**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
```

## View More Similar Problems

## Merging Communities

People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w

View Solution →## Components in a graph

There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu

View Solution →## Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

View Solution →## Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

View Solution →## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

View Solution →## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

View Solution →