**Nearest Bus Stop From a House - Google Top Interview Questions**

### Problem Statement :

You are given a two-dimensional integer matrix containing 0s, 1s, 2s, and 3s where 0 represents an empty cell 1 represents a wall 2 represents a house 3 represents a bus stop Return the shortest distance from any house to any bus stop. You can move up, down, left, and right but you can't move through a house or a wall cell. If there's no solution, return -1. Constraints n, m ≤ 250 where n and m are the number of rows and columns in matrix Example 1 Input matrix = [ [2, 1, 3, 0], [1, 1, 1, 1], [0, 3, 0, 0], [0, 0, 0, 2] ] Output 3 Explanation We can go from the house at matrix[3][3] to bus stop at matrix[2][1].

### Solution :

` ````
Solution in C++ :
int r, c;
int dp[250][250];
int solve(vector<vector<int>>& m) {
r = m.size();
c = m[0].size();
queue<pair<int, int>> q;
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
if (m[i][j] == 2) {
dp[i][j] = 0;
q.emplace(i, j);
} else {
dp[i][j] = 1e9;
}
}
}
while (q.size()) {
auto [x, y] = q.front();
q.pop();
if (m[x][y] == 3) return dp[x][y];
int dx[4]{-1, 0, 1, 0};
int dy[4]{0, 1, 0, -1};
for (int k = 0; k < 4; k++) {
int nx = x + dx[k];
int ny = y + dy[k];
if (nx < 0 || nx >= r || ny < 0 || ny >= c || m[nx][ny] == 1) continue;
if (dp[nx][ny] > 1 + dp[x][y]) {
dp[nx][ny] = 1 + dp[x][y];
q.emplace(nx, ny);
}
}
}
return -1;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] matrix) {
int N = matrix.length;
if (N == 0)
return 0;
int M = matrix[0].length;
int INF = 1000000;
ArrayDeque<int[]> bfs = new ArrayDeque<int[]>();
int[][] dist = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
dist[i][j] = INF;
if (matrix[i][j] == 2) {
// this is a house and a source in the bfs
dist[i][j] = 0;
bfs.add(new int[] {i, j});
}
}
}
int ans = INF;
int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
while (!bfs.isEmpty()) {
int[] cell = bfs.pollFirst();
int d = dist[cell[0]][cell[1]];
if (matrix[cell[0]][cell[1]] == 3) {
ans = d;
break;
}
for (int[] dir : dirs) {
int newR = cell[0] + dir[0];
int newC = cell[1] + dir[1];
if (newR >= 0 && newR < N && newC >= 0 && newC < M && matrix[newR][newC] % 3 == 0
&& dist[newR][newC] == INF) {
dist[newR][newC] = d + 1;
bfs.add(new int[] {newR, newC});
}
}
}
return (ans == INF ? -1 : ans);
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, A):
R, C = len(A), len(A[0])
queue = []
for r, row in enumerate(A):
for c, v in enumerate(row):
if v == 2:
queue.append((r, c))
dist = {loc: 0 for loc in queue}
for r, c in queue:
if A[r][c] == 3:
return dist[r, c]
for nr, nc in ((r - 1, c), (r, c - 1), (r + 1, c), (r, c + 1)):
if 0 <= nr < R and 0 <= nc < C and A[nr][nc] != 1 and (nr, nc) not in dist:
queue.append((nr, nc))
dist[nr, nc] = dist[r, c] + 1
return -1
```

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