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

  • Last Updated : 27 Dec, 2021

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.

listwithArbit

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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




// C program to populate arbit pointers to 
// next higher value using merge sort
#include<stdio.h>
#include<stdlib.h>
  
// Link list node 
struct Node
{
    int data;
    struct Node* next, *arbit;
};
  
// Function prototypes 
struct Node* SortedMerge(struct Node* a, 
                         struct Node* b);
void FrontBackSplit(struct Node* source,
                    struct Node** frontRef, 
                    struct Node** backRef);
  
/* Sorts the linked list formed by 
   arbit pointers (does not change 
   next pointer or data) */
void MergeSort(struct Node** headRef)
{
    struct Node* head = *headRef;
    struct Node* a, *b;
  
    // Base case -- length 0 or 1 
    if ((head == NULL) || 
        (head->arbit == NULL))
        return;
  
    // Split head into 'a' and 'b' 
    // sublists 
    FrontBackSplit(head, &a, &b);
  
    // Recursively sort the sublists 
    MergeSort(&a);
    MergeSort(&b);
  
    // answer = merge the two sorted 
    // lists together 
    *headRef = SortedMerge(a, b);
}
  
   for details of this function */
struct Node* SortedMerge(struct Node* a, 
                         struct Node* b)
{
    struct 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);
    }
    else
    {
        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(struct Node* source,
                    struct Node** frontRef, 
                    struct Node** backRef)
{
    struct Node* fast, *slow;
  
    if (source==NULL || 
        source->arbit==NULL)
    {
        // length < 2 cases 
        *frontRef = source;
        *backRef = NULL;
        return;
    }
  
    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(struct Node** head_ref, 
          int new_data)
{
    // Allocate node 
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));
  
    // Put in the data  
    new_node->data = new_data;
  
    // Link the old list off 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(struct Node *node, 
                    struct Node *anode)
{
    printf("Traversal using Next Pointer");
    while (node!=NULL)
    {
        printf("%d, ", node->data);
        node = node->next;
    }
  
    printf("Traversal using Arbit Pointer");
    while (anode!=NULL)
    {
        printf("%d, ", 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
struct Node* populateArbit(struct Node *head)
{
    // Copy next pointers to arbit pointers
    struct Node *temp = head;
    while (temp != NULL)
    {
        temp->arbit = temp->next;
        temp = temp->next;
    }
  
    // Do merge sort for arbitrary pointers
    MergeSort(&head);
  
    // Return head of arbitrary pointer 
    // linked list
    return head;
}
  
// Driver code
int main()
{
    // Start with the empty list 
    struct 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 
    struct Node *ahead = populateArbit(head);
  
    printf("Result Linked List is: ");
    printListafter(head, ahead);
  
    getchar();
    return 0;
}


Output: 

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

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


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