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Javascript Program For Counting Rotations In Sorted And Rotated Linked List

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Given a linked list of n nodes which is first sorted, then rotated by k elements. Find the value of k.

The idea is to traverse singly linked list to check condition whether current node value is greater than value of next node. If the given condition is true, then break the loop. Otherwise increase the counter variable and increase the node by node->next. Below is the implementation of this approach.


// Program for count number of rotations in
// sorted linked list.
/* Linked list node */
class Node
// Function that count number of
// rotation in singly linked list.
function countRotation(head)
    // declare count variable and assign it 1.
    let count = 0;
    // declare a min variable and assign to
    // data of head node.
    let min =;
    // check that while head not equal to null.
    while (head != null) {
        // if min value is greater then
        // then it breaks the while loop and
        // return the value of count.
        if (min >
        // head assign the next value of head.
        head =;
    return count;
// Function to push element in linked list.
function push(head,data)
    // Allocate dynamic memory for newNode.
    let newNode = new Node();
    // Assign the data into newNode. = data;
    // assign the address of
    // head node. = (head);
    // newNode become the headNode.
    (head) = newNode;
    return head;
// Display linked list.
function printList(node)
    while (node != null) {
        document.write(" ");
        node =;
// Driver functions
// Create a node and initialize with null
let head = null;
// push() insert node in linked list.
head = push(head, 12);
head = push(head, 11);
head = push(head, 8);
head = push(head, 5);
head = push(head, 18);
head = push(head, 15);
document.write("Linked list rotated elements: ");
// Function call countRotation()
document.write(countRotation(head) +"<br>");
// This code is contributed by unknown2108


15 18 5 8 11 12 
Linked list rotated elements: 2

Time Complexity: O(n), where n is the number of elements present in the linked list. This is because we are traversing the whole linked list in order to find the count.
Auxiliary Space: O(1), As we are not using any extra space.

Please refer complete article on Count rotations in sorted and rotated linked list for more details!

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Last Updated : 29 Dec, 2022
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