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

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;
        }
    }
}


############

class Solution {
    private List<Integer>[] g;
    private int[] f;
    private boolean[] vis;
    private int k;

    public boolean leadsToDestination(int n, int[][] edges, int source, int destination) {
        vis = new boolean[n];
        g = new List[n];
        k = destination;
        f = new int[n];
        Arrays.setAll(g, key -> new ArrayList<>());
        for (var e : edges) {
            g[e[0]].add(e[1]);
        }
        return dfs(source);
    }

    private boolean dfs(int i) {
        if (i == k) {
            return g[i].isEmpty();
        }
        if (f[i] != 0) {
            return f[i] == 1;
        }
        if (vis[i] || g[i].isEmpty()) {
            return false;
        }
        vis[i] = true;
        for (int j : g[i]) {
            if (!dfs(j)) {
                f[i] = -1;
                return false;
            }
        }
        f[i] = 1;
        return true;
    }
}

// OJ: https://leetcode.com/problems/all-paths-from-source-lead-to-destination/
// Time: O(V + E)
// Space: O(V + E)
class Solution {
    vector<unordered_set<int>> G;
    vector<int> state; // -1 unvisited, 0 visiting, 1 visited
    int dest;
    bool dfs(int u) {
        if (state[u] != -1) return state[u];
        if (G[u].empty()) {
            return u == dest;
        }
        state[u] = 0;
        for (int v : G[u]) {
            if (!dfs(v)) return false;
        }
        return state[u] = 1;
    }
public:
    bool leadsToDestination(int n, vector<vector<int>>& E, int source, int dest) {
        G.assign(n, {});
        state.assign(n, -1);
        this->dest = dest;
        for (auto &e : E) {
            G[e[0]].insert(e[1]);
        }
        return dfs(source);
    }
};

class Solution:
    def leadsToDestination(
        self, n: int, edges: List[List[int]], source: int, destination: int
    ) -> bool:
        @cache
        def dfs(i):
            if i == destination:
                return not g[i]
            if i in vis or not g[i]: # not g[i]: not destination but no more neigbours
                return False
            vis.add(i)
            for j in g[i]:
                if not dfs(j):
                    return False
            return True

        g = defaultdict(list)
        for a, b in edges:
            g[a].append(b)
        vis = set()
        return dfs(source)

func leadsToDestination(n int, edges [][]int, source int, destination int) bool {
	vis := make([]bool, n)
	g := make([][]int, n)
	f := make([]int, n)
	for _, e := range edges {
		g[e[0]] = append(g[e[0]], e[1])
	}
	var dfs func(int) bool
	dfs = func(i int) bool {
		if i == destination {
			return len(g[i]) == 0
		}
		if f[i] != 0 {
			return f[i] == 1
		}
		if vis[i] || len(g[i]) == 0 {
			return false
		}
		vis[i] = true
		for _, j := range g[i] {
			if !dfs(j) {
				f[i] = -1
				return false
			}
		}
		f[i] = 1
		return true
	}
	return dfs(source)
}

1059 - All Paths from Source Lead to Destination

Question

Algorithm

Code

All Problems

All Solutions

Welcome to Subscribe On Youtube

Question

Algorithm

Code

All Problems

All Solutions