Teaching Kids Programming – Clone (Deep Copy) a Undirected Connected Graph using Breadth First Search Algorithm


Teaching Kids Programming: Videos on Data Structures and Algorithms

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.

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class Node {
    public int val;
    public List<Node> 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.

clone-graph Teaching Kids Programming - Clone (Deep Copy) a Undirected Connected Graph using Breadth First Search Algorithm algorithms BFS python teaching kids programming youtube video

Clone a 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 a Graph using Breadth First Search Algorithm

A Graph is a collections of vertices and edges and can be noted as tex_8b33ec76ae3a9520de7d175e47ea8da5 Teaching Kids Programming - Clone (Deep Copy) a Undirected Connected Graph using Breadth First Search Algorithm algorithms BFS python teaching kids programming youtube video

Apart from Depth First Search Algorithm, we can use Breadth First Search Algorithm to traverse a Graph. And we can clone/copy a Node while we are visiting a Node in the first time, and also we need to copy the edge i.e. add the neighbour nodes to the current nodes’ neighbour list.

We use a queue to implement a Breadth First Search Algorithm.

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"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, neighbors):
        self.val = val
        self.neighbors = neighbors
"""
class Solution(object):
    def cloneGraph(self, node):
        if not node:
            return node
        q = deque([node])
        seen = {}
        seen[node] = Node(node.val, [])
        while q:
            cur = q.popleft()
            for n in cur.neighbors:
                if n not in seen:
                    seen[n] = Node(n.val, [])
                    q.append(n)
                seen[cur].neighbors.append(seen[n])
        return seen[node]
"""
# Definition for a Node.
class Node(object):
    def __init__(self, val, neighbors):
        self.val = val
        self.neighbors = neighbors
"""
class Solution(object):
    def cloneGraph(self, node):
        if not node:
            return node
        q = deque([node])
        seen = {}
        seen[node] = Node(node.val, [])
        while q:
            cur = q.popleft()
            for n in cur.neighbors:
                if n not in seen:
                    seen[n] = Node(n.val, [])
                    q.append(n)
                seen[cur].neighbors.append(seen[n])
        return seen[node]

The time complexity is O(N+M) where N is the number of the vertices in the Graph, and M is the number of the edges. This is the same as: Teaching Kids Programming – Clone (Deep Copy) a Undirected Connected Graph using Recursive Depth First Search Algorithm. The space complexity is O(N) as we are using a queue to store the graph vertices/nodes.

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