Microsoft·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Microsoft coding round, one algorithmic problem about palindromes. Pretty standard stuff but the edge cases are where it gets annoying.

Questions Asked (1)

Given a non-negative integer n, find the smallest palindromic integer that is strictly greater than n.

Algorithms & Data Structures

Author's notes

The basic idea clicked fast: mirror the left half onto the right.

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AI HintsAI Generated

Suggested Approach

Start by clarifying edge cases and constraints, then propose a solution that manipulates the digits of n to form the next palindrome. A common efficient approach is to mirror the left half onto the right half and adjust if necessary, ensuring the result is strictly greater than n.

Pro tip: Mention that you would handle single-digit and all-9s numbers as special cases, and discuss how to avoid integer overflow by working with strings or digit arrays.

1. Clarify requirements and edge cases

Ask about input size, whether n can be very large (e.g., beyond 64-bit), and confirm that the result must be strictly greater. Discuss edge cases like n=0, n=9, n=99, etc.

2. Convert to digits and identify the midpoint

Convert the number to a digit array or string to easily access and modify digits. Determine the middle index to split the number into left and right halves.

3. Mirror the left half to create a palindrome candidate

Copy the left half (including the middle digit if odd length) reversed onto the right half. This yields the smallest palindrome with the same left half.

4. Check if the candidate is greater than n

If the mirrored candidate is greater than n, it's the answer. If not, increment the left half (with carry) and re-mirror. Handle cases where incrementing increases the number of digits (e.g., 999 -> 1001).

5. Handle special cases and return result

For all-9s numbers, the answer is 1 followed by zeros and a 1 (e.g., 99 -> 101). For single-digit numbers, simply return n+1 if n<9, else 11. Convert the final digit array back to an integer or string.

Key Points to Mention

Time and space complexity: O(d) where d is the number of digits, as we only traverse the digits a constant number of times.
Handling of carry propagation when incrementing the left half, especially when it causes an extra digit.
Edge cases: n=0, n=9, n=99, n=999, and numbers like 1234 where mirroring gives 1221 which is less than n.
Avoiding integer overflow by using string or array manipulation instead of arithmetic on large integers.
The importance of strictly greater: if the mirrored palindrome equals n, we must increment to get the next palindrome.
Testing with examples: n=123 -> 131, n=999 -> 1001, n=1234 -> 1331, n=1 -> 2.

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