# Copy List with Random Pointer

A linked list of length `n` is given such that each node contains an additional random pointer, which could point to any node in the list, or `null`.

Construct a deep copy of the list. The deep copy should consist of exactly `n` brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the `next` and `random` pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

For example, if there are two nodes `X` and `Y` in the original list, where `X.random --> Y`, then for the corresponding two nodes `x` and `y` in the copied list, `x.random --> y`.

The linked list is represented in the input/output as a list of `n` nodes. Each node is represented as a pair of `[val, random_index]` where:

• `val`: an integer representing `Node.val`
• `random_index`: the index of the node (range from `0` to `n-1`) that the `random` pointer points to, or `null` if it does not point to any node.

Your code will only be given the `head` of the original linked list.

Example 1: Input:

``` head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
```

Output:

``` [[7,null],[13,0],[11,4],[10,2],[1,0]]
```

Example 2: Input:

``` head = [[1,1],[2,1]]
```

Output:

``` [[1,1],[2,1]]
```

Example 3: Input:

``` head = [[3,null],[3,0],[3,null]]
```

Output:

``` [[3,null],[3,0],[3,null]]
```

Constraints:

• `0 <= n <= 1000`
• `-10`4` <= Node.val <= 10`4
• `Node.random` is `null` or is pointing to some node in the linked list.

This problem was taken from https://leetcode.com/problems/copy-list-with-random-pointer/

# Brute force using tree traversal

```/**
* // Definition for a Node.
* function Node(val, next, random) {
*    this.val = val;
*    this.next = next;
*    this.random = random;
* };
*/

/**
* @return {Node}
*/

function dfsTraverse(node, visited={}) {
if(!node) {
return null;
}

// if a new node is created, return it. Otherwise you will fall into circular loops
if(node?.clone) {
return node?.clone;
}

var newNode = new Node(node?.val, null, null);
node.clone = newNode;
var next = dfsTraverse(node?.next);
var random = dfsTraverse(node?.random);

newNode.next = next;
newNode.random = random;
return newNode;
}

return result;
};

function Node(val, next, random) {
this.val = val;
this.next = next;
this.random = random;
}
;// [1,null],[2,0],[3,1]

var nodes = {};
nodes['1'] = new Node(1,null,null);
nodes['2'] = new Node(2,null,null);
nodes['3'] = new Node(3,null,null);

nodes['1'].next = nodes['2'];
nodes['1'].random = null;

nodes['2'].next = nodes['3'];
nodes['2'].random = nodes['1'];

nodes['3'].next = null;
nodes['3'].random = nodes['2'];

//console.log("root");
//console.log(nodes['7']);
var result = copyRandomList(nodes['1']);
console.log(result);

```

# More elegant solution

```/**
* // Definition for a Node.
* function Node(val, next, random) {
*    this.val = val;
*    this.next = next;
*    this.random = random;
* };
*/

/**
* @return {Node}
*/
const copy = new Map();

// add all new nodes and values for now
while (cur) {
copy.set(cur, new Node(cur.val));
cur = cur.next;
}

// iterate again and point curent node to the newly created nodes using the key
while (cur) {
copy.get(cur).next = copy.get(cur.next) || null;
copy.get(cur).random = copy.get(cur.random) || null;
cur = cur.next;
}

}

function Node(val, next, random) {
this.val = val;
this.next = next;
this.random = random;
};

// [1,null],[2,0],[3,1]

var nodes = {};
nodes['1'] = new Node(1,null,null);
nodes['2'] = new Node(2,null,null);
nodes['3'] = new Node(3,null,null);

nodes['1'].next = nodes['2'];
nodes['1'].random = null;

nodes['2'].next = nodes['3'];
nodes['2'].random = nodes['1'];

nodes['3'].next = null;
nodes['3'].random = nodes['2'];

var result = copyRandomList(nodes['1']);
console.log(result);

```

# The graph

The graph represent connections between airports. Nodes (vertices)  represent airports, and  edges represent flights. # BFS search (Breadth First Search)

Breadth-first Search (BFS) starts by pushing all of the direct children to a queue (first-in, first-out). It then visits each item in queue and adds the next layer of children to the back of the queue. Since one node could be visited multiple times causing infinite loop, we have to keep track of all visited nodes.

## Using JavaScript Map

```// DATA
const airports = 'PHX BKK OKC JFK LAX MEX EZE HEL LOS LAP LIM'.split(' ');

const routes = [
['PHX', 'LAX'],
['PHX', 'JFK'],
['JFK', 'OKC'],
['JFK', 'HEL'],
['JFK', 'LOS'],
['MEX', 'LAX'],
['MEX', 'BKK'],
['MEX', 'LIM'],
['MEX', 'EZE'],
['LIM', 'BKK'],
];

// The graph

}

}

// Create the Graph

function bfs(start, searchItem) {

const visited = new Set();
const queue = [start];

while (queue.length > 0) {

const airport = queue.shift(); // mutates the queue

for (const destination of destinations) {
if (destination === searchItem)  {
console.log(`BFS found a route to`, searchItem)
}

if (!visited.has(destination)) {
queue.push(destination);
}
}
}

}

bfs('PHX', 'BKK');```

## Using JavaScript object

```// Data
var airports = {
'PHX': ['LAX', 'JFK'],
'BKK': ['MEX', 'LIM'],
'OKC': ['JFK'],
'JFK': ['LAX', 'PHX', 'OKC', 'HEL', 'LOS', 'EZE'],
'LAX': ['PHX', 'JFK', 'MEX'],
'MEX': ['LAX', 'EZE', 'BKK', 'LIM'],
'EZE': ['JFK', 'MEX'],
'HEL': ['JFK'],
'LOS': ['JFK'],
'LAP': [],
'LIM': ['MEX', 'BKK']
};

function bfs(start, endDest) {
var queue = [start];
var visited = {};

while(queue.length > 0) {
var curentNodeVal = queue.shift();
var childNodes = airports[curentNodeVal];
for(const childNode of childNodes) {
if(childNode === endDest) {
console.log("BFS found a route to :", endDest);
}

if(!visited[childNode]) {
console.log(childNode);
visited[childNode] = true;
queue.push( childNode);
}
}
}
}

bfs('PHX', 'BKK');```

# DFS (Depth First Search)

Depth-first Search (DFS) will go as far into the graph as possible until it reaches a node without any children, at which point it backtracks and continues the process. The algorithm can be implemented with a recursive function that keeps track of previously visited nodes. If a node has not been visited, we call the function recursively.

## Using JavaScript Map

```// DATA
const airports = 'PHX BKK OKC JFK LAX MEX EZE HEL LOS LAP LIM'.split(' ');

const routes = [
['PHX', 'LAX'],
['PHX', 'JFK'],
['JFK', 'OKC'],
['JFK', 'HEL'],
['JFK', 'LOS'],
['MEX', 'LAX'],
['MEX', 'BKK'],
['MEX', 'LIM'],
['MEX', 'EZE'],
['LIM', 'BKK'],
];

// The graph

}

}

// Create the Graph

function dfs(start, visited = new Set()) {

console.log(start)

for (const destination of destinations) {

if (destination === 'BKK') {
console.log(`DFS found Bangkok`)
return;
}

if (!visited.has(destination)) {
dfs(destination, visited);
}

}

}

dfs('PHX');```

## Using JavaScript object

```// Data
var airports = {
'PHX': ['LAX', 'JFK'],
'BKK': ['MEX', 'LIM'],
'OKC': ['JFK'],
'JFK': ['LAX', 'PHX', 'OKC', 'HEL', 'LOS', 'EZE'],
'LAX': ['PHX', 'JFK', 'MEX'],
'MEX': ['LAX', 'EZE', 'BKK', 'LIM'],
'EZE': ['JFK', 'MEX'],
'HEL': ['JFK'],
'LOS': ['JFK'],
'LAP': [],
'LIM': ['MEX', 'BKK']
};

function dfs(start, endDest, visited = {}) {

visited[start] = true;
console.log(start);

const destinations = airports[start];

for(const destination of  destinations) {
if (destination === endDest) {
console.log(`DFS found route to`, endDest);
}

if(!visited[destination]) {
dfs(destination, endDest, visited);
}

}
}

dfs('PHX', 'BKK');```