LeetCode 133. Clone Graph

Description

https://leetcode.com/problems/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.

Example 4:

Input: adjList = [[2],[1]]
Output: [[2],[1]]

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.

Explanation

We can use both depth-first search or breadth-first search to build a clone graph with help from a hashmap which maps the original nodes to the clone nodes.

Python Solution

"""
# Definition for a Node.
class Node:
    def __init__(self, val = 0, neighbors = None):
        self.val = val
        self.neighbors = neighbors if neighbors is not None else []
"""

class Solution:
    def cloneGraph(self, node: 'Node') -> 'Node':
        if not node:
            return None
        visited = {}
        clone_root = self.helper(node, visited)
        
        
        return clone_root
                    
                    
    def helper(self, node, visited):        
        if not node:
            return None
        
        if node in visited:
            return visited[node]
        
        if not node.neighbors:
            return Node(node.val)
        
        
        clone = Node(node.val)
        
        visited[node] = clone
        
        for neighbor in node.neighbors:            
            clone_neighbor = self.helper(neighbor, visited)
            
            clone.neighbors.append(clone_neighbor)
        
        return clone
"""
# Definition for a Node.
class Node:
    def __init__(self, val = 0, neighbors = None):
        self.val = val
        self.neighbors = neighbors if neighbors is not None else []
"""

class Solution:
    def cloneGraph(self, node: 'Node') -> 'Node':
        if not node:
            return None
        
        visited = {}
        
        root = node
        queue = []        
        queue.append(node)
        visited[node] = Node(node.val)
        
        while queue:        

            current_node = queue.pop(0)

            for neighbor in current_node.neighbors:
                if neighbor not in visited:
                    queue.append(neighbor)
                    visited[neighbor] = Node(neighbor.val)

                visited[current_node].neighbors.append(visited[neighbor])

        return visited[root]
                
  • Time Complexity: O(N).
  • Space Complexity: O(N).

Leave a Reply

Your email address will not be published.