Clone Graph LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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 Solutions

Time:O(∣V∣+∣E∣)
Space: O(|V| + |E|)O(∣V∣+∣E∣)

C++

class Solution {
 public:
  Node* cloneGraph(Node* node) {
    if (!node)
      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.count(v)) {
          map[v] = new Node(v->val);
          q.push(v);
        }
        map[u]->neighbors.push_back(map[v]);
      }
    }

    return map[node];
  }
};

Java

 class Solution {
  public Node cloneGraph(Node node) {
    if (node == null)
      return null;

    Queue<Node> q = new ArrayDeque<>(Arrays.asList(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);
  }
}

Python

 
class Solution:
  def cloneGraph(self, node: 'Node') -> 'Node':
    if not node:
      return None

    q = 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]

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