Zest·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Two coding problems for a Data Scientist role at Zest. Nothing too wild but the meeting rooms problem took me longer than I'd like to admit.

Questions Asked (2)

Find the maximum depth of a binary tree, where depth is the number of nodes along the longest root-to-leaf path.

Algorithms & Data Structures

Author's notes

Went with recursion right away, base case returns 0 for null nodes and you just take the max of left and right subtree depths plus one.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the definition of depth (nodes vs edges) and the tree structure. Then present a recursive DFS solution that computes the maximum depth of left and right subtrees, and discuss iterative BFS as an alternative. Analyze time and space complexity, and mention edge cases like empty tree.

Pro tip: In interviews, always state the recursive relation and base case before coding; it shows structured thinking. Also, mention that BFS can be more memory-efficient for very deep trees, demonstrating awareness of trade-offs.

1. Clarify the problem

Confirm that depth is measured in nodes (root depth = 1) and that the tree may be empty. Ask if the tree is balanced or if there are constraints on recursion depth.

2. Choose an approach

Decide between recursive DFS (simpler, elegant) and iterative BFS (avoids recursion limit). Explain your choice based on tree size and potential stack overflow.

3. Derive the recursive relation

For DFS: maxDepth(root) = 1 + max(maxDepth(root.left), maxDepth(root.right)), with base case maxDepth(null) = 0. This clearly defines the solution.

4. Analyze complexity and edge cases

Time O(n) since each node visited once; space O(h) for recursion stack (h = height) or O(n) worst-case. Handle empty tree (return 0) and single node (return 1).

5. Implement and test

Write clean code (e.g., Python) and walk through a small example. If time permits, discuss iterative BFS using a queue and level counting.

Key Points to Mention

Definition of depth: number of nodes on the longest root-to-leaf path (root depth = 1).
Recursive DFS solution with base case and recurrence relation.
Iterative BFS solution using a queue to process level by level.
Time complexity O(n) and space complexity O(h) for DFS, O(w) for BFS (w = max width).
Edge cases: empty tree, skewed tree (linked list), balanced tree.
Trade-offs: recursion depth limit vs. memory usage of BFS.

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

Given a list of meeting time intervals, find the minimum number of conference rooms needed to hold all meetings without overlap. Meetings are half-open intervals so back-to-back meetings don't conflict.

Algorithms & Data Structures

Author's notes

This one tripped me up more than it should have.

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AI HintsAI Generated

Suggested Approach

Clarify that meetings are half-open intervals, then present the sweep line algorithm: create events for start (+1) and end (-1), sort them, and track the running sum to find the maximum concurrent meetings. Alternatively, use a min-heap of end times, adding a room when a meeting starts after the earliest end. Discuss time and space complexity, and mention edge cases like empty input or back-to-back meetings.

Pro tip: Emphasize that back-to-back meetings don't conflict because intervals are half-open, so when sorting events, process end events before start events at the same time to avoid overcounting. This shows attention to detail and understanding of interval semantics.

1. Clarify the problem and constraints

Confirm that intervals are half-open [start, end) and that back-to-back meetings are allowed. Ask about input size, whether intervals are sorted, and if the list can be empty.

2. Choose an approach

Decide between the sweep line (event sorting) and min-heap methods. Explain why both work and their trade-offs in terms of time and space complexity.

3. Walk through the algorithm

For sweep line: create events, sort with ends before starts at equal times, iterate and track current and max rooms. For heap: sort by start, push end times, pop if no overlap, track heap size.

4. Analyze complexity and edge cases

State time complexity O(n log n) due to sorting, space O(n). Discuss edge cases: empty list, single meeting, all overlapping, back-to-back meetings.

5. Test with examples

Run through a small example like [[0,30],[5,10],[15,20]] to verify the algorithm yields 2 rooms, and explain how the half-open property affects the result.

Key Points to Mention

Half-open intervals mean end == start is not an overlap, so process end events before start events at the same timestamp.
Sweep line algorithm: create +1 for start and -1 for end, sort events, track running sum and maximum.
Min-heap approach: sort meetings by start time, use a heap of end times; if the earliest end <= current start, reuse a room (pop), then push current end.
Time complexity O(n log n) and space O(n) for both approaches.
Edge cases: empty input returns 0, single meeting returns 1, all meetings overlapping returns n.
The problem is equivalent to finding the maximum number of overlapping intervals at any point.

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