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

Last Updated : 20 Mar, 2023
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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.  


// Javascript program to populate
// arbit pointers to next higher
// value using merge sort
var head;
// Link list node
class Node
    { = val;
        this.arbit = null; = null;
// Utility function to print
// result linked list
function printList(node, anode)
    "Traversal using Next Pointer<br/>");
    while (node != null)
        document.write( + ", ");
        node =;
    "<br/>Traversal using Arbit Pointer<br/>");
    while (anode != null)
        document.write( + ", ");
        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
function populateArbit(start)
    var temp = start;
    // Copy next pointers to arbit
    // pointers
    while (temp != null)
        temp.arbit =;
        temp =;
    // Do merge sort for arbitrary pointers
    // and return head of arbitrary pointer
    // linked list
    return MergeSort(start);
/* Sorts the linked list formed by arbit
   pointers (does not change next pointer
   or data) */
function MergeSort(start)
    // Base case -- length 0 or 1
    if (start == null ||
        start.arbit == null)
        return start;
    /* Split head into 'middle' and
       'nextofmiddle' sublists */
    var middle = getMiddle(start);
    var nextofmiddle = middle.arbit;
    middle.arbit = null;
    // Recursively sort the sublists
    var left = MergeSort(start);
    var right = MergeSort(nextofmiddle);
    /* answer = merge the two sorted lists
       together */
    var sortedlist = SortedMerge(left,
    return sortedlist;
// Utility function to get the
// middle of the linked list
function getMiddle(source)
    // Base case
    if (source == null)
        return source;
    var fastptr = source.arbit;
    var slowptr = source;
    // Move fastptr by two and slow ptr
    // by one. Finally slowptr will point
    // to middle node
    while (fastptr != null)
        fastptr = fastptr.arbit;
        if (fastptr != null)
            slowptr = slowptr.arbit;
            fastptr = fastptr.arbit;
    return slowptr;
function SortedMerge(a, b)
    var result = null;
    // Base cases
    if (a == null)
        return b;
    else if (b == null)
        return a;
    // Pick either a or b, and recur
    if ( <=
        result = a;
        result.arbit =
        SortedMerge(a.arbit, b);
        result = b;
        result.arbit = SortedMerge(a, b.arbit);
    return result;
// Driver code
/* Let us create the list shown above */
head = new Node(5); = new Node(10); = new Node(2); = new Node(3);
// Sort the above created Linked List
var ahead = populateArbit(head);
"Result Linked List is:<br/>");
printList(head, ahead);
// This code is contributed by gauravrajput1


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!

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