Searching for a target element in a rotated sorted array presents an interesting challenge that tests understanding of both array manipulation and binary search techniques. This article explores how to perform an efficient search by adapting the classic search algorithm for a sorted array that has undergone array rotation.

Understanding Rotated Sorted Arrays

A sorted array usually allows straightforward searching using binary search. However, when the array experiences an array shift—a process known as array rotation—the order is disrupted at a rotation point. Despite the rotation, parts of the array remain sorted, and a slight modification of the search technique can still enable logarithmic time search performance.

Example of a rotated sorted array:

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Original Sorted Array: [1, 2, 3, 4, 5, 6, 7]
After Rotation: [5, 6, 7, 1, 2, 3, 4]

Problem Statement

Given a rotated sorted array and a target value, the task is to find the element location or return -1 if it is not present. The goal is to maintain efficient search performance, ideally in logarithmic time.

Brute Force Approach: Why Not?

A basic search algorithm like linear search would check each element one by one. While simple, this has a time complexity of O(n), which is far less efficient than desired for larger arrays.

Optimized Approach: Modified Binary Search

To achieve an efficient search, we adapt the binary search by considering the characteristics introduced by the array rotation.

Steps:

1. Initialize two pointers: left at the start and right at the end of the array.
2. Perform a binary search by calculating the middle index.
3. Determine which side of the array is properly sorted.
4. Use pivot index analysis to decide whether to search the left or right segment.
5. Adjust pointers accordingly and continue until the target is found or pointers cross.

Key Concepts: Pivot Index and Rotation Point

The pivot index represents where the smallest value (i.e., the true start of the sorted array) now resides after the array shift. This index helps determine which half of the sorted array remains sorted during the search.

Finding the rotation point is crucial because it indicates the division between two sorted segments, enabling us to make informed decisions during each iteration of the binary search.

Binary Search Algorithm for Rotated Array

Here is the Python code for clarity:

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def search(nums, target):
    left, right = 0, len(nums) - 1
    
    while left <= right:
        mid = (left + right) // 2
        
        if nums[mid] == target:
            return mid
        
        # Determine if the left half is sorted
        if nums[left] <= nums[mid]:
            if nums[left] <= target < nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        else:
            # Right half is sorted
            if nums[mid] < target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1
    
    return -1

This code guarantees logarithmic time efficiency by eliminating half of the search space at each step.

Time Complexity and Space Complexity

• Time Complexity: O(log n) due to the nature of binary search.
• Space Complexity: O(1) as only a few pointers are used for element location tracking.

Common Mistakes to Avoid

• Failing to correctly identify the sorted half.
• Incorrectly moving the pointers (left and right) based on wrong conditions.
• Ignoring the special handling required when the array is not rotated at all.

Conclusion

Searching an element in a rotated sorted array combines the logical reasoning behind binary search with insights into the pivot index and rotation point. By carefully analyzing the array rotation and maintaining efficient search techniques, one can find the target value quickly, achieving optimal logarithmic time performance. Mastering this approach is crucial for coding interviews and real-world problems involving dynamic datasets.