1059. All Paths from Source Lead to Destination - Explanation

Problem Link

Description

Given the edges of a directed graph where edges[i] = [aᵢ, bᵢ] indicates there is an edge between nodes aᵢ and bᵢ, 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

Constraints:

• 1 <= n <= 10⁴
• 0 <= edges.length <= 10⁴
• edges.length == 2
• 0 <= aᵢ, bᵢ <= n - 1
• 0 <= source <= n - 1
• 0 <= destination <= n - 1
• The given graph may have self-loops and parallel edges.

Company Tags

Google4

1. Depth First Search

class Solution:
    
    # We don't use the state WHITE as such anywhere. Instead, the "null" value in the states array below is a substitute for WHITE.
    GRAY = 1
    BLACK = 2

    def leadsToDestination(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
        graph = self.buildDigraph(n, edges)
        return self.leadsToDest(graph, source, destination, [None] * n)
        
    def leadsToDest(self, graph, node, dest, states):
        
        # If the state is GRAY, this is a backward edge and hence, it creates a Loop.
        if states[node] != None:
            return states[node] == Solution.BLACK
        
        # If this is a leaf node, it should be equal to the destination.
        if len(graph[node]) == 0:
            return node == dest
        
        # Now, we are processing this node. So we mark it as GRAY.
        states[node] = Solution.GRAY
        
        for next_node in graph[node]:
            
            # If we get a `false` from any recursive call on the neighbors, we short circuit and return from there.
            if not self.leadsToDest(graph, next_node, dest, states):
                return False
        
        # Recursive processing done for the node. We mark it BLACK.
        states[node] = Solution.BLACK
        return True
        
    def buildDigraph(self, n, edges):
        graph = [[] for _ in range(n)]
        
        for edge in edges:
            graph[edge[0]].append(edge[1])
            
        return graph

Time & Space Complexity

• Time complexity:

  • Typically for an entire DFS over an input graph, it takes $O(V + E)$ where $V$ represents the number of vertices in the graph and likewise, $E$ represents the number of edges in the graph. In the worst case $E$ can be $O(V^2)$ in case each vertex is connected to every other vertex in the graph. However even in the worst case, we will end up discovering a cycle very early on and prune the recursion tree. If we were to traverse the entire graph, then the complexity would be $O(V^2)$ as the $O(E)$ part would dominate. However, due to pruning and backtracking in case of cycle detection, we end up with an overall time complexity of $O(V)$.

• Space complexity: $O(V + E)$

  • Where $O(E)$ is occupied by the adjacency list and $O(V)$ is occupied by the recursion stack and the color states.

Where $V$ represents the number of vertices in the graph and $E$ represents the number of edges in the graph.

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Why not Breadth-First Search?

From this Stack Overflow answer:

A BFS could be reasonable if the graph is undirected (be my guest at showing an efficient algorithm using BFS that would report the cycles in a directed graph!), where each cross edge defines a cycle (edge going from a node to an already visited node). If the cross edge is {v1, v2}, and the root (in the BFS tree) that contains those nodes is r, then the cycle is r ~ v1 - v2 ~ r (~ is a path, - a single edge), which can be reported almost as easily as in DFS.

The only reason to use a BFS would be if you know your (undirected) graph is going to have long paths and small path cover (in other words, deep and narrow). In that case, BFS would require proportionally less memory for its queue than DFS' stack (both still linear of course).

In all other cases, DFS is clearly the winner.