In this guide, you will get 133. Clone Graph LeetCode Solution with the best time and space complexity. The solution to Clone Graph problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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 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).
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.
Complexity Analysis
Time Complexity: O(|V| + |E|)
Space Complexity: O(|V| + |E|)
133. Clone Graph LeetCode Solution in C++
class Solution {
public:
Node* cloneGraph(Node* node) {
if (node == nullptr)
return nullptr;
queue<Node*> q{{node}};
unordered_map<Node*, Node*> map{{node, new Node(node->val)}};
while (!q.empty()) {
Node* u = q.front();
q.pop();
for (Node* v : u->neighbors) {
if (!map.contains(v)) {
map[v] = new Node(v->val);
q.push(v);
}
map[u]->neighbors.push_back(map[v]);
}
}
return map[node];
}
};
/* code provided by PROGIEZ */
133. Clone Graph LeetCode Solution in Java
class Solution {
public Node cloneGraph(Node node) {
if (node == null)
return null;
Queue<Node> q = new ArrayDeque<>(List.of(node));
Map<Node, Node> map = new HashMap<>();
map.put(node, new Node(node.val));
while (!q.isEmpty()) {
Node u = q.poll();
for (Node v : u.neighbors) {
if (!map.containsKey(v)) {
map.put(v, new Node(v.val));
q.offer(v);
}
map.get(u).neighbors.add(map.get(v));
}
}
return map.get(node);
}
}
// code provided by PROGIEZ
133. Clone Graph LeetCode Solution in Python
class Solution:
def cloneGraph(self, node: 'Node') -> 'Node':
if not node:
return None
q = collections.deque([node])
map = {node: Node(node.val)}
while q:
u = q.popleft()
for v in u.neighbors:
if v not in map:
map[v] = Node(v.val)
q.append(v)
map[u].neighbors.append(map[v])
return map[node]
# code by PROGIEZ
