Salesforce·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Salesforce software engineer interview that went pretty deep into algorithm design. The core problem was grouping anagrams, which sounds straightforward until they push you to optimize past the obvious sorting approach.

Questions Asked (1)

Given an array of strings, group all words that are anagrams of each other. Start with a sorting-based solution, then optimize to linear time using character frequency hashing. Walk through key construction, collision handling, and the time/space tradeoffs between the two approaches.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I started with the sort-each-word approach and got through it fine, O(n * m log m) and all that.

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AI HintsAI Generated

Suggested Approach

Start by explaining the sorting-based approach: sort each word and use the sorted string as a key in a hash map. Then optimize to linear time by using a character frequency count (e.g., a 26-length array or a tuple) as the key, avoiding the O(k log k) sort per word. Discuss how to handle collisions (e.g., using a delimiter or fixed-size array) and compare the time/space tradeoffs between the two methods.

Pro tip: Mention that in practice, the sorting approach is often faster for short strings due to lower constant factors, but the frequency approach is asymptotically better for long strings. Also, clarify that the frequency array must be converted to an immutable type (like a tuple) to be used as a dictionary key.

1. Clarify the problem and constraints

Ask about input size, string length, character set (e.g., lowercase English letters), and whether the output order matters. This helps determine the best approach and edge cases.

2. Present the sorting-based solution

Explain that you can sort each string and use the sorted version as a key in a hash map, grouping all strings with the same sorted key. Analyze time complexity: O(n * k log k) where n is number of strings and k is max length.

3. Optimize to linear time with frequency hashing

Describe using a character frequency count (e.g., an array of size 26 for lowercase letters) as the key. Convert the array to a tuple or string to make it hashable. This reduces time to O(n * k) since counting takes O(k) per word.

4. Discuss collision handling and key construction

For the frequency approach, ensure the key uniquely represents the character counts. For example, use a tuple of counts or a string with delimiters. Mention that using a fixed-size array avoids collisions if the character set is known.

5. Compare time and space tradeoffs

Sorting: O(n*k log k) time, O(n*k) space for keys. Frequency: O(n*k) time, O(n*k) space for keys. Note that frequency approach may use more space per key (e.g., 26 integers) but is asymptotically faster. Also mention that for small k, sorting might be faster in practice.

Key Points to Mention

Time complexity of sorting approach: O(n * k log k) where n is number of strings and k is max length.
Time complexity of frequency approach: O(n * k) because counting characters takes O(k) per string.
Space complexity: O(n * k) for storing keys in the hash map, plus the output.
Key construction: sorted string vs. frequency tuple/array; need immutable key for hashing.
Collision handling: using a delimiter or fixed-size array to avoid ambiguity.
Tradeoffs: sorting may be faster for short strings due to lower overhead; frequency approach scales better for long strings.

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