# Wildfire - Amazon Top Interview Questions

### Problem Statement :

```You are given a two-dimensional integer matrix representing a forest where a cell is:

0 if it's empty
1 if it's a tree
2 if it's a tree on fire
Every day, a tree catches fire if there is an adjacent (top, down, left, right) tree that's also on fire. Return the number of days it would take for every tree to be on fire. If it's not possible, return -1.

Constraints

0 ≤ n * m ≤ 200,000 where n and m are the number of rows and columns in matrix

Example 1

Input

matrix = [
[1, 1, 1],
[1, 2, 1],
[1, 1, 1]
]

Output

2

Explanation

On the first day fire will spread to everywhere except the corner trees and then the next day they will spread to the corner trees.

Example 2

Input

matrix = [
[1, 1, 1],
[0, 0, 0],
[1, 2, 1]
]

Output

-1

Explanation

Fire on bottom row can't move across the middle row because it's empty.```

### Solution :

```                        ```Solution in C++ :

int solve(vector<vector<int>>& matrix) {
int n = matrix.size();
if (n == 0) return 0;
int m = matrix[0].size();
int total = 0;

queue<pair<int, int>> q;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (matrix[i][j] != 0) total++;
if (matrix[i][j] == 2) q.push({i, j});
}
}

int dir[5] = {0, -1, 0, 1, 0};
int cnt = 0, days = 0;
while (!q.empty()) {
int sz = q.size();
cnt += sz;

for (int j = 0; j < sz; j++) {
int cx = q.front().first;
int cy = q.front().second;
q.pop();

for (int i = 0; i < 4; i++) {
int nx = cx + dir[i];
int ny = cy + dir[i + 1];

if (nx < 0 || ny < 0 || nx >= n || ny >= m || matrix[nx][ny] != 1) continue;
matrix[nx][ny] = 2;
q.push({nx, ny});
}
}
if (!q.empty()) days++;
}
return (cnt == total) ? days : -1;
}
// TC :O(N*M), SC :O(N*M)```
```

```                        ```Solution in Python :

class Solution:
def solve(self, matrix):
# basic check
if matrix == []:
return 0

# initialization of the variables
n, m, dist = len(matrix), len(matrix[0]), 0

# filling the queue with all the trees which are already on fire
d = deque((i, j, dist) for i in range(n) for j in range(m) if matrix[i][j] == 2)
dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]

# performing bfs
while d:
i, j, dist = d.popleft()
for x, y in dirs:
r, c = i + x, j + y
if 0 <= r < n and 0 <= c < m and matrix[r][c] == 1:
d.append((r, c, dist + 1))
# print('burnt matrix[{}][{}]'.format(r,c),'on day ',dist+1)
matrix[r][c] = 0

# checking if any tree is left unburnt
for i in range(n):
for j in range(m):
if matrix[i][j] == 1:
return -1
return dist```
```

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