3 Steps to Search in a Rotated Sorted Array with Duplicates

In computer science and programming, solving problems efficiently is a cornerstone of algorithm design. One classic problem is searching in a rotated sorted array. While the basic version of this problem assumes all elements are unique, a more advanced variant involves duplicates. In this blog, we’ll dive deep into understanding and solving the problem of searching in a rotated sorted array with duplicates.

Problem Statement

Given a rotated sorted array nums (which may contain duplicates), determine if a target value exists within the array.

Example

Input:

• nums = [2, 5, 6, 0, 0, 1, 2]
• target = 0

Output:

• true

Input:

• nums = [2, 5, 6, 0, 0, 1, 2]
• target = 3

Output:

• false

The array is considered rotated if some pivot exists such that the array is split into two sorted subarrays. For example:

• A non-rotated array: [1, 2, 3, 4, 5]
• Rotated: [4, 5, 1, 2, 3] or [2, 3, 4, 5, 1]

When duplicates are introduced, the problem becomes more challenging because the presence of repeated values can obscure the pivot point and make comparisons less reliable.

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Understanding the Problem

Challenges with Duplicates

1. Ambiguity in Comparisons: If duplicate values are present at critical positions (like the boundaries of the array), the comparison logic used to identify the sorted half of the array may fail.
2. Increased Complexity: With duplicates, simple binary search might not suffice, requiring additional steps to skip redundant values.

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Approach to Solve the Problem

We’ll use a variation of binary search, modified to handle duplicates. The algorithm works as follows:

Algorithm Steps

1. Initialize Pointers: Set low to 0 and high to len(nums) - 1.
2. Iterative Search:
   • While low <= high:
     • Compute the middle index: mid = (low + high) // 2.
     • Check if nums[mid] equals the target. If true, return true.
3. Determine the Sorted Region:
   • If nums[low] < nums[mid]: The left half is sorted.
   • If nums[mid] < nums[high]: The right half is sorted.
   • If nums[low] == nums[mid] == nums[high]: Shrink the range by incrementing low and decrementing high.
4. Adjust the Search Range:
   • If the target lies in the sorted region, adjust low or high accordingly.
   • Otherwise, explore the unsorted region.
5. Return Result:
   • If the target is not found after the loop, return false.

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Implementation in Python

Here’s a Python implementation of the algorithm:

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

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

        # Check if mid is the target
        if nums[mid] == target:
            return True

        # Handle duplicates by skipping redundant values
        if nums[low] == nums[mid] == nums[high]:
            low += 1
            high -= 1
        # Left half is sorted
        elif nums[low] <= nums[mid]:
            if nums[low] <= target < nums[mid]:
                high = mid - 1
            else:
                low = mid + 1
        # Right half is sorted
        else:
            if nums[mid] < target <= nums[high]:
                low = mid + 1
            else:
                high = mid - 1

    return False

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Complexity Analysis

1. Time Complexity:
   • In the worst case (due to duplicates), the algorithm may degrade to O(n). This occurs when all elements are duplicates, as the skipping logic (low++ and high--) processes each element individually.
   • In the average and best cases, the time complexity remains O(log n), similar to binary search.
2. Space Complexity:
   • The algorithm uses O(1) extra space since the search is conducted in place.

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Edge Cases to Consider

1. Empty Array:
   • Input: nums = [], target = 1
   • Output: false
2. All Duplicates:
   • Input: nums = [2, 2, 2, 2], target = 2
   • Output: true
3. Target Outside Range:
   • Input: nums = [1, 1, 3, 1], target = 4
   • Output: false
4. Minimal Rotation:
   • Input: nums = [1, 3, 1, 1, 1], target = 3
   • Output: true

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Visualizing the Process

Let’s take the input nums = [2, 5, 6, 0, 0, 1, 2] and target = 0.

Step-by-Step Execution

1. low = 0, high = 6, mid = 3.
   • nums[mid] = 0, matches the target.
   • Return true.

For a more complex example where duplicates are handled, the algorithm would carefully shrink the range and refine its search.

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Applications

1. Search Problems:
   • This problem is a classic example in competitive programming and coding interviews.
2. Real-World Scenarios:
   • Rotated arrays with duplicates occur in situations like cyclic scheduling, circular buffers, or fault-tolerant storage systems.

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Conclusion

Searching in a rotated sorted array with duplicates is a fascinating problem that demonstrates how algorithmic nuances adapt to handle edge cases like duplicates. By understanding the underlying principles and refining the binary search method, we can solve this problem efficiently.

With the provided Python implementation and detailed explanation, you’re now well-equipped to tackle this problem in interviews or practical applications.

Here are the Imp Links –

1. https://leetcode.com/problems/search-in-rotated-sorted-array-ii/
2. https://www.geeksforgeeks.org/search-an-element-in-a-sorted-and-pivoted-array/
3. https://www.programiz.com/dsa/binary-search
4. https://stackoverflow.com/questions/18562840/search-in-a-rotated-sorted-array-with-duplicates
5. https://codeforces.com/blog/entry/73890

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