Introduction

In this blog post, we will explore how to solve the LeetCode problem "133" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 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.

Explanation

• DFS Traversal: Use Depth-First Search (DFS) to traverse the original graph.
  • Cloning: Create a new node for each node encountered in the original graph and store the mapping between original and cloned nodes.
  • Neighbor Mapping: While traversing, map the neighbors of the original nodes to the corresponding cloned nodes.

• Time Complexity: O(N + E), where N is the number of nodes and E is the number of edges. Space Complexity: O(N), due to the space used by the hash map and the recursion stack during DFS.

Code

    # Definition for a Node.
class Node:
    def __init__(self, val, neighbors):
        self.val = val
        self.neighbors = neighbors

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

        visited = {}  # Map original node to cloned node

        def dfs(node):
            if node in visited:
                return visited[node]

            cloned_node = Node(node.val, [])
            visited[node] = cloned_node

            for neighbor in node.neighbors:
                cloned_neighbor = dfs(neighbor)
                cloned_node.neighbors.append(cloned_neighbor)

            return cloned_node

        return dfs(node)