How is it possible to do Binary Search...

Binary search on a doubly-linked list can be performed in $$O(n)$$ time, which might seem counterintuitive given that binary search on arrays typically operates in $$O(\log n)$$ time. The key difference lies in the nature of linked lists, which do not support random access. Here’s a detailed explanation of how binary search can be implemented on a doubly-linked list and why its time complexity is $$O(n)$$:

Implementation of Binary Search on a Doubly-Linked List

1. Initial Setup:

   • Start with pointers to the head and tail of the doubly-linked list.
   • Calculate the total number of nodes in the list, $$n$$.

2. Finding the Middle Element:

   • Use two pointers, slow and fast, to find the middle element. The slow pointer moves one step at a time, while the fast pointer moves two steps at a time.
   • When the fast pointer reaches the end of the list, the slow pointer will be at the middle node.

3. Comparison and Range Adjustment:

   • Compare the middle element with the target value.
   • If the middle element matches the target, return the node.
   • If the target is less than the middle element, adjust the search range to the first half of the list.
   • If the target is greater than the middle element, adjust the search range to the second half of the list.

4. Repeat:

   • Repeat the process of finding the middle element and adjusting the search range until the target is found or the search range collapses.

Time Complexity Analysis

• Finding the Middle Element: Each time you need to find the middle element, you must traverse half of the current list segment. This traversal takes $$O(n/2)$$ time in the first iteration, $$O(n/4)$$ in the second, and so on.
• Total Time Complexity: Summing up these traversals gives a series: $$O(n/2) + O(n/4) + O(n/8) + \ldots$$. This series converges to $$O(n)$$.

Thus, the overall time complexity of performing binary search on a doubly-linked list is $$O(n)$$ because each step involves traversing a portion of the list, and the sum of these traversals is linear in the number of nodes.

Example Code

Here is a simplified example in C++ to illustrate the concept:

struct Node {
    int data;
    Node* next;
    Node* prev;
};

Node* binarySearch(Node* head, int target) {
    if (!head) return nullptr...