LeetCode 1059. All Paths from Source Lead to Destination

Description

https://leetcode.com/problems/all-paths-from-source-lead-to-destination/

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

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.

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.

Python Solution

Use depth-first search to see if all paths end with the destination node.

class Solution:
    def leadsToDestination(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
   
        adjacency_list = [set() for i in range(n)]
        
        for edge in edges:            
            adjacency_list[edge[0]].add(edge[1])
            
        return self.helper(source, destination, adjacency_list, set())
        
                
        
    def helper(self, node, destination, adjacency_list, visited):
        if node in visited:
            return False

        if node != destination and not adjacency_list[node]:
            return False
        
        if node == destination and len(adjacency_list[node]) > 0:
            return False
        
        if node == destination:
            return True
        
        visited.add(node)
        
        
        for neighbor in adjacency_list[node]:   
            if not self.helper(neighbor, destination, adjacency_list, visited):
                return False
            
        visited.remove(node)
        
        
        return True
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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