Amazon·Data Scientist·Technical Phone Screen·Senior

Senior

May 2026

Summary

Live coding screen for a Data Scientist role at Amazon. One algorithmic problem, no frills, just you and a sorted array and the expectation that you know binary search without leaning on any standard library shortcuts.

Questions Asked (1)

Given a sorted array of distinct integers and a target value, return the index of the target if it exists, or the index where it should be inserted to keep the array sorted. You must implement binary search from scratch without using any built-in search utilities.

Algorithms & Data Structures

Author's notes

I knew the problem immediately but fumbled the pointer logic on my first pass.

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AI HintsAI Generated

Suggested Approach

Use binary search to find the target or the insertion point by maintaining low and high pointers. At each step, compare the middle element with the target and adjust the search range accordingly. Return low when the search ends, as it will be the index where the target should be inserted.

Pro tip: Emphasize that the algorithm runs in O(log n) time and O(1) space, which is optimal for this problem. Also, mention that handling edge cases like empty arrays or targets outside the range is straightforward with this approach.

1. Clarify the problem and constraints

Confirm that the array is sorted, contains distinct integers, and that you need to return the index if found or the insertion index otherwise. Ask about edge cases like empty array or target not present.

2. Initialize pointers

Set low = 0 and high = len(nums) - 1 to define the search space.

3. Perform binary search

While low <= high, compute mid = (low + high) // 2. If nums[mid] == target, return mid. If nums[mid] < target, set low = mid + 1; else set high = mid - 1.

4. Return insertion index

After the loop, low is the index where the target should be inserted to maintain sorted order. Return low.

5. Test with examples

Walk through examples like target present, target smaller than all, target larger than all, and empty array to verify correctness.

Key Points to Mention

Time complexity: O(log n) due to halving the search space each iteration.
Space complexity: O(1) as only a few variables are used.
Handling duplicates: Since the array has distinct integers, no special handling is needed.
Edge cases: empty array, target less than first element, target greater than last element.
Binary search invariant: low is the smallest index where target could be inserted, high is the largest index where target could be inserted minus one.
Avoiding overflow: use mid = low + (high - low) // 2 for safety in languages with fixed-size integers.

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