| comments | difficulty | edit_url | rating | source | tags | ||||
|---|---|---|---|---|---|---|---|---|---|
true |
Hard |
2346 |
Biweekly Contest 77 Q4 |
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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
We notice that if a stay time
We define the left boundary of binary search as
The key problem is how to determine whether a stay time
The time complexity is
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 lclass Solution {
private int[][] grid;
private boolean[][] fire;
private boolean[][] vis;
private final int[] dirs = {-1, 0, 1, 0, -1};
private int m;
private int n;
public int maximumMinutes(int[][] grid) {
m = grid.length;
n = grid[0].length;
this.grid = grid;
fire = new boolean[m][n];
vis = new boolean[m][n];
int l = -1, r = m * n;
while (l < r) {
int mid = (l + r + 1) >> 1;
if (check(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l == m * n ? 1000000000 : l;
}
private boolean check(int t) {
for (int i = 0; i < m; ++i) {
Arrays.fill(fire[i], false);
Arrays.fill(vis[i], false);
}
Deque<int[]> q1 = new ArrayDeque<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
q1.offer(new int[] {i, j});
fire[i][j] = true;
}
}
}
for (; t > 0 && !q1.isEmpty(); --t) {
q1 = spread(q1);
}
if (fire[0][0]) {
return false;
}
Deque<int[]> q2 = new ArrayDeque<>();
q2.offer(new int[] {0, 0});
vis[0][0] = true;
for (; !q2.isEmpty(); q1 = spread(q1)) {
for (int d = q2.size(); d > 0; --d) {
int[] p = q2.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;
q2.offer(new int[] {x, y});
}
}
}
}
return false;
}
private Deque<int[]> spread(Deque<int[]> q) {
Deque<int[]> nq = 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;
nq.offer(new int[] {x, y});
}
}
}
return nq;
}
}class Solution {
public:
int maximumMinutes(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
bool vis[m][n];
bool fire[m][n];
int dirs[5] = {-1, 0, 1, 0, -1};
auto spread = [&](queue<pair<int, int>>& q) {
queue<pair<int, int>> nq;
while (q.size()) {
auto [i, j] = q.front();
q.pop();
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && grid[x][y] == 0) {
fire[x][y] = true;
nq.emplace(x, y);
}
}
}
return nq;
};
auto check = [&](int t) {
memset(vis, false, sizeof(vis));
memset(fire, false, sizeof(fire));
queue<pair<int, int>> q1;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
q1.emplace(i, j);
fire[i][j] = true;
}
}
}
for (; t && q1.size(); --t) {
q1 = spread(q1);
}
if (fire[0][0]) {
return false;
}
queue<pair<int, int>> q2;
q2.emplace(0, 0);
vis[0][0] = true;
for (; q2.size(); q1 = spread(q1)) {
for (int d = q2.size(); d; --d) {
auto [i, j] = q2.front();
q2.pop();
if (fire[i][j]) {
continue;
}
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && !fire[x][y] && grid[x][y] == 0) {
if (x == m - 1 && y == n - 1) {
return true;
}
vis[x][y] = true;
q2.emplace(x, y);
}
}
}
}
return false;
};
int l = -1, r = m * n;
while (l < r) {
int mid = (l + r + 1) >> 1;
if (check(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l == m * n ? 1e9 : l;
}
};func maximumMinutes(grid [][]int) int {
m, n := len(grid), len(grid[0])
fire := make([][]bool, m)
vis := make([][]bool, m)
dirs := [5]int{-1, 0, 1, 0, -1}
for i := range fire {
fire[i] = make([]bool, n)
vis[i] = make([]bool, n)
}
l, r := -1, m*n
spread := func(q [][2]int) [][2]int {
nq := [][2]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
nq = append(nq, [2]int{x, y})
}
}
}
return nq
}
check := func(t int) bool {
for i := range fire {
for j := range fire[i] {
fire[i][j] = false
vis[i][j] = false
}
}
q1 := [][2]int{}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if grid[i][j] == 1 {
q1 = append(q1, [2]int{i, j})
fire[i][j] = true
}
}
}
for ; t > 0 && len(q1) > 0; t-- {
q1 = spread(q1)
}
q2 := [][2]int{{0, 0}}
vis[0][0] = true
for ; len(q2) > 0; q1 = spread(q1) {
for d := len(q2); d > 0; d-- {
p := q2[0]
q2 = q2[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
q2 = append(q2, [2]int{x, y})
}
}
}
}
return false
}
for l < r {
mid := (l + r + 1) >> 1
if check(mid) {
l = mid
} else {
r = mid - 1
}
}
if l == m*n {
return int(1e9)
}
return l
}function maximumMinutes(grid: number[][]): number {
const m = grid.length;
const n = grid[0].length;
const fire = Array.from({ length: m }, () => Array.from({ length: n }, () => false));
const vis = Array.from({ length: m }, () => Array.from({ length: n }, () => false));
const dirs: number[] = [-1, 0, 1, 0, -1];
let [l, r] = [-1, m * n];
const spread = (q: number[][]): number[][] => {
const nq: number[][] = [];
while (q.length) {
const [i, j] = q.shift()!;
for (let k = 0; k < 4; ++k) {
const [x, y] = [i + dirs[k], j + dirs[k + 1]];
if (x >= 0 && x < m && y >= 0 && y < n && !fire[x][y] && grid[x][y] === 0) {
fire[x][y] = true;
nq.push([x, y]);
}
}
}
return nq;
};
const check = (t: number): boolean => {
for (let i = 0; i < m; ++i) {
fire[i].fill(false);
vis[i].fill(false);
}
let q1: number[][] = [];
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (grid[i][j] === 1) {
q1.push([i, j]);
fire[i][j] = true;
}
}
}
for (; t && q1.length; --t) {
q1 = spread(q1);
}
if (fire[0][0]) {
return false;
}
const q2: number[][] = [[0, 0]];
vis[0][0] = true;
for (; q2.length; q1 = spread(q1)) {
for (let d = q2.length; d; --d) {
const [i, j] = q2.shift()!;
if (fire[i][j]) {
continue;
}
for (let k = 0; k < 4; ++k) {
const [x, y] = [i + dirs[k], j + dirs[k + 1]];
if (
x >= 0 &&
x < m &&
y >= 0 &&
y < n &&
!vis[x][y] &&
!fire[x][y] &&
grid[x][y] === 0
) {
if (x === m - 1 && y === n - 1) {
return true;
}
vis[x][y] = true;
q2.push([x, y]);
}
}
}
}
return false;
};
while (l < r) {
const mid = (l + r + 1) >> 1;
if (check(mid)) {
l = mid;
} else {
r = mid - 1;
}
}
return l === m * n ? 1e9 : l;
}