Use a modified binary search to find the minimum element in a rotated sorted array. Compare the middle element with the rightmost element to decide which half to search, ensuring O(log n) time. Handle edge cases like no rotation or duplicates if allowed.
Pro tip: Clarify upfront whether the array contains duplicates, as that affects the algorithm's complexity and correctness. Mentioning this shows attention to detail and prevents incorrect assumptions.
Restate the problem: a sorted array rotated at an unknown pivot, find the minimum element in O(log n). Confirm assumptions: array is sorted ascending, rotation is non-trivial (could be 0), and whether duplicates exist.
Use binary search by comparing the middle element with the rightmost element. If mid > right, the minimum is in the right half; otherwise, it's in the left half (including mid).
Initialize left=0, right=n-1. While left < right, compute mid = left + (right-left)//2. If arr[mid] > arr[right], set left = mid+1; else set right = mid. Return arr[left].
Time complexity is O(log n) because we halve the search space each iteration. Space is O(1). Discuss edge cases: array not rotated (minimum at index 0), single element, and duplicates (if present, worst-case O(n) but can be handled with modifications).
Walk through examples: [4,5,6,7,0,1,2] returns 0; [3,4,5,1,2] returns 1; [1,2,3] returns 1. Mention that the algorithm correctly handles these.
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