Bloomberg·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jul 2026

Summary

Bloomberg SWE interview with a sliding window problem that looked straightforward but had enough edge case depth to keep me second-guessing myself the whole time.

Questions Asked (1)

Given an array of positive integers and a target value T greater than 1, count the number of contiguous subarrays whose product is strictly less than T. Your solution must run in O(n) time and use O(1) extra space. Also discuss edge cases like T <= 1 and arrays containing 1s, and explain why your approach is correct.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

Sliding window clicked pretty fast for me, shrink from the left when the product goes over T and accumulate the count.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Use a sliding window (two pointers) to maintain a window whose product is strictly less than T, counting all valid subarrays ending at each right pointer. Handle edge cases upfront: if T <= 1, return 0; if the array contains 1s, the window expands naturally without breaking the product condition. Explain the O(n) time and O(1) space complexity and prove correctness by showing that every valid subarray is counted exactly once.

Pro tip: Emphasize that the sliding window works because all numbers are positive, so the product is monotonic with window size; this is a key insight that interviewers look for. Also, proactively mention that you avoid integer overflow by using division instead of multiplication when shrinking the window.

1. Clarify and handle edge cases

Check if T <= 1: since all numbers are positive integers, no subarray product can be < T, so return 0. Also note that 1s in the array do not affect the product, so they can be included without issue.

2. Initialize sliding window

Set left = 0, product = 1, and count = 0. Iterate right from 0 to n-1, multiplying product by nums[right].

3. Shrink window while product >= T

While product >= T and left <= right, divide product by nums[left] and increment left. This maintains the invariant that the window [left, right] has product < T.

4. Count valid subarrays

Add (right - left + 1) to count, representing all subarrays ending at right with product < T.

5. Return count and discuss complexity

After the loop, return count. Explain that each element is added and removed at most once, giving O(n) time and O(1) space.

Key Points to Mention

Sliding window technique with two pointers, leveraging the monotonic property of products with positive integers.
Edge case T <= 1: return 0 immediately because no positive integer product can be less than 1.
Handling 1s: they do not increase the product, so the window can include them without violating the condition.
Counting all valid subarrays ending at each right pointer: (right - left + 1) subarrays.
Time complexity O(n) because each element is processed at most twice (once by right, once by left).
Space complexity O(1) because only a few variables are used, regardless of input size.

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