Clone Graph

Problem

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.

Pseudocode

- to clone a connected undirected graph
  - need value of the node
  - array of the neighbouring nodes
  
- initialize a new node with node.val,
  - recurse through node.neighbor, and place that neighbor into array
  - use visited array, add to array if not visited,
    - if visited, just add to neighbor array without recursing (already been there)

Solution

// from solutions
var cloneGraph = function (node) {
  const visited = new Array();

  if (node == null) {
    return node;
  }

  return cloneHelper(node, visited);
};

var cloneHelper = function (node, visited) {
  // console.log(visited, node.val)
  const newNode = new Node(node.val);

  visited[node.val] = newNode;

  for (const neighbor of node.neighbors) {
    if (!visited[neighbor.val]) {
      const newNeighbor = cloneHelper(neighbor, visited);
      newNode.neighbors.push(newNeighbor);
    } else {
      const newNeighbor = visited[neighbor.val];
      newNode.neighbors.push(newNeighbor);
    }
  }
  return newNode;
};

Time and Space Complexity

Time

What did the code do
Total -

Space

What did the code do
Total -

PreviousBinary Tree Level Order Traversal NextEvaluate Reverse Polish Notation

Last updated 2 years ago

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).

- to clone a connected undirected graph - need value of the node - array of the neighbouring nodes - initialize a new node with node.val, - recurse through node.neighbor, and place that neighbor into array - use visited array, add to array if not visited, - if visited, just add to neighbor array without recursing (already been there)

// from solutions var cloneGraph = function (node) { const visited = new Array(); if (node == null) { return node; } return cloneHelper(node, visited); }; var cloneHelper = function (node, visited) { // console.log(visited, node.val) const newNode = new Node(node.val); visited[node.val] = newNode; for (const neighbor of node.neighbors) { if (!visited[neighbor.val]) { const newNeighbor = cloneHelper(neighbor, visited); newNode.neighbors.push(newNeighbor); } else { const newNeighbor = visited[neighbor.val]; newNode.neighbors.push(newNeighbor); } } return newNode; };