Find Peak Element

LeetCode 162. Binary search for any peak in an array where `nums[i] != nums[i+1]`. Edge-handling at boundaries is the differentiator. Requirements Return the index of any peak (element greater than both neighbors; out-of-bounds neighbors treated as -inf). O(log n) binary search: if nums[mid] < nums[mid+1], search right half; else search left half (inclusive). 2D variant follow-up: LeetCode 1901 — binary search on columns, scan column for max. Local-minimum variant: find a value smaller than its neighbors using an iterative approach. The interviewer may also ask you to minimize the number of conditional branches. Notes Common bug: handling the right boundary when mid + 1 is out of bounds. 2D follow-up appears in onsite rounds, not phone screens. For the local-minimum variant, clarify whether endpoints are eligible, whether comparisons are strict, and whether adjacent equal values are allowed. A strict local minimum need not exist when duplicates are allowed. A boundary-search implementation can accidentally revisit the same region. State the shrinking invariant clearly, dry-run both endpoints, and handle equality according to the agreed contract. Preparation Write the 1D binary search in <8 min with the boundary invariant proved. Drill LeetCode 1901 if you're aiming for E5+. Rework the iterative local-minimum variant with as few branch cases as possible, then test both boundaries and a minimum in the middle.