cloneGraph.js

//https://leetcode.com/problems/clone-graph/
// 133. Clone Graph
// Medium

// 1848

// 1384

// Add to List

// Share
// 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 val (int) and a list (List[Node]) of its neighbors.

// class Node {
//     public int val;
//     public List<Node> neighbors;
// }


// Test case format:

// For simplicity sake, 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.

// 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.
// Example 4:


// Input: adjList = [[2],[1]]
// Output: [[2],[1]]


// Constraints:

// 1 <= Node.val <= 100
// Node.val is unique for each node.
// Number of Nodes will not exceed 100.
// There is 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.


var cloneGraph = function(node) {
  //get all nodes
  const oldNodeArr = getAllNodes(node)


  //find a mapping between old nodes and new nodes
  const newNodeArr = cloneAndMapAllNodes(oldNodeArr)

  connectNewNodesBasedOnOldNodes(oldNodeArr, newNodeArr)
  //duplicate the edges

  return newNodeArr[0]
};

function connectNewNodesBasedOnOldNodes(oldNodeArr, newNodeArr) {
  oldNodeArr.forEach(
      (node, idx) =>{
          const newNode = newNodeArr[idx]
          node.neighbors.forEach((neighbor)=>{
              const neighborIdx = oldNodeArr.indexOf(neighbor)
              newNode.neighbors.push(newNodeArr[neighborIdx])
          })
      }
  )
}

function cloneAndMapAllNodes (oldNodeArr) {
  const newNodesArr = []
  oldNodeArr.forEach((node)=>{
      newNodesArr.push(new Node(node.val))
  })
  return newNodesArr
}

function cloneAndMapAllNodes (oldNodeArr) {
  const newNodesArr = []
  oldNodeArr.forEach((node)=>{
      newNodesArr.push(new Node(node.val))
  })
  return newNodesArr
}

function getAllNodes (head) {
  if(!head) return []
  const visited = new Set()
  visited.add(head)
  const queue = [head];
  const res = [];
  while(queue.length) {
      const curNode = queue.shift();
      res.push(curNode)

      curNode.neighbors.forEach((neighbor)=>{
          if (!visited.has(neighbor)) {
              visited.add(neighbor)
              queue.push(neighbor)
          }
      })
  }

  return res
}