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:
0represents grass,1represents fire,2represents 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.
Example 3:
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.lengthn == grid[i].length2 <= m, n <= 3004 <= m * n <= 2 * 104grid[i][j]is either0,1, or2.grid[0][0] == grid[m - 1][n - 1] == 0
class Solution:
def maximumMinutes(self, grid: List[List[int]]) -> int:
def spread(fire, q):
nf = deque()
while 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 not fire[x][y] and grid[x][y] == 0:
fire[x][y] = True
nf.append((x, y))
return nf
def check(t):
fire = [[False] * n for _ in range(m)]
f = deque()
for i, row in enumerate(grid):
for j, v in enumerate(row):
if v == 1:
fire[i][j] = True
f.append((i, j))
while t and f:
f = spread(fire, f)
t -= 1
if fire[0][0]:
return False
q = deque([(0, 0)])
vis = [[False] * n for _ in range(m)]
vis[0][0] = True
while q:
for _ in range(len(q)):
i, j = q.popleft()
if fire[i][j]:
continue
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 not fire[x][y]
and not vis[x][y]
and grid[x][y] == 0
):
if x == m - 1 and y == n - 1:
return True
vis[x][y] = True
q.append((x, y))
f = spread(fire, f)
return False
m, n = len(grid), len(grid[0])
left, right = -1, m * n
while left < right:
mid = (left + right + 1) >> 1
if check(mid):
left = mid
else:
right = mid - 1
return int(1e9) if left == m * n else leftclass Solution {
private static int[] dirs = {-1, 0, 1, 0, -1};
private int[][] grid;
private int m;
private int n;
public int maximumMinutes(int[][] grid) {
this.grid = grid;
m = grid.length;
n = grid[0].length;
int left = -1, right = m * n;
while (left < right) {
int mid = (left + right + 1) >> 1;
if (check(mid)) {
left = mid;
} else {
right = mid - 1;
}
}
return left == m * n ? (int) 1e9 : left;
}
private boolean check(int t) {
boolean[][] fire = new boolean[m][n];
Deque<int[]> f = new ArrayDeque<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
fire[i][j] = true;
f.offer(new int[] {i, j});
}
}
}
while (t-- > 0 && !f.isEmpty()) {
f = spread(fire, f);
}
if (fire[0][0]) {
return false;
}
Deque<int[]> q = new ArrayDeque<>();
boolean[][] vis = new boolean[m][n];
q.offer(new int[] {0, 0});
vis[0][0] = true;
while (!q.isEmpty()) {
for (int i = q.size(); i > 0; --i) {
int[] p = q.poll();
if (fire[p[0]][p[1]]) {
continue;
}
for (int k = 0; k < 4; ++k) {
int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && !vis[x][y]
&& grid[x][y] == 0) {
if (x == m - 1 && y == n - 1) {
return true;
}
vis[x][y] = true;
q.offer(new int[] {x, y});
}
}
}
f = spread(fire, f);
}
return false;
}
private Deque<int[]> spread(boolean[][] fire, Deque<int[]> q) {
Deque<int[]> nf = new ArrayDeque<>();
while (!q.isEmpty()) {
int[] p = q.poll();
for (int k = 0; k < 4; ++k) {
int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && grid[x][y] == 0) {
fire[x][y] = true;
nf.offer(new int[] {x, y});
}
}
}
return nf;
}
}class Solution {
public:
vector<int> dirs = {-1, 0, 1, 0, -1};
int maximumMinutes(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
int left = -1, right = m * n;
while (left < right) {
int mid = (left + right + 1) >> 1;
if (check(mid, grid))
left = mid;
else
right = mid - 1;
}
return left == m * n ? 1e9 : left;
}
bool check(int t, vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<vector<bool>> fire(m, vector<bool>(n));
queue<vector<int>> f;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
fire[i][j] = true;
f.push({i, j});
}
}
}
while (t-- && f.size()) f = spread(fire, f, grid);
queue<vector<int>> q;
vector<vector<bool>> vis(m, vector<bool>(n));
q.push({0, 0});
vis[0][0] = true;
while (!q.empty()) {
for (int i = q.size(); i > 0; --i) {
auto p = q.front();
q.pop();
if (fire[p[0]][p[1]]) continue;
for (int k = 0; k < 4; ++k) {
int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && !vis[x][y] && grid[x][y] == 0) {
if (x == m - 1 && y == n - 1) return true;
vis[x][y] = true;
q.push({x, y});
}
}
}
f = spread(fire, f, grid);
}
return false;
}
queue<vector<int>> spread(vector<vector<bool>>& fire, queue<vector<int>>& f, vector<vector<int>>& grid) {
queue<vector<int>> nf;
int m = grid.size(), n = grid[0].size();
while (!f.empty()) {
auto p = f.front();
f.pop();
for (int k = 0; k < 4; ++k) {
int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && grid[x][y] == 0) {
fire[x][y] = true;
nf.push({x, y});
}
}
}
return nf;
}
};func maximumMinutes(grid [][]int) int {
m, n := len(grid), len(grid[0])
dirs := []int{-1, 0, 1, 0, -1}
spread := func(fire [][]bool, q [][]int) [][]int {
nf := [][]int{}
for len(q) > 0 {
p := q[0]
q = q[1:]
for k := 0; k < 4; k++ {
x, y := p[0]+dirs[k], p[1]+dirs[k+1]
if x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && grid[x][y] == 0 {
fire[x][y] = true
nf = append(nf, []int{x, y})
}
}
}
return nf
}
check := func(t int) bool {
fire := make([][]bool, m)
vis := make([][]bool, m)
f := [][]int{}
for i, row := range grid {
fire[i] = make([]bool, n)
vis[i] = make([]bool, n)
for j, v := range row {
if v == 1 {
fire[i][j] = true
f = append(f, []int{i, j})
}
}
}
for t > 0 && len(f) > 0 {
f = spread(fire, f)
t--
}
if fire[0][0] {
return false
}
q := [][]int{{0, 0}}
vis[0][0] = true
for len(q) > 0 {
for i := len(q); i > 0; i-- {
p := q[0]
q = q[1:]
if fire[p[0]][p[1]] {
continue
}
for k := 0; k < 4; k++ {
x, y := p[0]+dirs[k], p[1]+dirs[k+1]
if x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && !vis[x][y] && grid[x][y] == 0 {
if x == m-1 && y == n-1 {
return true
}
vis[x][y] = true
q = append(q, []int{x, y})
}
}
}
f = spread(fire, f)
}
return false
}
left, right := -1, m*n
for left < right {
mid := (left + right + 1) >> 1
if check(mid) {
left = mid
} else {
right = mid - 1
}
}
if left == m*n {
return int(1e9)
}
return left
}