Finding the number of unique paths on a gird - c++

Question

I'm trying to understand how to solve the problem of finding all unique paths in a grid using dynamic programming:
A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below). How many possible unique paths are there?
I was looking at this article and I was wondering why in the below solution, the matrix is initialized at M_MAX + 2 and N_MAX + 2, and also why in the function signature of backtrack, why the last parameter is initialized with int mat[][N_MAX+2]
const int M_MAX = 100;
const int N_MAX = 100;
int backtrack(int r, int c, int m, int n, int mat[][N_MAX+2]) {
if (r == m && c == n)
return 1;
if (r > m || c > n)
return 0;
if (mat[r+1][c] == -1)
mat[r+1][c] = backtrack(r+1, c, m, n, mat);
if (mat[r][c+1] == -1)
mat[r][c+1] = backtrack(r, c+1, m, n, mat);
return mat[r+1][c] + mat[r][c+1];
}
int bt(int m, int n) {
int mat[M_MAX+2][N_MAX+2];
for (int i = 0; i < M_MAX+2; i++) {
for (int j = 0; j < N_MAX+2; j++) {
mat[i][j] = -1;
}
}
return backtrack(1, 1, m, n, mat);
}
Then in the author's bottom-up approach solution:
const int M_MAX = 100;
const int N_MAX = 100;
int dp(int m, int n) {
int mat[M_MAX+2][N_MAX+2] = {0};
mat[m][n+1] = 1;
for (int r = m; r >= 1; r--)
for (int c = n; c >= 1; c--)
mat[r][c] = mat[r+1][c] + mat[r][c+1];
return mat[1][1];
}
I don't know what the purpose of the line mat[m][n+1] = 1; serves.
I'm not familiar with Java, so I apologize if these boil down to syntactical or language-specific questions.

Answer 1

Firstly, notice that the author and the second solution both use 1-based indexing. So, of course, mat[M_MAX+1][N_MAX+1] would be quite justified.
Now, notice the logic the author is using.
mat[r][c] = mat[r+1][c] + mat[r][c+1];
Hence, to prevent r+1 or c+1 from going out of bounds when c = n+1 or r = m+1, instead of adding an if-statement like this:
if (r == m)
mat[r][c] = mat[r][c+1];
if (c == n)
mat[r][c] = mat[r+1][c];
He has decided to simply add an extra row or column with 0 value stored in it. Hence:
mat[M_MAX+2][N_MAX+2] = {0};
Finally, in a bottom up approach, one must initialize mat[m][n] to 1. Instead of doing that, knowing that mat[m][n] = mat[m+1][n] + mat[m][n+1];, he initialized :
mat[m][n+1] = 1; // mat[m+1][n] = 0;
Feel free to ask any questions in comments.