Search in Rotated Sorted Array — Interview Guide

• At each mid, one half is always sorted; identify it by comparing nums[mid] with nums[left].
• If the target lies in the sorted half, search there; otherwise search the other half.
• Handle duplicates carefully — when nums[left] == nums[mid] you cannot tell which side is sorted.

def search(nums, target):
    l, r = 0, len(nums) - 1
    while l <= r:
        m = (l + r) // 2
        if nums[m] == target: return m
        if nums[l] <= nums[m]:
            if nums[l] <= target < nums[m]: r = m - 1
            else: l = m + 1
        else:
            if nums[m] < target <= nums[r]: l = m + 1
            else: r = m - 1
    return -1

• Search with duplicates present (worst case degrades to O(n)).
• Find the rotation pivot (minimum element) first, then binary search the correct half.
• Return the index of the target if it exists, else the index where it would be inserted.