Answer: Lyft SWE Interview Question 2024 June | Binary Search | on-campus

Solution 1: Search in Rotated Sorted Array II — Binary Search with Duplicate Handling Approach This is an extension of the standard rotated search, but the presence of **duplicates** means we cannot always determine which half is sorted. For example, in nums = [1,0,1,1,1], nums[left] == nums[mid] == nums[right]. We must handle this edge case: Initialize left = 0 and right = nums.length - 1. Calculate mid = left + (right - left) / 2. If nums[mid] == target, return ...