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Java-based LeetCode algorithm problem solutions, regularly updated
package g2201_2300.s2258_escape_the_spreading_fire;
// #Hard #Array #Breadth_First_Search #Binary_Search #Matrix
// #2022_06_18_Time_33_ms_(77.40%)_Space_44.2_MB_(94.95%)
import java.util.ArrayDeque;
/**
* 2258 - Escape the Spreading Fire\.
*
* Hard
*
* 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.
*
* **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.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`
**/
public class Solution {
private int[][] setFire(int[][] grid, int[][] dir) {
int n = grid.length;
int m = grid[0].length;
ArrayDeque bfs = new ArrayDeque<>();
int[][] fire = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
fire[i][j] = Integer.MAX_VALUE;
if (grid[i][j] == 1) {
fire[i][j] = 0;
bfs.add(i * m + j);
}
if (grid[i][j] == 2) {
fire[i][j] = 0;
}
}
}
while (!bfs.isEmpty()) {
int rm = bfs.removeFirst();
int x = rm / m;
int y = rm % m;
for (int d = 0; d < 4; d++) {
int nx = x + dir[d][0];
int ny = y + dir[d][1];
if (nx >= 0 && ny >= 0 && nx < n && ny < m && fire[nx][ny] == Integer.MAX_VALUE) {
fire[nx][ny] = fire[x][y] + 1;
bfs.add(nx * m + ny);
}
}
}
return fire;
}
private boolean isPoss(int[][] fire, int[][] dir, int time) {
if (time >= fire[0][0]) {
return false;
}
int n = fire.length;
int m = fire[0].length;
ArrayDeque bfs = new ArrayDeque<>();
bfs.add(0);
boolean[][] isVis = new boolean[n][m];
isVis[0][0] = true;
while (!bfs.isEmpty()) {
int size = bfs.size();
while (size-- > 0) {
int rm = bfs.removeFirst();
int x = rm / m;
int y = rm % m;
if (x == n - 1 && y == m - 1) {
return true;
}
for (int d = 0; d < 4; d++) {
int nx = x + dir[d][0];
int ny = y + dir[d][1];
if (nx >= 0 && ny >= 0 && nx < n && ny < m && !isVis[nx][ny]) {
if (nx == n - 1 && ny == m - 1) {
if (time + 1 <= fire[nx][ny]) {
isVis[nx][ny] = true;
bfs.add(nx * m + ny);
}
} else {
if (time + 1 < fire[nx][ny]) {
isVis[nx][ny] = true;
bfs.add(nx * m + ny);
}
}
}
}
}
time++;
}
return false;
}
public int maximumMinutes(int[][] grid) {
int[][] dir = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
int[][] fire = setFire(grid, dir);
int lo = 0;
int hi = (int) 1e9;
while (lo <= hi) {
int mid = ((hi - lo) >> 1) + lo;
if (isPoss(fire, dir, mid)) {
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return hi;
}
}
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