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 any midpoint, at least one half of the array is 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, move hi to mid - 1. If not, move lo to mid + 1. Otherwise the right half is sorted, so run the mirror check. Return mid when you hit the target, and -1 when the loop ends. The common pitfall is the comparison. Use <= for the left-half check, or two-element windows break. Another is off-by-one on the range bounds. Test with [1] and target 0, and with a fully unrotated array. Distinct values mean you don't need duplicate handling. If your mind goes blank on the boundaries mid-assessment, StealthCoder can supply the working loop as a hedge.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.