Problem Statement

Given a reference of a node in a connected undirected graph.

Return a deep copy (clone) of the graph.

Each node in the graph contains a value (int) and a list (list[GraphNode]) of its neighbors.

class GraphNode:
    def __init__(self, val = 0, neighbors = None):
        self.val = val
        self.neighbors = neighbors if neighbors is not None else []

Additional information

• The number of nodes in the graph is in the range [0, 100].
• 1 <= Node.val <= 100
• Node.val is unique for each node.
• There are no repeated edges and no self-loops in the graph.
• The Graph is connected and all nodes can be visited starting from the given node.

Example 1:

Input: adjList = [[2,4],[1,3],[2,4],[1,3]]

Output: [[2,4],[1,3],[2,4],[1,3]]

Explanation: There are 4 nodes in the graph.
1st node (val = 1)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
2nd node (val = 2)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).
3rd node (val = 3)'s neighbors are 2nd node (val = 2) and 4th node (val = 4).
4th node (val = 4)'s neighbors are 1st node (val = 1) and 3rd node (val = 3).

Example 2:

Input: adjList = [[]]

Output: [[]]

Example 3:

Input: adjList = []

Output: []

Hash Map + DFS (Optimal)

Intuition

The difficulty in cloning the graph is that nodes reference each other cyclically. If Node 1 has Node 2 as a neighbor, and Node 2 has Node 1 as a neighbor, a naive recursive copy would loop forever.

We solve this with a Hash Map that maps each original node to its clone. Before recursing into a neighbor, we check if it's already been cloned. If so, we return the existing clone instead of creating a new one.

Algorithm

1. Initialize a hash map old_to_new.
2. Create a recursive function dfs(node):

• If node is None, return None.
• If node is already in old_to_new, return the existing clone.
• Create a new GraphNode(node.val) and store it in old_to_new.
• For each neighbor, recursively call dfs(neighbor) and append the result to the clone's neighbors.
• Return the clone.

3. Call dfs(node) and return the result.

def clone_graph(node: 'GraphNode') -> 'GraphNode':
    old_to_new = {}
    
    def dfs(node):
        if not node:
            return None
        if node in old_to_new:
            return old_to_new[node]
            
        copy = GraphNode(node.val)
        old_to_new[node] = copy
        
        for nei in node.neighbors:
            copy.neighbors.append(dfs(nei))
            
        return copy
        
    return dfs(node)

Time & Space Complexity

• Time complexity: O(V + E)

• Reason: We visit every node (V) and traverse every edge (E) exactly once.

• Space complexity: O(V)

• Reason: The hash map stores V nodes, and the recursion stack can go V levels deep in the worst case.