Use a modified binary search that handles rotation and duplicates by comparing the middle element with the target and the boundaries. When duplicates cause ambiguity (e.g., nums[left] == nums[mid] == nums[right]), shrink the search space by moving left and right pointers inward. This maintains O(log n) average time, though worst-case becomes O(n) due to duplicates.
Pro tip: Explicitly discuss the trade-off between worst-case O(n) and average O(log n) due to duplicates, and mention that if duplicates are rare, the algorithm performs well in practice. Also, clarify that you're returning a boolean, not the index, which simplifies the logic.
Confirm the array is sorted and rotated, may contain duplicates, and the function returns a boolean. Discuss edge cases like empty array, single element, and all duplicates.
Explain that you'll use two pointers (left, right) and a while loop. At each step, compute mid and compare nums[mid] with target. If equal, return true.
If nums[left] < nums[mid], the left half is sorted; check if target lies within it. If nums[left] > nums[mid], the right half is sorted; check similarly. If nums[left] == nums[mid] == nums[right], increment left and decrement right to reduce ambiguity.
State that average time is O(log n), but worst-case is O(n) due to duplicates. Space is O(1). Mention that this is optimal for the given constraints.
Walk through a few test cases: target present, absent, duplicates causing ambiguity, and edge cases. Verify the logic step by step.
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