### Introduction

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 sorted Singly Linked List and an integer X, and we need to find and tell if the list contains a given element X or not.

### Problem Statement Understanding

Letâs learn programming languages online and try to understand the problem statement with the help of examples.

If the given linked list is

- According to the problem statement, the given linked list is sorted, and we need to tell if the list contains element X or not.
- If element X is present we will print
**Element Found**else we print**Element not Found**. - We can see that X = 5 is present in the linked list, so we will print âElement Foundâ.

If the given sorted linked list is headâ1â3â5â7â9âNULL and X=8.

- So we can see that X = 8 is not present in the given linked list, so we will print
**Element not Found**.

Now I think from the above examples, the problem is clear. So letâs see how we can approach it.

A straightforward way to check if an element **X** is present in the linked list is to traverse the linked list and check if for any node **(nodeâdata == X)**:

- If
**nodeâdata == X**, it will mean that element**X**is present in the linked list. - Otherwise, if none of the nodeâs data is equal to X during the whole traversal, it will mean that element X is not present in the linked list.

Although the above linear traversal to check if any element X is present in the linked list or not will work fine, but here we are asked to use **Binary search**.

So letâs see how we can use binary search to check if an element X is present in a linked list or not.

### Approach

Here we will see how we can use binary search algorithm to check whether a certain element X is present in the given list or not.

Initially, the range of binary search will be the complete list, i.e., from head to the last node of the list. Binary search first compares the target element X with the middle element based on which it reduces the range for further search.

- If the middle element is equal to the target element X, then we have found our element.
- Else, if the target element X is less than the middle element, then the range of binary search will reduce, and our new range will be from start to middle element of the linked list.
- Else, If the target element X is greater than the middle element, the search range will reduce and our new search range will be from
**middleânext**to the end of the list.

The searching will go on until one of the following condition hits:

- We find element X.
- The search range collapses, and we donât have any element in the search range to check.

The time complexity of binary search is O(log n) which is much better than linear search which takes linear time O(n) to search an element, but for binary to work properly the given must be sorted. Its time complexity is less than O(n) because it does not check every element of the given list. The search reduces to half at every iteration.

Letâs see the algorithm.

### Algorithm

To find an element X in the singly linked list, we need to find the middle of the list and then divide the list into 2 halves and perform the operations on the favorable part of the list and repeat the above steps until we find the required key in the list.

- First, we need to find the middle node of the linked list, which we will find using the slow and fast pointer method.
- In slow and fast pointer method, we will run a while loop until the fast pointer or next of the fast pointer hits null and in each iteration, we will move fast by 2 nodes at a time (fast = fastânextânext) and slow one node at a time ( slow=slowânext).
- So when the pointers exit the while loop, the slow pointer will be pointing to the middle node of the linked list.

- Then, after finding the middle node of the list, we need to compare the value of the middle node of the list with the target element X.
- If the middle node value of the list is equal to target element X, so we return middle as the answer.
- If the middle node value is less than the target element X, we update the start pointer with middleânext.
- If the middle node value is greater than the element X, we update the last pointer to the middle node.

- We perform the above operations until the start node equals last node or the last node becomes NULL.
- Then we check if the node returned by the function is not NULL we print
**Element Found**else we print**Element not Found**.

### Dry Run

### Code Implementation:

#includeusing namespace std; struct Node { int data; struct Node* next; }; Node *newNode(int x) { struct Node* temp = new Node; temp->data = x; temp->next = NULL; return temp; } struct Node* find_middle_node(Node* start, Node* last) { if (start == NULL){ return NULL; } struct Node* slow = start; struct Node* fast = start -> next; while (fast != last) { fast = fast -> next; if (fast != last) { slow = slow -> next; fast = fast -> next; } } return slow; } struct Node* binarySearch(Node *head, int value){ struct Node* start = head; struct Node* last = NULL; do { Node* mid = find_middle_node(start, last); if (mid == NULL){ return NULL; } if (mid -> data == value){ return mid; } else if (mid -> data < value){ start = mid -> next; } else{ last = mid; } } while (last == NULL || last != start); return NULL; } int main() { Node *head = newNode(2); head->next = newNode(5); head->next->next = newNode(7); head->next->next->next = newNode(11); head->next->next->next->next = newNode(15); head->next->next->next->next->next = newNode(18); int value = 5; if (binarySearch(head, value) == NULL) printf("Element not Found\n"); else printf("Element Found"); return 0; }

#### Output

Element Found

**Time Complexity:** O(n), where n is the total number of nodes in the Singly Linked List though we are performing Binary search still in the worst case the complexity is O(n).

So, in this blog, we have tried to explain how you can search if an element is present in a singly linked list using binary search. 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.