day-115.cpp

/*
  All Paths From Source to Target


Given a directed, acyclic graph of N nodes.  Find all possible paths from node 0 to node N-1, and return them in any order.

The graph is given as follows:  the nodes are 0, 1, ..., graph.length - 1.  graph[i] is a list of all nodes j for which the edge (i, j) exists.

Example:
Input: [[1,2], [3], [3], []] 
Output: [[0,1,3],[0,2,3]] 
Explanation: The graph looks like this:
0--->1
|    |
v    v
2--->3
There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.
Note:

The number of nodes in the graph will be in the range [2, 15].
You can print different paths in any order, but you should keep the order of nodes inside one path.

*/

// Simple DFS solution to calculate all the paths in the graph
class Solution {
   public:
    void findPathDFS(vector<vector<int>>& graph, vector<vector<int>>& answer, vector<int>& current, int currNode) {
        current.push_back(currNode);
        if (currNode == graph.size() - 1)
            answer.push_back(current);
        else {
            for (int neighbourNode : graph[currNode]) findPathDFS(graph, answer, current, neighbourNode);
        }
        current.pop_back();
    }

    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        if (graph.size() == 0) return {};
        vector<vector<int>> answer;

        vector<int> current;

        findPathDFS(graph, answer, current, 0);

        return answer;
    }
};