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C++ Program For Pointing To Next Higher Value Node In A Linked List With An Arbitrary Pointer

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Given singly linked list with every node having an additional “arbitrary” pointer that currently points to NULL. Need to make the “arbitrary” pointer point to the next higher value node.


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A Simple Solution is to traverse all nodes one by one, for every node, find the node which has the next greater value of the current node and change the next pointer. Time Complexity of this solution is O(n2).

An Efficient Solution works in O(nLogn) time. The idea is to use Merge Sort for linked list
1) Traverse input list and copy next pointer to arbit pointer for every node. 
2) Do Merge Sort for the linked list formed by arbit pointers.

Below is the implementation of the above idea. All of the merger sort functions are taken from here. The taken functions are modified here so that they work on arbit pointers instead of next pointers.  


// C++ program to populate arbit pointers
// to next higher value using merge sort
#include <bits/stdc++.h>
using namespace std;
// Link list node
class Node
    int data;
    Node* next, *arbit;
// Function prototypes
Node* SortedMerge(Node* a, Node* b);
void FrontBackSplit(Node* source,
                    Node** frontRef,
                    Node** backRef);
/* Sorts the linked list formed by
   arbit pointers (does not change
   next pointer or data) */
void MergeSort(Node** headRef)
    Node* head = *headRef;
    Node* a, *b;
    /* Base case -- length 0
       or 1 */
    if ((head == NULL) ||
        (head->arbit == NULL))
    /* Split head into 'a' and
       'b' sublists */
    FrontBackSplit(head, &a, &b);
    // Recursively sort the sublists
    /* answer = merge the two sorted
       lists together */
    *headRef = SortedMerge(a, b);
   for details of this function */
Node* SortedMerge(Node* a, Node* b)
    Node* result = NULL;
    // Base cases
    if (a == NULL)
        return (b);
    else if (b == NULL)
        return (a);
    // Pick either a or b, and recur
    if (a->data <= b->data)
        result = a;
        result->arbit =
        SortedMerge(a->arbit, b);
        result = b;
        result->arbit =
        SortedMerge(a, b->arbit);
    return (result);
/* Split the nodes of the given list into
   front and back halves, and return the
   two lists using the reference parameters.
   If the length is odd, the extra node
   should go in the front list. Uses the
   fast/slow pointer strategy. */
void FrontBackSplit(Node* source,
                    Node** frontRef,
                    Node** backRef)
    Node* fast, *slow;
    if (source == NULL ||
        source->arbit == NULL)
        // length < 2 cases
        *frontRef = source;
        *backRef = NULL;
    slow = source,
    fast = source->arbit;
    /* Advance 'fast' two nodes, and
       advance 'slow' one node */
    while (fast != NULL)
        fast = fast->arbit;
        if (fast != NULL)
            slow = slow->arbit;
            fast = fast->arbit;
    /* 'slow' is before the midpoint
        in the list, so split it in
        two at that point. */
    *frontRef = source;
    *backRef = slow->arbit;
    slow->arbit = NULL;
/* Function to insert a node at
   the beginning of the linked list */
void push(Node** head_ref, int new_data)
    // Allocate node
    Node* new_node = new Node();
    // Put in the data
    new_node->data = new_data;
    // Link the old list of the
    // new node
    new_node->next = (*head_ref);
    new_node->arbit = NULL;
    // Move the head to point to the
    // new node
    (*head_ref) = new_node;
// Utility function to print result
// linked list
void printListafter(Node *node,
                    Node *anode)
    cout << "Traversal using Next Pointer";
    while (node!=NULL)
        cout << node->data << ", ";
        node = node->next;
    printf("Traversal using Arbit Pointer");
    while (anode!=NULL)
        cout << anode->data << ", ";
        anode = anode->arbit;
// This function populates arbit pointer
// in every node to the next higher value.
// And returns pointer to the node with 
// minimum value
Node* populateArbit(Node *head)
    // Copy next pointers to arbit
    // pointers
    Node *temp = head;
    while (temp != NULL)
        temp->arbit = temp->next;
        temp = temp->next;
    // Do merge sort for arbitrary
    // pointers
    // Return head of arbitrary pointer
    // linked list
    return head;
// Driver code
int main()
    // Start with the empty list
    Node* head = NULL;
    // Let us create the list shown
    // above
    push(&head, 3);
    push(&head, 2);
    push(&head, 10);
    push(&head, 5);
    // Sort the above created Linked List
    Node *ahead = populateArbit(head);
    cout << "Result Linked List is: ";
    printListafter(head, ahead);
    return 0;
// This is code is contributed by rathbhupendra


Result Linked List is:
Traversal using Next Pointer
5, 10, 2, 3,
Traversal using Arbit Pointer
2, 3, 5, 10,

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

Space Complexity: O(1). We are not using any extra space.

Please refer complete article on Point to next higher value node in a linked list with an arbitrary pointer for more details!

Last Updated : 03 Apr, 2023
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