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I knew this was a binary search problem pretty quickly but spent too long second-guessing the edge cases, like what happens when the array has duplicates or is already sorted.
Clarify that the array is sorted in ascending order and contains distinct elements, then explain that the number of rotations equals the index of the minimum element. Use binary search to find the minimum element in O(log n) time by comparing the middle element with the rightmost element.
Pro tip: Explicitly state that if the array is not rotated (i.e., already sorted), the answer is 0, and handle edge cases like empty array or single element gracefully. Also, mention that if duplicates are allowed, the problem becomes trickier and may require a modified binary search with O(n) worst-case time.
Confirm that the array is sorted in ascending order, contains distinct elements, and is rotated some number of times (possibly zero). Discuss edge cases: empty array, single element, and already sorted array.
Explain that rotating a sorted array shifts elements, so the minimum element ends up at the rotation point. The number of rotations is exactly the index of the minimum element.
Use binary search: compare 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). Narrow down until left == right.
Write clean code with proper loop conditions and return the index. Test with examples: [4,5,6,7,0,1,2] returns 4, [1,2,3,4] returns 0, [2,1] returns 1.
State time complexity O(log n) and space O(1). If duplicates are allowed, explain that binary search may degrade to O(n) and outline a modified approach.
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