This problem requires a careful binary search approach over a rotated array that can contain duplicates. Start by comparing the middle element with the boundaries to decide which half to explore. Adjust for cases where duplicates obscure the sorted order to ensure correct detection of the target without linear scans.
Problem Statement
You are given an integer array nums sorted in non-decreasing order, which may contain duplicate values. The array is rotated at an unknown pivot index such that the resulting array starts from nums[k] followed by nums[k+1] to nums[n-1], then nums[0] to nums[k-1]. Your task is to determine if a given integer target exists in this rotated array.
Return true if the target is present in nums, and false otherwise. For example, nums = [2,5,6,0,0,1,2] rotated from a sorted array may have target = 0, which should return true, or target = 3, which should return false. Handle duplicates carefully to maintain efficiency in the binary search over the valid answer space.
Examples
Example 1
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
Example details omitted.
Example 2
Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false
Example details omitted.
Constraints
- 1 <= nums.length <= 5000
- -104 <= nums[i] <= 104
- nums is guaranteed to be rotated at some pivot.
- -104 <= target <= 104
Solution Approach
Modified Binary Search
Perform a standard binary search but handle rotation by comparing mid with left and right boundaries. If duplicates cause ambiguity, increment left or decrement right to reduce the search space.
Handling Duplicate Elements
When nums[left] == nums[mid] == nums[right], we cannot determine which half is sorted, so shrink the boundaries incrementally. This ensures the search space gradually reduces without skipping potential target positions.
Decision Logic for Sorted Halves
After resolving duplicates, identify the sorted half and check if the target lies within it. Adjust pointers accordingly to continue binary search in the promising segment, maintaining O(log n) time in the best scenario and O(n) in the worst-case with many duplicates.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Best-case time complexity is O(log n) when duplicates are minimal, but worst-case can degrade to O(n) due to duplicate values. Space complexity is O(1) as the algorithm operates in-place with only pointers.
What Interviewers Usually Probe
- Expect discussion on handling duplicates during binary search in rotated arrays.
- Check for clear reasoning about deciding which half is sorted despite rotations.
- Notice whether candidate handles worst-case linear scan scenario due to identical elements.
Common Pitfalls or Variants
Common pitfalls
- Failing to handle duplicates correctly, leading to infinite loops or missed targets.
- Assuming strictly increasing array properties, which breaks with repeated elements.
- Overcomplicating the rotation logic instead of shrinking the search space carefully.
Follow-up variants
- Search in Rotated Sorted Array without duplicates, which simplifies the binary search logic.
- Find the minimum element in a rotated sorted array with duplicates.
- Count occurrences of a target in a rotated sorted array allowing duplicates.
How GhostInterview Helps
- GhostInterview highlights boundary conditions where duplicates affect rotation detection.
- It provides step-by-step decision-making for which array half to search.
- Guides on avoiding infinite loops by adjusting pointers incrementally during ambiguous cases.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
Why is a simple binary search not enough for Search in Rotated Sorted Array II?
Because duplicates can make it impossible to determine the sorted half, requiring boundary adjustments to maintain correctness.
What is the worst-case time complexity for this problem?
O(n), which occurs when duplicates force the search to reduce boundaries one step at a time.
Can this algorithm work if there are no duplicates?
Yes, it simplifies to standard rotated binary search, achieving O(log n) consistently.
How do I know which half of the array is sorted?
Compare nums[left], nums[mid], and nums[right]; after resolving duplicates, the sorted half can be identified to decide where to continue searching.
Does GhostInterview provide code solutions for Search in Rotated Sorted Array II?
It guides through the stepwise reasoning and edge-case handling rather than giving direct code, ensuring understanding of binary search adjustments.
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