# Question

Formatted question description: https://leetcode.ca/all/1059.html

```
1059. All Paths from Source Lead to Destination
Given the edges of a directed graph, and two nodes source and destination of this graph,
determine whether or not all paths starting from source eventually end at destination, that is:
At least one path exists from the source node to the destination node
If a path exists from the source node to a node with no outgoing edges, then that node is equal to destination.
The number of possible paths from source to destination is a finite number.
Return true if and only if all roads from source lead to destination.
Example 1:
Input: n = 3, edges = [[0,1],[0,2]], source = 0, destination = 2
Output: false
Explanation: It is possible to reach and get stuck on both node 1 and node 2.
Example 2:
Input: n = 4, edges = [[0,1],[0,3],[1,2],[2,1]], source = 0, destination = 3
Output: false
Explanation: We have two possibilities: to end at node 3, or to loop over node 1 and node 2 indefinitely.
Example 3:
Input: n = 4, edges = [[0,1],[0,2],[1,3],[2,3]], source = 0, destination = 3
Output: true
Example 4:
Input: n = 3, edges = [[0,1],[1,1],[1,2]], source = 0, destination = 2
Output: false
Explanation: All paths from the source node end at the destination node, but there are an infinite number of paths, such as 0-1-2, 0-1-1-2, 0-1-1-1-2, 0-1-1-1-1-2, and so on.
Example 5:
Input: n = 2, edges = [[0,1],[1,1]], source = 0, destination = 1
Output: false
Explanation: There is infinite self-loop at destination node.
Note:
The given graph may have self loops and parallel edges.
The number of nodes n in the graph is between 1 and 10000
The number of edges in the graph is between 0 and 10000
0 <= edges.length <= 10000
edges[i].length == 2
0 <= source <= n - 1
0 <= destination <= n - 1
```

# Algorithm

There are 2 cases it should return false.

- case 1: it encounters a node that has no outgoing edges, but it is not destination.
- case 2: it has cycle.

Otherwise, it returns true.

Could iterate graph with BFS. When indegree of a node becomes negative, then ther is cycle.

Time Complexity: `O(n+e)`

. e = edges.length.

Space: `O(n+e)`

.

# Code

Java

```
public class All_Paths_from_Source_Lead_to_Destination {
class Solution_bfs {
public boolean leadsToDestination(int n, int[][] edges, int source, int destination) {
Set<Integer>[] graph = new Set[n]; // node => set of its next nodes
for (int i = 0; i < n; i++) {
graph[i] = new HashSet<Integer>();
}
int[] inDegrees = new int[n];
for (int[] edge : edges) {
graph[edge[0]].add(edge[1]);
inDegrees[edge[1]]++;
}
LinkedList<Integer> q = new LinkedList<Integer>();
q.add(source);
while (!q.isEmpty()) {
int cur = q.poll();
if (graph[cur].size() == 0 && cur != destination) {
return false;
}
for (int nei : graph[cur]) {
if (inDegrees[nei] < 0) { // i.e. a cycle
return false;
}
inDegrees[nei]--;
q.add(nei);
}
}
return true;
}
}
class Solution_dfs {
public boolean leadsToDestination(int n, int[][] edges, int source, int destination) {
Map<Integer, Set<Integer>> graph = new HashMap<>();
for (int[] edge : edges) {
Set<Integer> neighbours = graph.getOrDefault(edge[0], new HashSet<>());
neighbours.add(edge[1]);
graph.put(edge[0], neighbours);
}
if (graph.get(destination) != null) return false;
boolean[] isVisited = new boolean[n];
isVisited[source] = true;
return dfs(graph, isVisited, source, destination);
}
private boolean dfs(Map<Integer, Set<Integer>> graph, boolean[] isVisited, int source, int destination) {
if (source == destination) {
return true;
}
Set<Integer> neighbours = graph.getOrDefault(source, new HashSet<>());
if (neighbours.size() == 0) return false;
for (int neib : neighbours) {
if (isVisited[neib]) return false; // cycle spotted
isVisited[neib] = true;
if (!dfs(graph, isVisited, neib, destination)) return false;
isVisited[neib] = false;
}
return true;
}
}
}
```