The problem is we have to count number of paths between a source and a destination in a graph. The brute force approach is using backtracking which is $O(n!)$.
Is there any better solution to this problem?
Note that we have to count paths so we can't visit the same node again in a path.
I think this problem is np-hard and is similar to finding hamiltonian path in a graph. So, I can use Held-Karp algorithm to solve it in $O(n^2*2^n)$ using dynamic programming. Am I right?
Let $A$ denote the adjacency matrix, $A^k$ its $k$th power, and $(A^k)_{ij}$ the entry of $A^k$ at row $i$, column $j$. Then $(A^k)_{ij}$ is a count of the number of paths of length $k$ from node $i$ to node $j$.
Let $M = \text{Id} + A + A^2 + A^3 + \cdots$. Then $M_{st}$ is a count of the number of paths (of any length) from source $s$ to destination $t$. So, we just need to compute the matrix $M$. The matrix $M$ can be computed using the matrix identity
$$M = (\text{Id} - A)^{-1}.$$
Thus, with a matrix subtraction and inversion, then looking at the $u,j$ entry of the resulting inverse, you can obtain a count of the number of such paths.
This includes paths where a node can be repeated. If you want to count the number of simple paths, then the problem is #P-complete for general graphs (and thus likely has no efficient algorithm), or can be done in linear time for directed acyclic graphs by topological sorting and then using dynamic programming.
