Binary Search — Rightmost Index of Duplicate

In a sorted array with duplicate values, find the rightmost (last) occurrence index of a target value using binary search. Asked as the coding portion of an OCI IC4 phone screen after a 30-minute resume walk-through. Requirements Input: a sorted array of integers (ascending) that may contain duplicates, and a target value. Output: the largest index i such that arr[i] == target. Return -1 if the target is not present. Must use binary search — linear scan is not accepted. Notes Standard right-boundary binary search: lo, hi = 0, n - 1 result = -1 while lo <= hi: mid = (lo + hi) // 2 if arr[mid] == target: result = mid lo = mid + 1 # keep searching right elif arr[mid] < target: lo = mid + 1 else: hi = mid - 1 return result The key inversion vs the textbook "first occurrence" version: on equality, push lo = mid + 1 (move right) instead of hi = mid - 1 (move left). Common bug: forgetting to update result on equality before advancing the pointer — the loop exits without ever recording the matching index. For finding both boundaries simultaneously, run the algorithm twice (left + right) and return (left, right), the count being right - left + 1. Equivalent to LeetCode 34 ("Find First and Last Position of Element in Sorted Array"). Preparation Implement both the leftmost and rightmost binary-search variants from scratch in under 10 minutes each. Walk through the algorithm on [1, 2, 2, 2, 3], target=2 → return index 3. Trace lo, hi, mid after each iteration. This round was 30 minutes for resume + 30 for the problem — get coding done in under 15 minutes so any follow-up (counting occurrences, generalising to non-integer arrays) fits comfortably.