Clone graph

Problem Description

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[Node]) of its neighbors.

class Node {
    public int val;
    public List<Node> neighbors;
}

Test case format:

For simplicity, each node's value is the same as the node's index (1-indexed). For example, the first node with val == 1, the second node with val == 2, and so on. The graph is represented in the test case using an adjacency list.

An adjacency list is a collection of unordered lists used to represent a finite graph. Each list describes the set of neighbors of a node in the graph.

The given node will always be the first node with val = 1. You must return the copy of the given node as a reference to the cloned graph.

Examples

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: [[]]
Explanation: Note that the input contains one empty list. The graph consists of only one node with val = 1 and it does not have any neighbors.

Example 3:

Input: adjList = []
Output: []
Explanation: This an empty graph, it does not have any nodes.
 

Constraints

• 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.

---

Intuition

Use a Depth-First Search (DFS) approach to traverse the original graph. During the traversal:

• Create a new node for each encountered node and maintain a mapping between the original nodes and their corresponding clones.
• When processing a node's neighbors, check if the clone of the neighbor has already been created. If so, use the clone from the mapping; otherwise, recursively explore the neighbor to create its clone.

Approach

1. Define a Helper Function dfs:

   The dfs function performs the following tasks:

   • Parameters: Takes a node cur from the original graph and a mapping mp which stores the relationship between original nodes and their clones.
   • Steps:
     1. Create a New Node:
        • Create a new clone node with the same value as cur.
     2. Add Mapping:
        • Add an entry in the mapping mp to link the original node cur with the newly created clone.
     3. Initialize Neighbors:
        • Initialize an empty vector neighbour to store the cloned neighbors of cur.
     4. Process Neighbors:
        • Iterate through the neighbors of cur:
          • If the Neighbor is Already Cloned: If mp.find(it) != mp.end(), use the clone from the mapping.
          • If the Neighbor is Not Cloned: Recursively call dfs on the neighbor to get its clone.
     5. Set Neighbors:
        • Set the neighbours of the clone to the neighbour vector.
     6. Return Clone:
        • Return the clone.

2. Define the cloneGraph Function:

   This function initializes the cloning process:

   • Steps:
     1. Initialize the Mapping:
        • Create an unordered map mp to store the mapping between original nodes and their clones.
     2. Check for Null Node:
        • Check if the node is NULL, and if so, return NULL.
     3. Call dfs:
        • Call the dfs function with the original node and the mapping mp.

Code in Different Languages

C++

class Solution {
public:
Node* dfs(Node* cur,unordered_map<Node*,Node*>& mp)
    {
        vector<Node*> neighbour;
        Node* clone=new Node(cur->val);
        mp[cur]=clone;
            for(auto it:cur->neighbors)
            {
                if(mp.find(it)!=mp.end())   //already in map
                {
                    neighbour.push_back(mp[it]);    
                }
                else
                    neighbour.push_back(dfs(it,mp));
            }
            clone->neighbors=neighbour;
            return clone;
    }
    Node* cloneGraph(Node* node) {
        unordered_map<Node*,Node*> mp;
        if(node==NULL)
            return NULL;
            
        return dfs(node,mp);
    }};

Java

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> neighbors;
    public Node() {
        val = 0;
        neighbors = new ArrayList<Node>();
    }
    public Node(int _val) {
        val = _val;
        neighbors = new ArrayList<Node>();
    }
    public Node(int _val, ArrayList<Node> _neighbors) {
        val = _val;
        neighbors = _neighbors;
    }
}
*/

class Solution {

    HashMap<Integer, Node> visited = new HashMap<>();

    public Node cloneGraph(Node node) {
        Node ans = new Node();
        if(node == null){
            return null;
        }
        ans.val = node.val;
        visited.put(node.val, ans);
        for(int i=0; i<node.neighbors.size(); i++){
            if(!visited.containsKey(node.neighbors.get(i).val)){
                cloneGraph(node.neighbors.get(i));
            }
        }

        for(int i=0; i<node.neighbors.size(); i++){
            ans.neighbors.add(visited.get(node.neighbors.get(i).val));
        }

        return ans;
    }
}

Python

class Solution:
    def cloneGraph(self, node: 'Node') -> 'Node':
        if not node: return node
        
        q, clones = deque([node]), {node.val: Node(node.val, [])}
        while q:
            cur = q.popleft() 
            cur_clone = clones[cur.val]            

            for ngbr in cur.neighbors:
                if ngbr.val not in clones:
                    clones[ngbr.val] = Node(ngbr.val, [])
                    q.append(ngbr)
                    
                cur_clone.neighbors.append(clones[ngbr.val])
                
        return clones[node.val]

Complexity Analysis

• Time Complexity: O(V + E)

  • V is the number of nodes and E is the number of edges. The DFS traverses each node and edge once.

• Space Complexity: O(V)

  • Space is used for the mapping between nodes and their clones, which requires O(V) space.