SEARCH IN ROTATED SORTED ARRAY

 

Search in Rotated Sorted Array – Modified Binary Search

Introduction

In this problem, we are given a sorted array that has been rotated at some unknown index. Our task is to find a target element in this array in O(log n) time.

Since the array is not fully sorted anymore, a normal binary search will not work directly. So we modify the binary search logic.

Problem Statement

Given:

  • A sorted array that is rotated

  • A target value

Return:

  • Index of the target if found

  • Otherwise return -1

Example

Input:

nums = [4,5,6,7,0,1,2], target = 0

Output:

4

Solution Code

class Solution {
    public int search(int[] nums, int target) {
        int start = 0, end = nums.length - 1;

        while (start <= end) {
            int mid = start + (end - start) / 2;

            if (nums[mid] == target) return mid;

            
            if (nums[mid] >= nums[start]) {
                if (target >= nums[start] && target < nums[mid]) {
                    end = mid - 1;
                } else {
                    start = mid + 1;
                }
            } 
            
            else {
                if (target > nums[mid] && target <= nums[end]) {
                    start = mid + 1;
                } else {
                    end = mid - 1;
                }
            }
        }

        return -1;
    }
}

Approach Used: Modified Binary Search

Idea

Even though the array is rotated:

  • At least one half will always be sorted

We use this property to decide where to search.

How It Works

  1. Find mid

  2. Check if nums[mid] == target = return index

  3. Identify sorted half:

    • If nums[start] <= nums[mid] = left half is sorted

    • Else = right half is sorted

  4. Decide where target lies:

    • If target is inside sorted half = search there

    • Else = search in the other half

Why This Approach Was Chosen

  • Maintains O(log n) time

  • Efficient even after rotation

  • Uses properties of sorted array

Alternative Approach

  • Linear Search

    for(int i = 0; i < nums.length; i++)

     Time Complexity: O(n)

Why not used?
Does not meet the required O(log n) constraint

Time and Space Complexity

  • Time Complexity: O(log n)

  • Space Complexity: O(1)

Important Observation

  • One half of the array is always sorted

  • Rotation does not completely destroy sorted order

  • That’s the key idea behind solving this

Edge Cases

  • Array with one element

  • Target not present

  • Rotation at index 0 (normal sorted array)

Conclusion

By slightly modifying binary search, we can efficiently search in a rotated sorted array. The key insight is identifying the sorted half in each step and narrowing the search accordingly.

Comments

Post a Comment

Popular posts from this blog

THE DELIVERY MAN

  The Delivery Man – Python List Operations  Introduction Lists in Python are used to store multiple items in a single variable. They are flexible, allowing us to add, remove, update, and manipulate elements easily. In this task, different list operations are performed to simulate a delivery man's workflow. 1. Creating a list and accessing an item items = ["Notebook", "Pencil", "Eraser", "Ruler", "Marker"] print(items[2]) Approach: Indexing Why: Lists use zero-based indexing, so the third item is accessed using index 2. 2. Adding a new item items.append("Glue Stick") print(items) Approach: append() method Why: It adds an item at the end of the list, which is ideal for adding new deliveries. 3. Inserting an item items.insert(2, "Highlighter") print(items) Approach: insert() method Why: Used when we need to add an item at a specific position instead of the end. 4. Removing an item items.remove("Ruler...
Read more

EC2 LAUNCHING

  EC2 Step 1: Launch EC2 Instance Go to: AWS Console --> EC2 --> Launch Instance Configure: Name : my-web-server AMI : Amazon Linux 2 (easy for beginners) Instance Type : t2.micro (free tier) Step 2: Key Pair Create or select a key pair. Step 3: Network Settings  Allow these: SSH (port 22) --> for login  HTTP (port 80) --> for web access Step 4: Launch Instance Click Launch Instance Step 5: Connect to Instance Use: EC2 --> Connect --> EC2 Instance Connect (easiest) OR via terminal: with this command. for this you have to download the pem files while selecting the key pair.  ssh -i your-key.pem ec2-user@<public-ip> Step 6: Install Web Server For Amazon Linux: sudo yum update -y sudo yum install httpd -y Step 7: Start Web Server sudo systemctl start httpd sudo systemctl enable httpd Step 8: Add a Web Page echo "Hello from EC2" | sudo tee /var/www/html/index.html Step 9: Get Public IP Go to EC2 d...
Read more

SORT A LINKED LIST USING MERGE SORT

  Sort a Linked List using Merge Sort Introduction Sorting a linked list efficiently is a common problem in data structures. Unlike arrays, linked lists do not allow direct access to elements, so algorithms like quicksort are not ideal. Merge Sort is the best choice for linked lists because: It does not require random access It works efficiently with node-based structures Problem Statement Given the head of a linked list: Sort the list in ascending order Return the sorted linked list Example Input: 10 -> 30 -> 20 -> 40 -> 60 -> 50 Output: 10 -> 20 -> 30 -> 40 -> 50 -> 60 Solution Code class Solution: def mergeSort(self, head): if not head or not head.next: return head mid = self.getMiddle(head) next_to_mid = mid.next mid.next = None left = self.mergeSort(head) right = self.mergeSort(next_to_mid) return self.merge(left, right) def getMiddle(self,...
Read more