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Oracle·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Oracle SWE coding round, one algorithmic question the whole time. The interviewer was pretty focused on complexity analysis afterward which I wasn't fully ready for.

Questions Asked (1)

Q1

Given an integer n, find the largest number less than or equal to n whose digits are monotone increasing (each digit is less than or equal to the next).

Algorithms & Data Structures
Author's notes

I got the greedy idea pretty quickly.

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AI HintsAI Generated

Suggested Approach

First clarify the problem and edge cases, then propose a greedy algorithm that scans digits from left to right to find the first violation of monotonicity. When a violation is found, decrement the digit at the violation point, set all subsequent digits to 9, and backtrack to fix any new violations. Finally, handle leading zeros and return the result.

Pro tip: Mention that you can avoid string conversion by using arithmetic operations, but string manipulation is often clearer and less error-prone in an interview. Also, test with edge cases like n=0, n=10, and n=1234 to ensure correctness.

1. Clarify and Define

Confirm that digits are compared from left to right (most significant to least) and that monotone increasing means each digit is <= the next. Discuss edge cases such as n < 10, n with repeated digits, and n = 0.

2. Identify Violation

Scan the digits from left to right to find the first position where a digit is greater than the next digit. If no violation, n itself is the answer.

3. Adjust and Backtrack

At the violation, decrement the offending digit, set all following digits to 9, and then backtrack to ensure the prefix remains monotone increasing. Repeat if necessary.

4. Handle Leading Zeros

After adjustments, remove any leading zeros that may have been introduced (e.g., if the first digit becomes 0).

5. Return and Test

Convert the resulting digit array back to an integer and return it. Walk through a few examples to verify correctness.

Key Points to Mention

  • Greedy approach: fix the leftmost violation first to maximize the number.
  • Time complexity: O(d) where d is the number of digits, as we may scan and backtrack at most twice.
  • Space complexity: O(d) for storing digits, or O(1) if using arithmetic operations.
  • Edge cases: n=0, n=10, n=1234, n=332, n=120.
  • Alternative: using arithmetic operations (modulo and division) instead of string conversion.
  • Proof of correctness: after adjustment, the number is the largest monotone increasing number <= n.

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