Clone Graph LeetCode Solution – Queslers

Problem – Clone Graph

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.

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.

Clone Graph LeetCode Solution in 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 clone and stored in map
                {
                    neighbour.push_back(mp[it]);    //directly push back the clone node from map to neigh
                }
                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;
        if(node->neighbors.size()==0)   //if only one node present no neighbors
        {
            Node* clone= new Node(node->val);
            return clone; 
        }
        return dfs(node,mp);
    }
};

Clone Graph LeetCode Solution in Python

    def cloneGraph(self, node: 'Node') -> 'Node':
        if not node:
            return 
        nodeCopy = Node(node.val, [])
        dic = {node: nodeCopy}
        queue = collections.deque([node])
        while queue:
            node = queue.popleft()
            for neighbor in node.neighbors:
                if neighbor not in dic: # neighbor is not visited
                    neighborCopy = Node(neighbor.val, [])
                    dic[neighbor] = neighborCopy
                    dic[node].neighbors.append(neighborCopy)
                    queue.append(neighbor)
                else:
                    dic[node].neighbors.append(dic[neighbor])
        return nodeCopy

Clone Graph LeetCode Solution in Java

public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if (node == null) return null;
        
        UndirectedGraphNode newNode = new UndirectedGraphNode(node.label); //new node for return
        HashMap<Integer, UndirectedGraphNode> map = new HashMap(); //store visited nodes
        
        map.put(newNode.label, newNode); //add first node to HashMap
        
        LinkedList<UndirectedGraphNode> queue = new LinkedList(); //to store **original** nodes need to be visited
        queue.add(node); //add first **original** node to queue
        
        while (!queue.isEmpty()) { //if more nodes need to be visited
            UndirectedGraphNode n = queue.pop(); //search first node in the queue
            for (UndirectedGraphNode neighbor : n.neighbors) {
                if (!map.containsKey(neighbor.label)) { //add to map and queue if this node hasn't been searched before
                    map.put(neighbor.label, new UndirectedGraphNode(neighbor.label));
                    queue.add(neighbor);
                }
                map.get(n.label).neighbors.add(map.get(neighbor.label)); //add neighbor to new created nodes
            }
        }
        
        return newNode;
    }
}

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