Java Program For Pointing Arbit Pointer To Greatest Value Right Side Node In A Linked List

Given singly linked list with every node having an additional “arbitrary” pointer that currently points to NULL. We need to make the “arbitrary” pointer to the greatest value node in a linked list on its right side.

A Simple Solution is to traverse all nodes one by one. For every node, find the node which has the greatest value on the right side and change the next pointer. The Time Complexity of this solution is O(n²).

An Efficient Solution can work in O(n) time. Below are the steps.  

1. Reverse the given linked list.
2. Start traversing the linked list and store the maximum value node encountered so far. Make arbit of every node to point to max. If the data in the current node is more than the max node so far, update max.
3. Reverse modified linked list and return head.

Following is the implementation of the above steps.  

Java

// Java program to point arbit pointers
// to highest value on its right
class GfG{

// Link list node
static class Node
{
    int data;
    Node next, arbit;
}

/* Function to reverse the
   linked list */
static Node reverse(Node head)
{
    Node prev = null,
         current = head, next = null;
    while (current != null)
    {
        next = current.next;
        current.next = prev;
        prev = current;
        current = next;
    }
    return prev;
}

// This function populates arbit pointer
// in every node to the greatest value
// to its right.
static Node populateArbit(Node head)
{
    // Reverse given linked list
    head = reverse(head);

    // Initialize pointer to maximum
    // value node
    Node max = head;

    // Traverse the reversed list
    Node temp = head.next;
    while (temp != null)
    {
        // Connect max through arbit
        // pointer
        temp.arbit = max;

        // Update max if required
        if (max.data < temp.data)
            max = temp;

        // Move ahead in reversed list
        temp = temp.next;
    }

    // Reverse modified linked list
    // and return head.
    return reverse(head);
}

// Utility function to print result
// linked list
static void printNextArbitPointers(Node node)
{
    System.out.println("Node    " +
                       "Next Pointer    " +
                       "Arbit Pointer");
    while (node != null)
    {
        System.out.print(node.data +
                         "        ");

        if (node.next != null)
            System.out.print(node.next.data +
                             "        ");
        else
            System.out.print("NULL" +
                             "        ");

        if (node.arbit != null)
            System.out.print(node.arbit.data);
        else
            System.out.print("NULL");

        System.out.println();
        node = node.next;
    }
}

/* Function to create a new node
   with given data */
static Node newNode(int data)
{
    Node new_node = new Node();
    new_node.data = data;
    new_node.next = null;
    return new_node;
}

// Driver code
public static void main(String[] args)
{
    Node head = newNode(5);
    head.next = newNode(10);
    head.next.next = newNode(2);
    head.next.next.next = newNode(3);

    head = populateArbit(head);

    System.out.println(
           "Resultant Linked List is: ");
    printNextArbitPointers(head);
}
}
// This code is contributed by Prerna Saini.

Output:

Resultant Linked List is: 
Node    Next Pointer    Arbit Pointer
5               10              10
10              2               3
2               3               3
3               NULL            NULL

Time Complexity: O(n), where n is the number of nodes in the linked list. 

Space Complexity: O(1). The algorithm uses a constant amount of extra space to store temporary variables during the traversal of the linked list.

Recursive Solution:
We can recursively reach the last node and traverse the linked list from the end. The recursive solution doesn’t require reversing the linked list. We can also use a stack in place of recursion to temporarily hold nodes. Thanks to Santosh Kumar Mishra for providing this solution. 

Java

// Java program to point arbit pointers
// to highest value on its right
class GfG
{
    // Link list node
    static class Node
    {
        int data;
        Node next, arbit;
    }

    static Node maxNode;

    // This function populates arbit pointer
    // in every node to the greatest value
    // to its right.
    static void populateArbit(Node head)
    {
        // if head is null simply return
        // the list
        if (head == null)
            return;

        /* if head->next is null it means we
           reached at the last node just update
           the max and maxNode */
        if (head.next == null)
        {
            maxNode = head;
            return;
        }

        /* Calling the populateArbit to the
           next node */
        populateArbit(head.next);

        /* updating the arbit node of the current
           node with the maximum value on the
           right side */
        head.arbit = maxNode;

        /* if current Node value id greater then
           the previous right node then update it */
        if (head.data > maxNode.data)
            maxNode = head;

        return;
    }

    // Utility function to print result
    // linked list
    static void printNextArbitPointers(Node node)
    {
        System.out.println("Node    " +
                           "Next Pointer    " +
                           "Arbit Pointer");
        while (node != null)
        {
            System.out.print(node.data +
                             "            ");

            if (node.next != null)
                System.out.print(node.next.data +
                                 "                ");
            else
                System.out.print("NULL" +
                                 "            ");

            if (node.arbit != null)
                System.out.print(node.arbit.data);
            else
                System.out.print("NULL");

            System.out.println();
            node = node.next;
        }
    }

    /* Function to create a new node
       with given data */
    static Node newNode(int data)
    {
        Node new_node = new Node();
        new_node.data = data;
        new_node.next = null;
        return new_node;
    }

    // Driver code
    public static void main(String[] args)
    {
        Node head = newNode(5);
        head.next = newNode(10);
        head.next.next = newNode(2);
        head.next.next.next = newNode(3);

        populateArbit(head);

        System.out.println(
               "Resultant Linked List is: ");
        printNextArbitPointers(head);
    }
}
// This code is contributed by shubham96301

Output:

Resultant Linked List is: 
Node    Next Pointer    Arbit Pointer
5               10              10
10              2               3
2               3               3
3               NULL            NULL

Time complexity: The time complexity of the populateArbit() function is O(n), where n is the number of nodes in the linked list.

Space complexity: The space complexity of the program is O(n), where n is the number of nodes in the linked list. 

Please refer complete article on Point arbit pointer to greatest value right side node in a linked list for more details!