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: pick mid, then at least one side of it is always sorted. Compare nums[lo] to nums[mid]. If nums[lo] <= nums[mid], the left half is sorted. Check whether target falls inside [nums[lo], nums[mid]). If yes, set hi = mid - 1, otherwise lo = mid + 1. If the left half isn't sorted, the right half is, so run the mirror check. The common pitfall is the comparison operators. Use <= on nums[lo] <= nums[mid] because lo and mid can be the same index with a one or two element window. Another trap is looking for the pivot first and doing two searches. It works, but it's more code and more bugs. Values are distinct, so no duplicate handling is needed. Target must be O(log n), so a linear scan fails the requirement. If the boundary checks slip away mid-assessment, StealthCoder is the hedge that gives you a working solution on the spot.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.