Two Sigma·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Two Sigma data scientist interview with a coding problem that looks like a simple palindrome question until you actually read the constraints. The O(n) requirement rules out the naive increment-and-check approach entirely, which I did not fully appreciate until I was already mid-explanation.

Questions Asked (1)

Given a positive integer K, find the smallest palindrome strictly greater than K. Your solution must run in time proportional to the number of digits, not by incrementing one at a time.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I started by thinking about mirroring the left half onto the right half, which gets you partway there.

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AI HintsAI Generated

Suggested Approach

Treat the number as a string of digits and construct the smallest palindrome greater than K by mirroring the left half onto the right half. If the mirrored result is not greater, increment the middle digit(s) and re-mirror, handling carries and digit growth. This yields an O(n) solution where n is the number of digits.

Pro tip: Explicitly discuss edge cases like all 9s (e.g., 999 → 1001) and numbers where the middle digit is 9, as these often trip up candidates. Also, mention that you can avoid string manipulation by using arithmetic, but string-based is simpler and still O(n).

1. Convert to string and split

Convert K to a string and split it into left half, middle (if odd length), and right half. This allows easy mirroring.

2. Mirror left half to form palindrome

Create a palindrome by mirroring the left half onto the right half. If the length is odd, keep the middle digit as is.

3. Check if palindrome > K

Compare the constructed palindrome with K. If it's greater, return it. If not, proceed to increment.

4. Increment middle and handle carry

Increment the middle digit (or the rightmost digit of the left half if even length) by 1, propagating carry leftwards. If carry overflows, increase the number of digits (e.g., 999 → 1001).

5. Re-mirror and return

After incrementing, mirror the left half again to form the new palindrome and return it.

Key Points to Mention

Time complexity O(n) where n is the number of digits, achieved by direct construction rather than iteration.
Handling of odd vs even length numbers when mirroring.
Edge cases: all 9s, numbers like 12321 (already palindrome), and numbers where increment causes carry (e.g., 12921 → 13031).
Space complexity O(n) for the string representation, which is optimal for this problem.
Trade-off between string manipulation and arithmetic approaches; string is simpler and less error-prone.
Testing with small examples to verify correctness, e.g., K=123 → 131, K=999 → 1001.

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