Return Any Strict Peak Index in Logarithmic Time

### Problem Find any peak in a zero-indexed array and return its index. A peak is strictly greater than each adjacent in-bounds value. Treat the conceptual values beyond either end as negative infinity, so an endpoint can be a peak. If there are several peaks, any one is valid. Use `O(log n)` time. ```cpp int findPeakElement(vector<int>& nums); ``` ### Examples ```text Input: nums = [2,7,3,8,4] Valid output: 1 ``` Index 3 is also valid; no leftmost or largest-peak rule is imposed. ```text Input: nums = [-2147483648] Output: 0 ``` ### Constraints & Assumptions - Length is between 1 and 1000. - Every value fits signed 32-bit range. - Adjacent entries are unequal. ### What a Strong Answer Covers - A valid strict local peak, including endpoint cases and multiple valid choices. - A logarithmic search invariant using the unequal-neighbor guarantee. - Boundary handling without an overflowing finite negative-infinity sentinel. ### Follow-up Questions - Why is a singleton peak even at the minimum signed integer value? - What choices are accepted when several local peaks exist?