LinkedIn·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

LinkedIn SWE interview with a classic rotated array search problem. Pretty standard algorithmic round but the O(log n) constraint is what makes it interesting, you can't just scan linearly and call it a day.

Questions Asked (1)

You're given a sorted integer array that's been rotated at some unknown pivot. Given a target value, return its index or -1 if it's not there. All values are distinct. Must run in O(log n).

Algorithms & Data Structures

Author's notes

The O(log n) requirement is basically telling you binary search, but the rotation breaks the usual assumption that one side is always smaller.

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AI HintsAI Generated

Suggested Approach

Use a modified binary search that determines which half is sorted at each step. Compare the target with the sorted half to decide which half to search next. This maintains O(log n) time complexity.

Pro tip: Clearly explain the condition for identifying the sorted half and how it guides the search. Mention edge cases like empty array or single element to show thoroughness.

1. Initialize pointers

Set left and right pointers to the start and end of the array.

2. Binary search loop

While left <= right, compute mid and check if it's the target.

3. Identify sorted half

Determine if the left half (from left to mid) is sorted by comparing nums[left] and nums[mid].

4. Decide search direction

If left half is sorted, check if target lies within its range; if so, search left, else search right. Otherwise, do the symmetric check for the right half.

5. Return result

If loop ends without finding target, return -1.

Key Points to Mention

Time complexity O(log n) due to halving search space each iteration.
Space complexity O(1) using iterative approach.
Handling of duplicates is not needed as all values are distinct.
Edge cases: empty array, single element, target not present.
The algorithm works by leveraging the fact that at least one half is always sorted.
Comparison logic: use nums[left] <= nums[mid] to check if left half is sorted.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.