Javascript 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.

Javascript

<script>
// Javascript program to point arbit pointers
// to highest value on its right

// Link list node
class Node
{
    constructor(val)
    {
        this.data = val;
        this.arbit = null;
        this.next = null;
    }
}

// Function to reverse the linked list
function reverse(head)
{
    var 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.
function populateArbit(head)
{
    // Reverse given linked list
    head = reverse(head);

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

    // Traverse the reversed list
    var 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
function printNextArbitPointers(node)
{
    document.write("Node     " +
                   "Next Pointer    " +
                   "Arbit Pointer<br/>");
    while (node != null)
    {
        document.write(node.data +
                       "          ");
        if (node.next != null)
            document.write(node.next.data +
                           "                    ");
         else
             document.write("NULL" +
                            "          ");

         if (node.arbit != null)
             document.write(node.arbit.data);
         else
             document.write("NULL");

         document.write("<br/>");
         node = node.next;
    }
}

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

// Driver code
var head = newNode(5);
head.next = newNode(10);
head.next.next = newNode(2);
head.next.next.next = newNode(3);
head = populateArbit(head);
document.write(
         "Resultant Linked List is: <br/>");
printNextArbitPointers(head);
// This code is contributed by umadevi9616
</script>

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).

Space complexity : O(1).

Recursive Solution:
We can recursively reach the last node and traverse the linked list from the end. The recursive solution doesn’t require reversing of 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. 

Javascript

<script>
// Javascript program to point arbit pointers
// to highest value on its right

// Link list node
class Node
{
    constructor()
    {
         this.data = 0;
         this.arbit = null;
         this.next = null;
    }
}

var maxNode;

// This function populates arbit pointer
// in every node to the greatest value
// to its right.
function populateArbit(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
function printNextArbitPointers(node)
{
    document.write(
             "Node Next Pointer Arbit Pointer<br/>");
    while (node != null)
    {
        document.write(node.data +
                       "     ");

        if (node.next != null)
            document.write(node.next.data +
                           "     ");
         else
             document.write("NULL" +
                            "   ");

         if (node.arbit != null)
             document.write(node.arbit.data);
         else
             document.write("NULL");

         document.write("<br/>");
         node = node.next;
    }
}

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

// Driver code
var head = newNode(5);
head.next = newNode(10);
head.next.next = newNode(2);
head.next.next.next = newNode(3);
populateArbit(head);
document.write(
         "Resultant Linked List is: <br/>");
printNextArbitPointers(head);
// This code contributed by gauravrajput1
</script>

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 no of nodes in given linked list

Auxiliary Space: O(N)

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