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Salesforce·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Salesforce SWE interview with a string manipulation problem. Pretty standard coding round, nothing too crazy, but the palindrome twist kept me second-guessing my edge cases the whole time.

Questions Asked (1)

Q1

Given a string, can you determine whether swapping exactly two of its characters (at most once) would make it a palindrome?

Algorithms & Data Structures
Author's notes

My first instinct was to just check character frequencies and call it a day, but that's not quite right.

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AI HintsAI Generated

Suggested Approach

Use a two-pointer technique to find mismatched character pairs from both ends of the string. If there are zero mismatches, check if the string has a duplicate character to swap; if there are exactly two mismatches, verify that swapping them makes the string a palindrome; otherwise, return false.

Pro tip: Clarify edge cases upfront, such as strings of length 1 or 2, and explicitly state that swapping identical characters is allowed only if a duplicate exists. This shows attention to detail and prevents incorrect assumptions.

1. Clarify the problem

Confirm that 'at most once' means you can perform zero or one swap, and that swapping two identical characters is allowed only if such a pair exists. Ask about input constraints (e.g., length, character set) to guide your solution.

2. Use two pointers to find mismatches

Initialize left and right pointers at the start and end of the string. Move them inward while characters match, and record the indices where they differ.

3. Analyze the mismatches

If there are no mismatches, check if the string contains any duplicate character (e.g., using a set or frequency count). If yes, return true; otherwise, false. If there are exactly two mismatches, check if swapping the characters at those indices makes the string a palindrome. If there are more than two mismatches, return false.

4. Verify the swap

For the two-mismatch case, after swapping, ensure the entire string is a palindrome by continuing the two-pointer check or by verifying that the swapped characters match their counterparts.

5. Discuss complexity and edge cases

State that the time complexity is O(n) and space is O(1) (or O(n) if using a set for duplicates). Mention edge cases like empty string, single character, and strings with all identical characters.

Key Points to Mention

  • Two-pointer technique for palindrome checking
  • Handling zero mismatches: need a duplicate character to swap
  • Handling exactly two mismatches: swap and verify
  • Time and space complexity analysis
  • Edge cases: empty string, length 1, length 2, all same characters
  • Clarifying that swapping identical characters is allowed only if a duplicate exists

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