Search in Rotated Sorted Array

In this task, I worked on finding a target element in a rotated sorted array using an efficient approach.
A rotated sorted array is one that was originally sorted but then rotated at some pivot point.
For example: [4, 5, 6, 7, 0, 1, 2]

What I Did
I created a function search that:
Takes an array nums and a target value
Returns the index of the target if found
Returns -1 if the target does not exist

How I Solved It
Instead of using a linear search , I used a modified binary search.

Approach
I used two pointers:

• low starting from index 0
• high starting from index n - 1
• Then I repeatedly calculated the middle index mid.

Logic Behind It
At each step:

1. Check if Middle is Target If nums[mid] == target, return mid
2. Identify the Sorted Half

Case 1: Left Half is Sorted
If nums[low] <= nums[mid]
Check if target lies between low and mid:

YES → move high = mid - 1
NO → move low = mid + 1

Case 2: Right Half is Sorted
Otherwise, the right half is sorted
Check if target lies between mid and high:

YES → move low = mid + 1
NO → move high = mid - 1

How It Works
Detects which side is properly sorted
Narrows down the search space accordingly
Repeats until it finds the target or exhausts the range

Time Complexity
O(log n)

CODE:

class Solution:
    def search(self, nums, target):
        low, high = 0, len(nums) - 1

        while low <= high:
            mid = (low + high) // 2  

            if nums[mid] == target:
                return mid

            if nums[low] <= nums[mid]:
                if nums[low] <= target < nums[mid]:
                    high = mid - 1
                else:
                    low = mid + 1
            else:
                if nums[mid] < target <= nums[high]:
                    low = mid + 1
                else:
                    high = mid - 1

        return -1