Salesforce·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Salesforce SWE coding round, pretty standard sliding window problem. Nothing too wild but the edge cases tripped me up more than I expected.

Questions Asked (1)

Given two strings, find all starting indices in the first string where an anagram of the second string begins.

Algorithms & Data Structures

Author's notes

I knew sliding window was the move pretty fast, but I fumbled the incremental update part.

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AI HintsAI Generated

Suggested Approach

Use a sliding window of length equal to the second string over the first string, maintaining character frequency counts to check for anagrams in O(n) time. Compare the frequency maps at each step, updating counts as the window slides. Return all starting indices where the frequency maps match.

Pro tip: Mention that you can optimize by tracking the number of matching characters instead of comparing full frequency maps each time, reducing constant factors. Also, clarify edge cases like empty strings or when the second string is longer than the first.

1. Clarify and Validate Input

Confirm assumptions: strings may contain any characters, case sensitivity, and that an anagram must be a contiguous substring. Check if the second string is longer than the first; if so, return an empty list.

2. Choose Data Structures

Use a frequency array (size 26 for lowercase letters) or a hash map for character counts. Initialize counts for the second string and the first window of the first string.

3. Slide the Window

Iterate through the first string, updating the window's frequency counts by adding the new character and removing the old one. Compare the window's counts with the target counts.

4. Optimize Comparison

Instead of comparing full frequency maps each time, maintain a count of how many characters have matching frequencies. When this count equals the number of distinct characters in the target, record the start index.

5. Return Results

Collect all valid starting indices in a list and return it. Discuss time and space complexity: O(n) time and O(1) space (since alphabet size is fixed).

Key Points to Mention

Sliding window technique to achieve linear time complexity
Frequency counting using arrays or hash maps
Handling edge cases: empty strings, second string longer than first, non-alphabetic characters
Optimization by tracking match count instead of full map comparison
Time complexity: O(n) where n is length of first string; space complexity: O(1) for fixed alphabet
Potential follow-up: what if the strings contain Unicode characters? Use a hash map instead of fixed array.

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