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

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  • Last Updated : 04 Aug, 2022
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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.

listwithArbit1

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.

C++




// C++ program to point arbit pointers
// to highest value on its right
#include<bits/stdc++.h>
using namespace std;
 
// Link list node
struct Node
{
    int data;
    Node* next, *arbit;
};
 
/* Function to reverse the
   linked list */
Node* reverse(Node *head)
{
    Node *prev = NULL,
         *current = head, *next;
    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.
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
void printNextArbitPointers(Node *node)
{
    printf("Node    Next Pointer    Arbit Pointer");
    while (node!=NULL)
    {
        cout << node->data <<
                "        ";
 
        if (node->next)
            cout << node->next->data <<
                    "        ";
        else cout << "NULL" << "        ";
 
        if (node->arbit)
            cout << node->arbit->data;
        else cout << "NULL";
 
        cout << endl;
        node = node->next;
    }
}
 
/* Function to create a new node with
   given data */
Node *newNode(int data)
{
    Node *new_node = new Node;
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
 
// Driver code
int main()
{
    Node *head = newNode(5);
    head->next = newNode(10);
    head->next->next = newNode(2);
    head->next->next->next = newNode(3);
 
    head = populateArbit(head);
 
    printf("Resultant Linked List is: ");
    printNextArbitPointers(head);
 
    return 0;
}


Output: 

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

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. 

C++




// C++ program to point arbit pointers
// to highest value on its right
#include <bits/stdc++.h>
using namespace std;
 
// Link list node
struct Node {
    int data;
    Node *next, *arbit;
};
 
// This function populates arbit pointer
// in every node to the greatest value
// to its right.
void populateArbit(Node* head)
{
    // using static maxNode to keep track
    // of maximum orbit node address on
    // right side
    static Node* maxNode;
 
    // 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
void printNextArbitPointers(Node* node)
{
    printf("Node    Next Pointer    Arbit Pointer");
    while (node != NULL) {
        cout << node->data << "        ";
 
        if (node->next)
            cout << node->next->data << "        ";
        else
            cout << "NULL"
                 << "        ";
 
        if (node->arbit)
            cout << node->arbit->data;
        else
            cout << "NULL";
 
        cout << endl;
        node = node->next;
    }
}
 
/* Function to create a new node
   with given data */
Node* newNode(int data)
{
    Node* new_node = new Node;
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
 
// Driver code
int main()
{
    Node* head = newNode(5);
    head->next = newNode(10);
    head->next->next = newNode(2);
    head->next->next->next = newNode(3);
 
    populateArbit(head);
 
    printf("Resultant Linked List is: ");
    printNextArbitPointers(head);
 
    return 0;
}


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 a linked list.

Auxiliary Space: O(1) since using constant space for variables

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


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