Salesforce·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Salesforce coding screen with one array problem that looks straightforward but has a few edge cases worth thinking through carefully.

Questions Asked (1)

Given an integer array, for each index produce two binary strings: one marking whether the element has appeared before that index, and one marking whether it appears again after that index. Return both strings.

Algorithms & Data Structures

Author's notes

Pretty clean problem once you realize it's just two passes and a frequency map.

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AI HintsAI Generated

Suggested Approach

Use two passes with a hash set: first left-to-right to mark seen elements, then right-to-left to mark elements that will appear again. Build the binary strings by appending '1' or '0' based on set membership before updating the set.

Pro tip: Clarify the exact output format and edge cases (empty array, single element) upfront, and mention that the solution runs in O(n) time and O(n) space, which is optimal for this problem.

1. Clarify requirements and edge cases

Confirm the definition of 'appeared before' and 'appears again after', and ask about empty arrays, single elements, and whether the strings should be returned as a pair or list.

2. Design the two-pass algorithm

Plan to use a hash set to track seen elements. First pass left-to-right builds the 'before' string; second pass right-to-left builds the 'after' string.

3. Implement the first pass

Initialize an empty set. For each element, check if it's in the set; append '1' if yes, '0' if no; then add the element to the set.

4. Implement the second pass

Clear the set. Iterate from right to left; for each element, check if it's in the set; append '1' if yes, '0' if no; then add the element to the set. Reverse the resulting string or build it in reverse order.

5. Analyze complexity and test

State that time complexity is O(n) and space is O(n). Walk through a small example to verify correctness, including duplicates.

Key Points to Mention

Hash set for O(1) membership checks
Two-pass approach: left-to-right for 'before', right-to-left for 'after'
Time complexity O(n) and space complexity O(n)
Handling duplicates correctly by updating the set after checking
Edge cases: empty array, single element, all duplicates
Building strings efficiently (e.g., using list of characters and join)

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