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Apple·Software Engineer·Onsite - Coding / Algorithms·Intermediate

Intermediate
Jun 2026

Summary

Apple SWE interview with three algorithmic problems back to back. The questions ranged from coordinate compression to grid connectivity to data structure design, so it was a pretty broad spread for a single session.

Questions Asked (3)

Q1

Given n inclusive time intervals representing users being online, return a list of disjoint time segments with the count of concurrent users in each segment. Timestamps can be very large, so a naive array approach won't work.

Algorithms & Data Structures
Author's notes

The large timestamp constraint is the whole point of the problem.

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AI HintsAI Generated

Suggested Approach

Use a sweep line algorithm: create events for interval starts (+1) and ends (-1), sort them by time, and sweep through to maintain a running count of active users. Output segments between consecutive event times where the count is non-zero, merging adjacent segments with the same count.

Pro tip: Clarify whether intervals are inclusive and how to handle zero-count gaps; explicitly state that you'll skip segments with count 0 and merge consecutive segments with equal counts to ensure disjointness.

1. Clarify interval semantics and edge cases

Confirm if intervals are inclusive [start, end] and how to treat touching intervals (e.g., [1,2] and [2,3] overlap at 2). Discuss handling of zero-length intervals and large timestamps.

2. Create and sort events

For each interval, create a start event (+1) at start time and an end event (-1) at end time (or end+1 if inclusive). Sort events by time, processing starts before ends at the same timestamp if intervals are inclusive.

3. Sweep to compute counts

Iterate through sorted events, maintaining a running count. At each event time, record the count before applying the event's delta, then apply the delta. This gives the count for the segment starting at that time.

4. Build output segments

For each consecutive pair of event times, if the count between them is >0, add a segment [time_i, time_{i+1}) with that count. Merge adjacent segments with the same count to ensure disjointness.

5. Analyze complexity and test

State time complexity O(n log n) due to sorting, space O(n). Walk through a small example to verify correctness, including edge cases like no overlap or all overlapping.

Key Points to Mention

  • Sweep line algorithm with events (+1 for start, -1 for end)
  • Sorting events by time, with tie-breaking rule for inclusive intervals
  • Maintaining a running count and outputting segments only when count > 0
  • Merging adjacent segments with identical counts to ensure disjointness
  • Time complexity O(n log n) and space O(n)
  • Handling large timestamps by avoiding array-based approaches

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

Q2

Fill an R x C grid with given counts of vegetable types such that every cell is filled and all cells of each type form a single 4-directionally connected region. Return any valid grid or indicate it's impossible.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This one surprised me more than it should have.

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AI HintsAI Generated

Suggested Approach

Start by checking necessary conditions: total cells must equal sum of counts, and each count must be at most half the total cells (or a tighter bound based on grid dimensions). Then use a constructive algorithm like spiral filling or row-by-row serpentine filling to assign cells to types, ensuring each type forms a single connected region. If no construction works, return impossible.

Pro tip: Mention that for a single connected region, the count of any type cannot exceed the number of cells in a spanning tree of the grid (R*C - (R-1)*(C-1) = R+C-1) if other types are present, but more practically, use a BFS/DFS to verify connectivity after construction and backtrack if needed. Also, note that Apple values clean, efficient code and edge-case handling.

1. Validate necessary conditions

Check that sum of counts equals R*C and that no count exceeds the maximum possible for a connected region (e.g., R*C - (number of other types) if others must be connected). Also consider parity or shape constraints.

2. Choose a construction strategy

Select a filling pattern that naturally creates connected regions, such as serpentine row-by-row filling, spiral filling, or using a priority queue to always extend the smallest region.

3. Implement and verify connectivity

After filling, run BFS/DFS from one cell of each type to ensure all cells of that type are reachable. If not, adjust the filling order or backtrack.

4. Handle edge cases and impossibility

If no valid grid is found after reasonable attempts, return impossible. Consider cases like single row/column, one type dominating, or counts that force disconnection.

5. Analyze complexity and trade-offs

Discuss time/space complexity of your approach (e.g., O(R*C) for filling and verification) and trade-offs between greedy construction and backtracking.

Key Points to Mention

  • Necessary condition: sum of counts equals R*C.
  • Maximum count for a connected region: at most R*C - (number of other types) if others must also be connected, but tighter bounds exist (e.g., for a single type, max is R*C).
  • Use of BFS/DFS to verify connectivity of each type.
  • Construction strategies: serpentine, spiral, or region-growing with priority queue.
  • Handling impossibility: when counts violate necessary conditions or construction fails.
  • Complexity analysis: O(R*C) time and space for filling and verification.

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

Q3

Design a data structure supporting insert, getMin, getMax, popMin, and popMax on a multiset of integers. Discuss time complexity and when a bucket/counting approach is better or worse.

Algorithms & Data StructuresTechnical Trade-offsSystem Design
Author's notes

Pretty standard answer here: a sorted structure like a balanced BST or two heaps (one min, one max) with lazy deletion gets you O(log n) on everything.

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AI HintsAI Generated

Suggested Approach

Start by clarifying requirements: multiset means duplicates allowed, and operations should be efficient. Propose a balanced BST (e.g., Red-Black Tree) or two heaps with lazy deletion, discussing O(log n) for all operations. Then compare with a bucket/counting approach (e.g., array of counts) which gives O(1) for insert and O(1) amortized for getMin/getMax if the range is small, but O(range) for popMin/popMax in worst case, and discuss trade-offs based on data distribution and constraints.

Pro tip: Mention that in practice, for bounded integer ranges, a Fenwick tree over counts can support all operations in O(log U) where U is the universe size, and that Apple often values practical optimizations for constrained environments.

1. Clarify requirements and constraints

Ask about the expected number of operations, the range of integers, memory constraints, and whether the multiset is static or dynamic. This determines the best data structure.

2. Propose a general-purpose solution

Describe a balanced BST (e.g., Red-Black Tree) or a combination of a min-heap and max-heap with lazy deletion, achieving O(log n) for all operations.

3. Analyze time and space complexity

For the balanced BST, all operations are O(log n) time and O(n) space. For two heaps with lazy deletion, insert is O(log n), getMin/getMax O(1), popMin/popMax O(log n) amortized.

4. Discuss bucket/counting approach

If the integer range is small (e.g., 0 to 10^6), use an array of counts. Insert is O(1), getMin/getMax can be O(1) with maintained pointers, but popMin/popMax may require scanning, leading to O(range) worst-case.

5. Compare trade-offs and conclude

Summarize when each approach is better: bucket/counting for small, dense ranges with frequent inserts and infrequent pops; balanced BST for large or sparse ranges with frequent pops.

Key Points to Mention

  • Balanced BST (e.g., Red-Black Tree) provides O(log n) for all operations and handles duplicates naturally.
  • Two heaps with lazy deletion: min-heap and max-heap, with a hash map to track valid elements; O(log n) insert, O(1) getMin/getMax, O(log n) amortized pop.
  • Bucket/counting approach: array of counts indexed by value; O(1) insert, O(1) getMin/getMax if min/max pointers maintained, but O(range) for popMin/popMax in worst case.
  • Fenwick tree (Binary Indexed Tree) over counts can support all operations in O(log U) where U is the universe size, balancing efficiency and memory.
  • Time complexity comparison: balanced BST O(log n) vs. bucket O(1) insert but O(range) pop; space complexity: O(n) vs. O(range).
  • When to use bucket/counting: small, dense integer range, frequent inserts, infrequent pops, memory not a concern; when to avoid: large or sparse range, frequent pops, memory constrained.

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