Find the Difference (Hash Map/XOR)
Given two strings $s$ and $t$, where $t$ contains all the letters of $s$ plus one extra letter, find the extra letter. Solve efficiently using XOR or a frequency map.
Why Interviewers Ask This
Spotify asks this question to evaluate a candidate's ability to optimize space complexity while maintaining linear time performance. They specifically look for knowledge of bitwise operations like XOR, which demonstrates deep understanding of binary arithmetic and memory efficiency—critical traits for building high-throughput streaming services where resource constraints matter.
How to Answer This Question
1. Clarify the problem by confirming input constraints, such as string length limits or character encoding, ensuring you understand that t is strictly s with one added character.
2. Propose two distinct solutions: a frequency map approach using a hash table, explaining its O(n) time and O(1) space (since alphabet size is fixed), followed by the optimal XOR approach.
3. Walk through the XOR logic step-by-step, demonstrating how identical characters cancel out (A XOR A = 0), leaving only the unique character.
4. Analyze both approaches, explicitly comparing their trade-offs to show why XOR is superior for this specific constraint regarding auxiliary space.
5. Implement the code cleanly, handling edge cases like empty strings or single-character inputs while verbalizing your thought process throughout.
Key Points to Cover
- Demonstrating awareness of space complexity optimization beyond basic correctness
- Explaining the mathematical property of XOR that makes cancellation possible
- Comparing multiple algorithmic approaches before selecting the optimal one
- Verbalizing the thought process clearly rather than just coding silently
- Handling edge cases like null inputs or single-character strings proactively
Sample Answer
I would start by clarifying that we are dealing with ASCII characters and that the order doesn't matter, only the presence of the extra character. First, I'd consider a brute-force sort, but that takes O(n log n) time, which isn't ideal. Next, I'd propose a Hash Map solution where we count frequencies of every character in s, then iterate through t to find the one with a count exceeding expectations. This gives us O(n) time but requires O(k) extra space for the map, where k is the alphabet size.
However, since Spotify values efficiency, I'd push further with the XOR approach. The key insight here is that XORing a number with itself results in zero, and XORing with zero returns the number. If we XOR all characters from both strings together, every character present in s will appear twice (once in s, once in t) and cancel out. Only the extra character in t remains because it has no pair. This reduces space complexity to O(1) while keeping time at O(n).
For example, if s is 'abcd' and t is 'abcde', XORing a^b^c^d^a^b^c^d^e simplifies to (a^a)^(b^b)^(c^c)^(d^d)^e, which equals 0^0^0^0^e, resulting in 'e'. I would implement this iteratively to avoid recursion overhead, ensuring the solution is robust and performant.
Common Mistakes to Avoid
- Ignoring the space complexity requirement and defaulting to sorting or hashing without justification
- Failing to explain the logic behind XOR, simply stating it works without showing the math
- Overlooking edge cases where the strings might be empty or contain only one character
- Not discussing the trade-offs between the Hash Map and XOR methods during the explanation phase
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