# Clone Graph

Clarify the problem:

- The problem requires us to clone an undirected graph.
- The graph is represented using an adjacency list, where each node in the graph is a Node object with a unique value and a list of neighbors.
- We need to implement a function that takes the reference to a node in the original graph and returns a deep copy of the graph.

Analyze the problem:

- Input: A reference to a node in the original graph.
- Output: A deep copy of the graph.
- Constraints:
- The graph is an undirected graph.
- The graph is represented using an adjacency list.
- Each node in the graph has a unique value.

Design an algorithm:

- We can use depth-first search (DFS) to traverse the graph and create a deep copy.
- Create an empty dictionary to store the mapping between original nodes and their clones.
- Implement a recursive helper function,
`dfs`

, that takes a node in the original graph as input:- Create a clone of the current node and add it to the dictionary.
- Iterate through the neighbors of the current node:
- If the neighbor is already in the dictionary, add its clone to the clone's neighbors.
- Otherwise, recursively call the
`dfs`

function on the neighbor and add its clone to the clone's neighbors.

- Call the
`dfs`

function on the input node to start the DFS traversal. - Return the clone of the input node.

Explain your approach:

- We will implement a function called
`cloneGraph`

to solve the problem. - The function will take a node in the original graph as input.
- We will create an empty dictionary to store the mapping between original nodes and their clones.
- We will use a recursive helper function,
`dfs`

, to perform the DFS traversal and create the deep copy of the graph. - The
`dfs`

function will create a clone of each node encountered and add it to the dictionary. - For each neighbor of a node, if the neighbor is already in the dictionary, we add its clone to the clone's neighbors. Otherwise, we recursively call the
`dfs`

function on the neighbor and add its clone to the clone's neighbors. - Finally, we call the
`dfs`

function on the input node and return its clone.

- We will implement a function called
Write clean and readable code:

python`class Node: def __init__(self, val=0, neighbors=None): self.val = val self.neighbors = neighbors if neighbors is not None else [] def cloneGraph(node): if not node: return None clone_dict = {} def dfs(node): clone = Node(node.val) clone_dict[node] = clone for neighbor in node.neighbors: if neighbor in clone_dict: clone.neighbors.append(clone_dict[neighbor]) else: clone.neighbors.append(dfs(neighbor)) return clone return dfs(node)`

Test your code:

- To test the code, we need to create a graph and verify that the cloned graph is correct.
- Let's create a graph with the following structure:lua

`1 -- 2 | | 4 -- 3`

- We will use the node with value 1 as the input to the
`cloneGraph`

function and compare the cloned graph with the original graph. `# Create the original graph node1 = Node(1) node2 = Node(2) node3 = Node(3) node4 = Node(4) node1.neighbors = [node2, node4] node2.neighbors = [node1, node3] node3.neighbors = [node2, node4] node4.neighbors = [node1, node3] # Clone the graph clone = cloneGraph(node1) # Check if the cloned graph is correct assert clone.val == 1 assert clone.neighbors[0].val == 2 assert clone.neighbors[1].val == 4 assert clone.neighbors[0].neighbors[0].val == 1 assert clone.neighbors[0].neighbors[1].val == 3 assert clone.neighbors[1].neighbors[0].val == 1 assert clone.neighbors[1].neighbors[1].val == 3 assert clone.neighbors[0].neighbors[0] is clone assert clone.neighbors[1].neighbors[0] is clone assert clone.neighbors[0].neighbors[1] is clone.neighbors[1] assert clone.neighbors[1].neighbors[1] is clone.neighbors[0]`

We have tested the code with a sample graph and verified that the cloned graph matches the original graph.

Optimize if necessary:

- The current solution has a time complexity of O(N), where N is the total number of nodes in the graph.
- This is because we perform a DFS traversal to visit each node exactly once.
- The space complexity is also O(N) because we use a dictionary to store the mapping between original nodes and their clones.

Handle error cases:

- The code assumes that the input node is not None.
- If the input node is None, the function returns None as well.
- The code does not explicitly handle other error cases, such as invalid input or cyclic graphs. Handling such cases can be added as needed.

Discuss complexity analysis:

- Let N be the total number of nodes in the graph.
- The time complexity of the solution is O(N) because we perform a DFS traversal to visit each node exactly once.
- The space complexity is also O(N) because we use a dictionary to store the mapping between original nodes and their clones.

python