133.Clone_Graph(Medium).md

133. Clone Graph (Medium)

Date and Time: Aug 9, 2024, 14:28 (EST)

Link: https://leetcode.com/problems/clone-graph/

Question:

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.

Walk-through:

Similar to 138. Copy List with Random Pointer, we use a hashmap{} to store the old node's copy.

1. We run dfs to check if a node's neighbor or a node itself is in the hashmap or not, which means they have created a copy or not.

2. We create a copy of current node by its node.val and run dfs on all the original node's neighbors and append their copy to the new node's neighbors.

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 []
"""

from typing import Optional
class Solution:
    def cloneGraph(self, node: Optional['Node']) -> Optional['Node']:
        hashmap = {}
        def dfs(node):
            if node in hashmap:
                return hashmap[node]
            hashmap[node] = Node(node.val)
            # Appending original node's neighbor into new node's neighbors
            for neighbor in node.neighbors:
                hashmap[node].neighbors.append(dfs(neighbor))
            return hashmap[node]
        return dfs(node) if node else None

Time Complexity: $O(E + V)$ we loop over all edges and vertices.
Space Complexity: $O(V)$, hashmap stores all vertices.

Java Solution:

C++ Solution:

Runtime and Memory comparison

Language	Runtime	Memory
Python3	ms	MB
Java	ms	MB
C++	ms	MB