
return True
if fireMinute[x][y] != -1 and fireMinute[x][y] <= minute:
continue
if seen[x][y]:
continue
q.append((x, y))
seen.add((x, y))
return False # code by PROGIEZ
