**Labyrinthian Possibilities - Amazon Top Interview Questions**

### Problem Statement :

You are given an N by M matrix of 0s and 1s. Starting from the top left corner, how many ways are there to reach the bottom right corner? Mod the result by 10 ** 9 + 7. You can only move right and down. 0 represents an empty space while 1 represents a wall you cannot walk through. The top left corner and bottom right corner will always be 0. Constraints 1 ≤ n, m ≤ 250 where n and m are the number of rows and columns in matrix. Example 1 Input matrix = [ [0, 0, 1], [0, 0, 1], [1, 0, 0] ] Output 2 Explanation There are two ways to get to the bottom right: Right, down, down, right Down, right, down, right Example 2 Input matrix = [ [0, 0, 0], [0, 0, 0], [0, 0, 0] ] Output 6 Explanation There are 6 ways here: Right, right, down, down Down, down, right, right Right, down, right, down Down, right, down, right Right, down, down, right Down, right, right, down Example 3 Input matrix = [ [0, 0, 0], [1, 1, 1], [0, 0, 0] ] Output 0 Explanation There is a wall in the middle preventing us from getting to the bottom right. Example 4 Input matrix = [ [0, 0, 0, 0], [1, 1, 1, 0], [0, 0, 0, 0] ] Output 1 Example 5 Input matrix = [ [0, 0, 0, 0, 0], [1, 1, 1, 0, 0], [0, 0, 0, 0, 0] ] Output 3

### Solution :

` ````
Solution in C++ :
int dp[252][252];
int findsum(vector<vector<int>>& matrix, int i, int j) {
if (i == matrix.size() - 1 && j == matrix[i].size() - 1 && matrix[i][j] == 0) {
return dp[i][j] = 1;
}
if (dp[i][j] != -1) return dp[i][j];
while (i < matrix.size() && j < matrix[i].size() && matrix[i][j] == 0) {
return dp[i][j] = (findsum(matrix, i + 1, j) + findsum(matrix, i, j + 1)) % 1000000007;
}
return 0;
}
int solve(vector<vector<int>>& matrix) {
memset(dp, -1, sizeof(dp));
return findsum(matrix, 0, 0);
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] matrix) {
int MOD = ((int) (Math.pow(10, 9)) + 7);
int m = matrix.length;
int n = matrix[0].length;
int[][] dp = new int[m][n];
dp[0][0] = matrix[0][0] == 1 ? 0 : 1;
if (dp[0][0] == 0)
return 0;
for (int i = 1; i < m; i++) {
if (matrix[i][0] != 1) {
dp[i][0] = dp[i - 1][0];
} else {
dp[i][0] = 0;
}
}
for (int i = 1; i < n; i++) {
if (matrix[0][i] != 1) {
dp[0][i] = dp[0][i - 1];
} else {
dp[0][i] = 0;
}
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (matrix[i][j] != 1) {
dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MOD;
} else {
dp[i][j] = 0;
}
}
}
return dp[m - 1][n - 1];
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, matrix):
if len(matrix) == 0:
return 0
matrix[0][0] = -1
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == 1:
continue
if i > 0 and matrix[i - 1][j] != 1:
matrix[i][j] += matrix[i - 1][j]
if j > 0 and matrix[i][j - 1] != 1:
matrix[i][j] += matrix[i][j - 1]
return (-1 * matrix[-1][-1]) % 1000000007
```

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