Search an element in a Doubly Linked List


Learn the most efficient way to search an element in a doubly linked list.

The linked list is one of the most important concepts and data structures to learn while preparing for interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview.

Problem Statement

In this problem, we are given a Doubly Linked List, and we need to search for a node with value X in the list and return its position.

Problem Statement Understanding:

Let’s try to understand the problem statement with the help of an example.

If the given doubly linked list is head ↔ 4 ↔ 5 ↔ 7 ↔ 6 ↔ 2 ↔ 1 ↔ 9 and X = 7.

  • According to the problem statement, we will have to search for the node with value X in the list, and finally, after finding the node, we have to return its position in the list.
  • We can clearly see that node with value 7 is present in the list, and this node is at 3rd position (considering one based indexing) in the list.
  • So finally, we will output 3, as 3 is the position of a node with value X = 7 in the given list.

Input: head ↔ 4 ↔ 5 ↔ 7 ↔ 6 ↔ 2 ↔ 1 ↔ 9, X(element to be searched) = 7
Output: 3

Also, there can be multiple nodes with value X in the list, so that's why we will return the position of the first node with value X in the list.

Note: If there is no node with the value of X in the list, the output will be -1.

Now I think the problem statement is clear, so let’s see how we can approach it.

Let’s move to the approach section.


In order to find an element X in the given doubly linked list:

  • We need to traverse the list until either the node is null or we have found the node with value X which we were searching for.
  • Also, we need a variable position that will keep the count of the elements we have visited so far.
  • Then finally, we will return the location of the element X in our list which will be given by the variable position.


  • First, we need to create a variable position that will keep track of number of nodes traversed.
  • Then using a pointer, say, temp, we will traverse the given list till the next node of the temp is null, or we found the element we were searching for.
  • Then we will check if the current node's data is equal to the element we were searching for or not.
    • If == X, our position variable will give the location of the element to be searched for and we will output it.
    • Else, it means that there is no node with value X in the given list, so we will output -1.

Dry Run

Code Implementation

using namespace std;

struct DLLNode {
	int data;
	DLLNode* next;
	DLLNode* prev;

void push(DLLNode** head, int new_data){
	DLLNode* new_node = (DLLNode*)malloc(sizeof(struct DLLNode));

	new_node->data = new_data;
	new_node->prev = NULL;
	new_node->next = (*head);

	if ((*head) != NULL) {
		(*head)->prev = new_node;

	(*head) = new_node;

int search_the_element(DLLNode** head, int X){
	DLLNode* temp = *head;
	int position = 0;

	while (temp->data != X && temp->next != NULL) {
		temp = temp->next;

	if (temp->data != X)
		return -1;

	return (position + 1);

int main()
	DLLNode* head = NULL;
	int X = 7;
	push(&head, 9);
	push(&head, 1);
	push(&head, 2);
	push(&head, 6);
	push(&head, 7);
    push(&head, 5);
	push(&head, 4);

public class Prepbytes {

    static class Node {
        int data;
        Node next;
        Node prev;

    static Node push(Node head_ref, int new_data) {
        Node new_node = new Node(); = new_data;
        new_node.prev = null; = head_ref;

        if (head_ref != null) {
            head_ref.prev = new_node;
        head_ref = new_node;
        return head_ref;

    static int search(Node head_ref, int x) {
        Node temp = head_ref;

        int position = 1;

        while ( != x && != null) {

            temp =;

        if ( != x)
            return -1;
        return position;

    public static void main(String[] args) {
        Node head = null;
        int X = 7;
        head = push(head, 9);
        head = push(head, 1);
        head = push(head, 2);
        head = push(head, 6);
        head = push(head, 7);
        head = push(head, 5);
        head = push(head, 4);
        System.out.print(search(head, X));



Time Complexity: O(n), where n is the total number of nodes in the Doubly Linked List.

So, in this blog, we have tried to explain how you can search a value in a doubly linked list. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.

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