# Clone a linked list with next and random pointer in O(1) space

### Introduction

One of the most crucial data structures to learn while preparing for interviews is the linked list. In a coding interview, having a thorough understanding of Linked Lists might be a major benefit.

### Problem Statement

In this problem, we are given a linked list of length N having two pointers in each node. The first pointer points to the next node of the list, and the second one is a random pointer which can point to any node of the linked list, or NULL. Now we are asked to write a program that clones the given list in O(1) space.

Note: The cloned linked list should consist of exactly N new nodes, where the value of each node is the same as the corresponding value in original linked list and the next and random pointers of the new nodes should point to new nodes in the cloned linked list in such a way that the pointers in the original linked list and cloned linked state represent the same list state. Finally, you have to return the head of the cloned linked list.

### Problem statement understanding

First, we need to understand what cloning a linked list means.

• Cloning means copying the linked list without changing the structure and the data inside it. Addresses of the nodes will change as new nodes will be created for the cloned list. From the above figure, we can see that the addresses of the cloned nodes are different, but the structure and data of the linked list are the same.

• So basically, first, we have to clone a linked list and then return the head of the cloned linked list as output.
##### Example Output: A new linked list that is a copy of the original list Now I think from the above examples, the problem statement is clear. So let's see how we will approach it. Any Ideas?

• If not, it's okay. We will see in the next section thoroughly how we can approach this problem.

Let’s move to the next section.

### Approach

Our approach will be simple:

• We will make a copy of the given linked list so that the next pointer of every node will be simply pointing to the copy of the next node.
• But for the random pointer, we will find out which node number it is pointing at in the original linked list.
• For example, in the original linked list the random pointer of 1st node is pointing to the 3rd node so we can see that in the cloned list also the random pointer of the first node is pointing to the third node.
• For doing this in the O(n) time complexity. We will use a clever technique. What we will do is that we will make a copy of the node and put it between the node and node->next. In this way we will observe that we can do the assignment of random pointers in O(1) time for every n node.
• Finally, after random pointer linking, we will separate the cloned list from the sequence of original and copied nodes and output the cloned linked list.

Let's move to the algorithm section to see the above approach step by step in detail.

### Algorithm

• We will first create a copy of 1st node and insert it between 1st node and 2nd in the original linked list.
• Similarly, we will create a copy of 2nd and will insert it between 2nd and 3rd node. We will continue in this manner and add the copy of node Xth between Xth node and (X+1)th node. The example given below will help you to understand this step better.
• If A -> B -> C is our original linked list, then the linked list after the above steps of adding a copy of Xth between Xth and (X+1)th node will be be A -> A’ -> B -> B’ -> C -> C’.
• Now we will link the random pointers of the newly created nodes in the following manner:
• original->next->random = original->random->next.
• Why does this work? This works because original->next is simply a copy of the original and original->random->next is simply a copy of the random.
• Now we will restore the original and cloned linked lists in the below manner by using a single loop.
• original->next = original->next->next
• copy->next = copy->next->next
• Taking the previous example only, A -> B -> C is the original list and the cloned list will be A’ -> B’ -> C’.
• Make sure that original->next is NULL and finally return the head of the cloned list.

### Dry Run        ### Code Implementation

```#include
using namespace std;

/* Structure of a node */
struct Node
{
int data;
Node *next,*random;
Node(int a)
{
data = a;
next = NULL;
random = NULL;
}
};

/* This function is used to print the linked list */
{
while (ptr != NULL)
{
cout << "Data = " << ptr->data << ", Random  = " << ptr->random->data << endl;
ptr = ptr->next;
}
}

/* This function will return the cloned linked list */
{
while (curr != NULL)
{
temp = curr->next;
curr->next = new Node(curr->data);
curr->next->next = temp;
curr = temp;
}

while (curr != NULL)
{
if(curr->next)
curr->next->random = curr->random ?
curr->random->next : curr->random;
curr = curr->next?curr->next->next:curr->next;
}

temp = copy;

while ((original !=NULL) && (copy != NULL))
{
original->next = original->next? original->next->next : original->next;
copy->next = copy->next?copy->next->next:copy->next;
original = original->next;
copy = copy->next;
}
return temp;
}

int main()
{

cout << "Original linked list : \n";
cout << "\nCloned linked list : \n";
return 0;
}

```

#### Output

Data = 1, Random = 5
Data = 3, Random = 7
Data = 5, Random = 3
Data = 7, Random = 1