Squarespace·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Apr 2026

Summary

Squarespace software engineer round, one algorithmic problem that looks straightforward until you realize there's a subtle design question buried in it. Pretty clean interview overall, nothing too wild.

Questions Asked (1)

Given an unsorted list of integers that may contain duplicates, and an unsorted list of intervals, count how many integers fall within every interval simultaneously.

Algorithms & Data Structures

Author's notes

The key move is reducing all the intervals down to a single range: take the max of all the left endpoints and the min of all the right endpoints.

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AI HintsAI Generated

Suggested Approach

Clarify that 'every interval simultaneously' means the intersection of all intervals, then compute that intersection. Sort the list of integers and use binary search to count how many fall within the intersection bounds, handling duplicates naturally.

Pro tip: Mention that if the intersection is empty, the answer is zero, and that sorting the integers once allows O(log n) counting per query if the problem is extended to multiple interval sets.

1. Clarify the problem

Confirm that 'every interval simultaneously' means the intersection of all intervals. Ask about edge cases: empty intervals, empty list, or intervals with no overlap.

2. Compute the intersection

Find the maximum of all lower bounds and the minimum of all upper bounds. If max_lower > min_upper, the intersection is empty and the answer is 0.

3. Sort the integers

Sort the list of integers in ascending order. This enables efficient counting of elements within a range using binary search.

4. Count using binary search

Use binary search to find the first index where the value is >= max_lower and the last index where the value is <= min_upper. The count is the difference between these indices plus one.

5. Analyze complexity and edge cases

State the time complexity: O(n log n) for sorting plus O(log n) for binary search, and O(1) extra space. Discuss handling duplicates and empty inputs.

Key Points to Mention

Intersection of intervals: max of lower bounds and min of upper bounds.
Sorting the integers to enable binary search.
Using binary search (e.g., bisect_left and bisect_right) to count elements in a range.
Time complexity: O(n log n) due to sorting, space complexity O(1) or O(n) depending on sort.
Handling duplicates: binary search counts them correctly.
Edge cases: empty list, empty intervals, non-overlapping intervals.

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