15.Copy List with Random Pointer

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

Return a deep copy of the list.

Example

Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]

Solution I :(Using Extra Space)

Algorithm: 1. Create all nodes in copy linked list using next pointers. 2. Store the next pointer mappings of original linked list. 4. Make the random pointer of copy linked list point to original Linked List. 5. Make the Next pointer of the original Linked List point to copy linked list. 6. Set the random pointer of copy linked list to original values using node->random = node->random->random->next 7.Change next pointer of all nodes in original linked list to point to the corresponding node in linked list.

class Solution
{
public:
    Node *copyRandomList(Node *head)
    {
        
        if(head == NULL){return NULL;}

        vector<Node *> v;

        Node *head1 = NULL;
        Node *p = head;
        Node *q = NULL;
        Node *t = NULL;

        // Creating a copy of original ll
        while (p != NULL)
        {
            Node *temp = new Node(p->val);
            if (head1 == NULL)
            {
                head1 = temp;
                t = head1;
            }
            else
            {
                t->next = temp;
                t = t->next;
            }
            p = p->next;
            // Store the original address
            v.push_back(p);
        }

        //changing random pointer
        p = head;
        t = head1;
        while (p != NULL)
        {
            t->random = p;
            p = p->next;
            t = t->next;
        }

        //changing next pointer
        p = head;
        q = head;
        t = head1;
        while (p != NULL)
        {
            p = p->next;
            q->next = t;
            q = p;
            t = t->next;
        }

        //assigning random pointer its correct value
        t = head1;
        while (t != NULL)
        {
            if (t->random->random != NULL)
            {
                t->random = t->random->random->next;
            }
            else
            {
                t->random = NULL;
            }

            t = t->next;
        }

         // Restore the original next pointer
        p = head;
        for (int i = 0; i < v.size(); i++)
        {
            p->next = v[i];
            p = p->next;
        }

        return head1;
    }
};

Time Complexity: O(n) , Space Complexity: O(n)

Solution II: (Using Constant Space)

Algorithm :

• Create the copy of node 1 and insert it between node 1 & node 2 in original Linked List, create a copy of 2 and insert it between 2 & 3. Continue in this fashion, add the copy of N after the Nth node.

• Now copy the random link:

 original->next->random= original->random->next

• Now restore the original and copy linked lists in this fashion in a single loop.

original->next = original->next->next; 
copy->next = copy->next->next;

class Solution
{
public:
    Node *copyRandomList(Node *head)
    {

        if (head == NULL)
        {
            return head;
        }

        Node *p = head;
        Node *q = head;

        //Inserting a copy in between original values
        while (p != NULL)
        {
            q = p->next;
            Node *temp = new Node(p->val);
            p->next = temp;
            temp->next = q;
            p = q;
        }

        //Copying the random pointers
        p = head;
        while (p)
        {
            if (p->random)
            {
                p->next->random = p->random->next;
            }
            else
            {
                p->next->random = NULL;
            }

            if (p->next)
            {
                p = p->next->next;
            }
            else
            {
                p = p->next;
            }
        }

        //Separting original and random pointers
        Node *orgi = head;
        Node *copy = head->next;
        Node *temp = copy;

        while (orgi && copy)
        {
            if (orgi->next)
            {
                orgi->next = orgi->next->next;
            }
            else
            {
                orgi->next = NULL;
            }

            if (copy->next)
            {
                copy->next = copy->next->next;
            }
            else
            {
                copy->next = NULL;
            }

            orgi = orgi->next;
            copy = copy->next;
        }

        return temp;
    }
};

Time Complexity: O(n)