Yes. At least one half remains normally sorted; its bounds reveal whether it can contain the target.

Rotation breaks global order once, but any middle point leaves at least one ordinary sorted half. Use that half as evidence, not the visual position of the target.

What the function promises: Return the target index or -1. Values are distinct and were produced by rotating a strictly increasing list.

def rotated_index(values, target):
    left, right = 0, len(values) - 1
    while left <= right:
        middle = (left + right) // 2
        if values[middle] == target:
            return middle
        if values[left] <= values[middle]:
            if values[left] <= target < values[middle]:
                right = middle - 1
            else:
                left = middle + 1
        elif values[middle] < target <= values[right]:
            left = middle + 1
        else:
            right = middle - 1
    return -1

assert rotated_index([4,5,6,7,0,1,2], 0) == 4
assert rotated_index([4,5,6,7,0,1,2], 3) == -1

Work and storage: O(log n) time and O(1) space under the distinct-value contract.

When to recognize it: At least one half remains normally sorted; its bounds reveal whether it can contain the target.

Check before applying it: With duplicates, equal endpoints can make the useful half ambiguous and force linear shrinking.

When left, middle, and right are equal, shrink an endpoint; worst-case time becomes O(n).