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

Intermediate
May 2026

Summary

Salesforce SWE coding round, two algorithm problems back to back. Nothing too wild conceptually but the second one has some tricky edge cases that'll bite you if you only think about it naively.

Questions Asked (2)

Q1

Given an integer array, construct two binary strings indicating for each element whether it has appeared before it (to the left) and whether it appears again after it (to the right).

Algorithms & Data Structures
Author's notes

Pretty approachable once you realize it's just two passes with a hash set.

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AI HintsAI Generated

Suggested Approach

Use two passes with a hash set: first pass left-to-right to mark if each element has been seen before, second pass right-to-left to mark if each element will be seen again. Build the two binary strings accordingly, ensuring O(n) time and O(n) space.

Pro tip: Clarify the definition of 'appeared before' and 'appears again'—whether it's based on value or index—and confirm the output format (e.g., '1' for true, '0' for false). Also, mention edge cases like empty array or single element.

1. Understand the problem

Restate the problem in your own words and ask clarifying questions about the definition of 'appeared before' and 'appears again' (value-based vs. index-based) and the expected output format.

2. Choose data structures

Select a hash set to track seen elements for O(1) lookups. Use two passes: one left-to-right for the 'before' string, one right-to-left for the 'after' string.

3. Implement the two passes

In the first pass, for each element, check if it's in the set; if yes, append '1' to the before string, else '0', then add the element to the set. In the second pass, do the same from right to left for the after string, then reverse it.

4. Handle edge cases and verify

Test with empty array, single element, all duplicates, and all unique elements. Ensure the strings are of correct length and characters are '0' or '1'.

5. Analyze complexity

State that time complexity is O(n) and space complexity is O(n) due to the hash set and output strings. Discuss potential optimizations if needed.

Key Points to Mention

  • Hash set for O(1) membership checks
  • Two-pass approach for efficiency
  • Time and space complexity analysis
  • Edge cases: empty array, single element, duplicates
  • Output format: binary strings with '1' for true, '0' for false
  • Clarifying questions to ensure correct interpretation

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

Q2

You have a binary string where every second, all occurrences of '01' are simultaneously replaced with '10'. How many steps until the string stabilizes?

Algorithms & Data Structures
Author's notes

This one is deceptively annoying.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem and edge cases, then derive an efficient solution by observing that each '1' moves right past '0's, and the stabilization time equals the maximum number of '0's any '1' must cross. Use a single pass to compute this maximum, explaining the reasoning clearly.

Pro tip: Mention that a naive simulation is O(n^2) and would fail for large inputs, then present the O(n) insight—this shows you think about scalability and optimization, which interviewers value.

1. Clarify the problem

Confirm that replacements happen simultaneously each second, and that the string stabilizes when no '01' remains. Ask about input size and constraints.

2. Explore with examples

Walk through small examples (e.g., '01', '0011', '0101') to observe the pattern and verify your understanding of the process.

3. Derive the key insight

Recognize that each '1' moves right past '0's, and the total time is the maximum number of '0's that any '1' must cross. This can be computed by scanning left to right, counting zeros and ones.

4. Design an efficient algorithm

Use a single pass: maintain a count of zeros seen so far and the maximum steps for any '1'. When encountering a '1', update the maximum with the current zero count; when encountering a '0', increment the zero count.

5. Analyze complexity and edge cases

State that the algorithm runs in O(n) time and O(1) space. Test edge cases like all zeros, all ones, and alternating patterns.

Key Points to Mention

  • Simultaneous replacement means each '1' can move at most one position per second.
  • The stabilization time is determined by the maximum number of '0's to the left of any '1'.
  • A single left-to-right scan can compute the answer in O(n) time.
  • Naive simulation is O(n^2) and inefficient for large strings.
  • Edge cases: empty string, no '01' initially, all '0's or all '1's.
  • The problem is equivalent to counting inversions where '1' precedes '0', but only the maximum per '1' matters.

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