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Microsoft·Data Scientist·Technical Phone Screen·Senior

Senior
Jun 2026

Summary

Microsoft data science technical phone screen that went pretty deep into algorithms territory. The problem looked like a search question on the surface but kept expanding into edge cases and proofs, which I was not fully ready for.

Questions Asked (4)

Q1

You have a sorted non-decreasing integer array of unknown length. You can only access it through an API that returns +infinity for out-of-bounds indices. Implement a lower_bound function that returns the smallest index where the value is >= target, or -1 if no such index exists. Your solution must run in O(log n) time.

Algorithms & Data StructuresAPI & IntegrationsTechnical Trade-offs
Author's notes

I knew exponential search was the move but fumbled explaining why doubling the index works for bounding n.

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AI HintsAI Generated

Suggested Approach

First, find the upper bound of the array by exponentially increasing the index (1, 2, 4, 8, ...) until the API returns +infinity. Then perform a binary search on the range [0, high] to find the lower bound. If the target is greater than all elements, return -1.

Pro tip: Clarify the API's behavior: it returns +infinity for out-of-bounds, so you can safely compare with target. Also, discuss edge cases like empty array, target smaller than all elements, and target larger than all elements.

1. Understand the problem and constraints

Restate the problem: sorted array, unknown length, API access, O(log n) time. Clarify that lower_bound returns the first index where value >= target, or -1 if none.

2. Find the search bounds

Use exponential search to find a high index such that the API returns +infinity. Start with index 1, double until out-of-bounds. This takes O(log n) time.

3. Binary search for lower bound

Perform binary search on the range [0, high] to find the smallest index where value >= target. If no such index (i.e., target > last element), return -1.

4. Handle edge cases and return result

Check if the array is empty (API returns +infinity at index 0). If target is greater than all elements, return -1. Otherwise, return the found index.

Key Points to Mention

  • Exponential search to find the upper bound in O(log n) time.
  • Binary search to find the lower bound within the identified range.
  • Handling of +infinity from the API as a sentinel for out-of-bounds.
  • Time complexity analysis: O(log n) for both exponential and binary search.
  • Edge cases: empty array, target smaller than all elements, target larger than all elements.
  • Space complexity: O(1) iterative approach.

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

Q2

How does your approach handle edge cases: an empty array, all values less than the target, all values greater than or equal to the target, and duplicate values?

Algorithms & Data Structures
Author's notes

Went fine.

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AI HintsAI Generated

Suggested Approach

Walk through each edge case systematically, explaining how your algorithm handles it without errors and maintains correctness. Emphasize the importance of defining clear termination conditions and invariants, and mention how you would test these cases.

Pro tip: Relate edge cases to real-world data scenarios (e.g., empty datasets, skewed distributions) to show practical awareness, and mention that handling edge cases is crucial for robust production code.

1. Clarify the algorithm and its assumptions

Briefly state the algorithm you are using (e.g., binary search for finding a target or insertion point) and its preconditions, such as sorted input.

2. Analyze empty array case

Explain that an empty array should return a sentinel value (e.g., -1 or None) or raise an appropriate exception, and ensure the algorithm checks for this upfront.

3. Handle all values less than target

Describe how the algorithm should return the insertion point (e.g., length of array) or indicate not found, depending on the problem, and ensure no out-of-bounds access.

4. Handle all values greater than or equal to target

Explain that the algorithm should return the first index (0) if all values are >= target, and ensure the search space is correctly narrowed.

5. Address duplicate values

Discuss how duplicates affect the result (e.g., finding first/last occurrence) and how to modify the algorithm to handle them, such as continuing the search after finding a match.

Key Points to Mention

  • Importance of defining clear termination conditions to avoid infinite loops.
  • Use of invariants to ensure correctness across edge cases.
  • Testing strategy: unit tests for each edge case.
  • Time and space complexity remains unchanged for edge cases.
  • Handling duplicates may require adjusting the comparison or search logic.
  • Real-world relevance: edge cases often occur in production data.

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

Q3

Prove the correctness of your exponential search plus binary search solution, and give a tight upper bound on the total number of API calls made.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is where I started sweating.

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AI HintsAI Generated

Suggested Approach

First, clearly state the algorithm: exponential search to find a range where the target lies, then binary search within that range. Then prove correctness by showing the exponential phase maintains the invariant that the target is within the current bounds, and the binary phase correctly narrows down to the target. Finally, derive the tight upper bound on API calls by analyzing the worst-case number of probes in each phase.

Pro tip: Emphasize that the bound is tight by constructing a worst-case input where the target is at the last possible position, and mention that the constant factors matter in practice for API call optimization.

1. Describe the algorithm

Explain exponential search: start with bound=1, double it until the element at bound is >= target or bound exceeds array size. Then binary search in the range [bound/2, min(bound, n-1)].

2. Prove correctness of exponential phase

Show that after each doubling, if the target exists, it lies within the current search interval. Use induction: initially interval [0,1] contains target if it's at index 0 or 1; if not, double bound and the target must be > previous bound, so it's in the new interval.

3. Prove correctness of binary search phase

State that binary search on a sorted subarray correctly finds the target if it exists, using the standard loop invariant that the target is within the current low-high range.

4. Analyze API call count

Count calls: exponential phase makes O(log k) calls where k is the target index (or n if not found). Binary search makes O(log k) calls. Total O(log k). For worst-case, k = n, so O(log n).

5. Derive tight bound

Show that the total number of calls is at most 2*ceil(log2(k+1)) + ceil(log2(k+1)) + O(1) = 3*ceil(log2(k+1)) + O(1). For k = n, this is Θ(log n). Provide a specific worst-case example achieving this bound.

Key Points to Mention

  • Exponential search doubles the bound until finding a range that contains the target.
  • Correctness proof uses loop invariants: exponential phase invariant that target is in [bound/2, bound] (or beyond), binary phase invariant that target is in [low, high].
  • API call count: exponential phase makes ⌈log2(k+1)⌉ calls (where k is target index), binary search makes ⌈log2(k+1)⌉ calls, total 2⌈log2(k+1)⌉ + O(1).
  • Tight bound: Θ(log k) calls, and worst-case when k = n gives Θ(log n).
  • Mention that the bound is tight by constructing an input where target is at the last position.
  • Discuss trade-offs: exponential search is better than binary search when target is near the beginning, but same asymptotic worst-case.

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

Q4

How would your approach change if the array might have been rotated once, first without duplicates and then with duplicates?

Algorithms & Data StructuresTechnical Trade-offsAdaptability & Ambiguity
Author's notes

Classic rotated binary search extension.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem: the array is sorted but rotated, and you need to search for a target. Explain how binary search adapts by identifying the sorted half, and then discuss how duplicates break the ability to determine which half is sorted, requiring a linear scan in the worst case. Emphasize the trade-off between time complexity and correctness, and mention that with duplicates the worst-case time becomes O(n) while without duplicates it remains O(log n).

Pro tip: Acknowledge that in real-world data science, duplicates are common, so you should always consider edge cases like all elements being the same. Also, mention that if the array is rotated multiple times, the same logic applies as long as it's sorted and rotated.

1. Clarify the problem and constraints

Confirm that the array is sorted and then rotated, and that we need to find a target or determine if it exists. Ask about duplicates and whether the rotation is exactly once.

2. Explain the no-duplicates case

Describe the modified binary search: at each step, determine which half is sorted by comparing mid with low and high. Then decide which half to search based on the target's value relative to the sorted half.

3. Explain the duplicates case

Show that when duplicates exist, comparing mid with low/high may not reveal the sorted half (e.g., all equal). In such cases, we cannot decide and must fall back to linear scan or shrink the search space by one.

4. Analyze time complexity

State that without duplicates, the time complexity is O(log n). With duplicates, the worst-case time complexity degrades to O(n) because we may need to scan linearly when many duplicates are present.

5. Discuss trade-offs and alternatives

Mention that if the array is known to have few duplicates, we can still achieve near O(log n) on average. Also, consider if preprocessing (like removing duplicates) is feasible, but note it changes the problem.

Key Points to Mention

  • Modified binary search for rotated sorted array
  • Identifying the sorted half by comparing mid with endpoints
  • Impact of duplicates on determining the sorted half
  • Worst-case time complexity: O(log n) without duplicates, O(n) with duplicates
  • Handling edge cases like all elements equal or target not present
  • Trade-off between time complexity and simplicity of implementation

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