Apple·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Apple SWE interview, got a classic binary search variant. Not a lot of context to share but the problem itself was enough to keep me busy.

Questions Asked (1)

Given a sorted array that has been rotated at some unknown pivot, search for a target value and return its index.

Algorithms & Data Structures

Author's notes

I knew this problem but still fumbled the edge cases mid-implementation.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Use a modified binary search that determines which half of the array is sorted at each step, then decides whether the target lies in that sorted half. This achieves O(log n) time by eliminating half the search space each iteration.

Pro tip: Clarify upfront whether the array contains duplicates, as that changes the algorithm's complexity and edge cases. Also, mention that you can find the pivot first and then do a standard binary search, but the single-pass approach is more elegant and efficient.

1. Clarify assumptions and edge cases

Ask about duplicates, array size, and whether the rotation is by a known amount. Confirm that the array is sorted in ascending order before rotation.

2. Initialize pointers and loop

Set left = 0 and right = n-1. While left <= right, compute mid and compare arr[mid] with target.

3. Determine sorted half

Check if the left half (arr[left] <= arr[mid]) is sorted. If so, check if target lies within that range; if yes, move right = mid-1, else left = mid+1. Otherwise, the right half must be sorted, so check if target lies there and adjust pointers accordingly.

4. Handle duplicates (if allowed)

If arr[left] == arr[mid] == arr[right], decrement right and increment left to skip duplicates, which may degrade to O(n) in worst case.

5. Return result

If target is found, return mid. If the loop ends without finding it, return -1.

Key Points to Mention

Time complexity: O(log n) for distinct elements, O(n) worst-case with duplicates.
Space complexity: O(1) iterative approach.
The key insight: at least one half of the array is always sorted.
Handling edge cases: empty array, single element, target not present, duplicates.
Comparison with alternative: finding pivot first then binary search (two passes vs one pass).
Importance of using <= in comparisons to avoid infinite loops.

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