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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