Binary search on answers, custom comparators, and sorting internals still asked about.
4 questions in Data Structures & Algorithms.
Binary search for the rotation point by comparing the middle element with the right end. If it is larger, the break lies to the right; otherwise it is at mid or before it. That is O(log n). With duplicates an equal comparison says nothing, so you drop the right end, and the worst case becomes O(n).
It is still binary search on the answer, over the gap rather than the positions, but the predicate flips. A gap that works means every smaller gap works, so you want the last feasible value, which needs lo = mid with a ceiling midpoint. Sort first, check each gap greedily, and the cost is O(n log n + n log R).
When you cannot construct the answer but can cheaply test a candidate, and that test is monotone, binary search the answer's numeric range instead of the input. The feasibility check here is greedy day-packing, giving O(n log S) time, and the whole implementation hinges on one loop invariant. It also connects monotonicity to the point an interviewer is testing.
Search a rotated sorted array in O(log n) by finding which half is sorted at each midpoint, then keeping the half where the target can legally fall. With duplicates, equal endpoints can hide the pivot, so the worst case becomes linear.