Apple·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Phone screen for a Data Scientist role at Apple, one algorithm question the whole time. The problem wasn't obscure but the space constraint made it genuinely tricky and I wasn't fully prepared for the follow-up on complexity trade-offs.

Questions Asked (1)

Write a function that finds the smallest missing positive integer in an unsorted array, with O(n) time and O(1) space. Then walk through the complexity trade-offs of your approach.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I knew the naive version instantly, sort the array and scan.

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AI HintsAI Generated

Suggested Approach

Use the array itself as a hash table by placing each number in its correct index (e.g., value v at index v-1) through cyclic swaps, then scan for the first index where the value doesn't match. This achieves O(n) time and O(1) space. After coding, discuss trade-offs: the in-place approach modifies the input, has a higher constant factor, and is less readable than O(n) space solutions.

Pro tip: At Apple, interviewers value clean, bug-free code and clear communication. Before coding, explicitly state that you'll modify the input array and confirm that's acceptable; then walk through edge cases like empty array, all negatives, and duplicates.

1. Clarify requirements and edge cases

Confirm that the array can be modified and that O(1) space means no additional data structures. Discuss edge cases: empty array, all non-positive numbers, duplicates, and large values.

2. Explain the in-place hashing idea

Describe the invariant: for an array of length n, the smallest missing positive is in [1, n+1]. Place each value v (1 ≤ v ≤ n) at index v-1 by swapping.

3. Implement the cyclic sort

Iterate through the array; while the current value is in range and not already in its correct position, swap it with the element at its target index. Handle duplicates by skipping if the target already has the correct value.

4. Scan for the missing positive

After rearranging, scan the array from left to right. The first index i where nums[i] != i+1 gives the missing positive i+1. If all match, return n+1.

5. Analyze complexity and trade-offs

Explain that each element is swapped at most once, giving O(n) time. Space is O(1) since only a few variables are used. Discuss trade-offs: input mutation, higher constant factor, and reduced readability compared to O(n) space solutions.

Key Points to Mention

The smallest missing positive must be in the range [1, n+1] for an array of length n.
In-place cyclic sort places each number at its correct index (value v at index v-1).
Time complexity is O(n) because each element is visited at most twice (once for swapping, once for scanning).
Space complexity is O(1) as no extra data structures are used.
Trade-offs: modifies input, higher constant factor, and less readable than using a hash set (O(n) space).
Edge cases: empty array, all negatives, duplicates, and values larger than n.

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