Search in Rotated Sorted Array

Given an integer array nums that was sorted in strictly increasing order and then rotated at an unknown pivot, and an integer target, return the index of target.
Return -1 when target does not appear in nums.
All values in nums are distinct. Your solution must run in O(log n) time.

Function
searchRotatedArray(nums: int[], target: int) → int

Examples
Example 1
nums = [4,5,6,7,0,1,2]
target = 0
return = 4
The target 0 appears at index 4.
Example 2
nums = [4,5,6,7,0,1,2]
target = 3
return = -1
The target 3 is absent, so the result is -1.
Example 3
nums = [1]
target = 0
return = -1
The only array value is 1, so 0 is absent.

Constraints
1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
nums contains distinct values.
nums was sorted in strictly increasing order and rotated at an unknown pivot.
-10^9 <= target <= 10^9

The trick: at every step of binary search, at least one half of the current range is sorted. Compare nums[lo] with nums[mid]. If nums[lo] <= nums[mid], the left half is sorted. Check if target falls inside it (nums[lo] <= target < nums[mid]). If yes, move hi to mid-1, otherwise move lo to mid+1. If the left isn't sorted, the right half is, so run the mirrored check. Values are distinct, so you skip the duplicate headache. The common pitfalls are using strict < instead of <= on the sorted check, which breaks two-element ranges, and getting the boundary inclusivity wrong on the target range. Also test the single-element array and an unrotated array. If you freeze on the boundary conditions during the live OA, StealthCoder is the hedge that hands you the clean version while you keep typing.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.