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;
        }
    }
}

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