Last updated on January 5th, 2025 at 01:08 am
Here, we see a Clone Graph LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.
List of all LeetCode Solution
Topics
Breadth-First Search, Depth-First Search, Graph
Companies
Facebook, Google, Uber, Pocketgems
Level of Question
Medium
Clone Graph LeetCode Solution
Table of Contents
1. 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[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.
2. Coding Pattern Used in Solution
The provided code in all four languages (C++, Java, JavaScript, and Python) follows the Graph Traversal pattern. Specifically, it uses Depth-First Search (DFS) in the C++, Java, and Python implementations, and Breadth-First Search (BFS) in the JavaScript implementation. This pattern is used to traverse and clone a graph.
3. Code Implementation in Different Languages
3.1 Clone Graph 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())
{
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;
if(node->neighbors.size()==0)
{
Node* clone= new Node(node->val);
return clone;
}
return dfs(node,mp);
}
};
3.2 Clone Graph Java
class Solution {
public Node cloneGraph(Node node) {
if (node == null) {
return null;
}
Map<Node, Node> visited = new HashMap<>();
return cloneGraphHelper(node, visited);
}
private Node cloneGraphHelper(Node node, Map<Node, Node> visited) {
Node copy = new Node(node.val);
visited.put(node, copy);
for (Node neighbor : node.neighbors) {
if (visited.containsKey(neighbor)) {
copy.neighbors.add(visited.get(neighbor));
} else {
Node neighborCopy = cloneGraphHelper(neighbor, visited);
copy.neighbors.add(neighborCopy);
}
}
return copy;
}
}
3.3 Clone Graph JavaScript
var cloneGraph = function(node) {
let start = node;
if (start === null) return null;
const vertexMap = new Map();
const queue = [start]
vertexMap.set(start, new Node(start.val));
while (queue.length > 0) {
const currentVertex = queue.shift();
for (const neighbor of currentVertex.neighbors) {
if (!vertexMap.has(neighbor)) {
vertexMap.set(neighbor, new Node(neighbor.val))
queue.push(neighbor);
}
vertexMap.get(currentVertex).neighbors.push(vertexMap.get(neighbor));
}
}
return vertexMap.get(start);
};
3.4 Clone Graph Python
class Solution(object):
def cloneGraph(self, node):
if not node:
return None
cloned = {}
stack = [node]
cloned[node] = Node(node.val)
while stack:
curr = stack.pop()
for neighbor in curr.neighbors:
if neighbor not in cloned:
cloned[neighbor] = Node(neighbor.val)
stack.append(neighbor)
cloned[curr].neighbors.append(cloned[neighbor])
return cloned[node]
4. Time and Space Complexity
| Time Complexity | Space Complexity | |
| C++ | O(V + E) | O(V) |
| Java | O(V + E) | O(V) |
| JavaScript | O(V + E) | O(V) |
| Python | O(V + E) | O(V) |
where,
V: Number of vertices (nodes) in the graph.
E: Number of edges in the graph.
- The code is designed to clone a graph using either DFS (C++, Java, Python) or BFS (JavaScript).
- It uses a
mapto track already cloned nodes, ensuring no node is cloned more than once. - The time complexity is O(V + E), and the space complexity is O(V) for all implementations.