Longest Common Subsequence (LCS)
Given two strings $text1$ and $text2$, return the length of their longest common subsequence. Use Dynamic Programming.
Why Interviewers Ask This
Airbnb asks the Longest Common Subsequence problem to evaluate a candidate's ability to recognize overlapping subproblems and apply dynamic programming efficiently. They are testing whether you can transition from brute-force recursion to an optimized DP solution, demonstrating logical structuring skills essential for building scalable infrastructure that handles complex data matching tasks.
How to Answer This Question
1. Clarify constraints: Immediately ask about string lengths and character sets to determine if O(N*M) space is acceptable or if space optimization is needed.
2. Define the state: Explicitly state your DP table definition, such as dp[i][j] representing the LCS length of text1[0...i] and text2[0...j].
3. Establish the recurrence relation: Explain the logic clearly—if characters match, add 1 to the diagonal; otherwise, take the max of the top or left cell.
4. Walk through an example: Use small strings like 'abc' and 'ac' to trace the table filling process verbally before writing code.
5. Optimize and analyze: Discuss time complexity O(N*M) and space complexity, mentioning how to reduce space to O(min(N,M)) if asked, mimicking Airbnb's focus on resource efficiency.
Key Points to Cover
- Explicitly define the DP state and recurrence relation before coding
- Demonstrate understanding of why brute force fails due to exponential redundancy
- Walk through a concrete trace with a small example to show logical flow
- Correctly identify time and space complexities as O(N*M)
- Connect the algorithmic pattern to real-world data comparison use cases
Sample Answer
I would start by defining the problem as finding the maximum length of a sequence present in both strings without changing relative order. First, I'd validate inputs and check edge cases like empty strings. Then, I'd propose a 2D DP table where dp[i][j] stores the LCS length for the first i characters of text1 and j characters of text2.
The core logic relies on two scenarios: if text1[i-1] equals text2[j-1], we extend the previous common subsequence by adding one, so dp[i][j] = dp[i-1][j-1] + 1. If they don't match, we carry forward the best result found so far by taking the maximum of excluding the current character from either string: dp[i][j] = max(dp[i-1][j], dp[i][j-1]).
For instance, with 'abcd' and 'acdf', the table fills up showing matches at 'a', 'c', and 'd'. After filling the grid, the bottom-right cell holds the answer. This approach runs in O(N*M) time and uses O(N*M) space. Given Airbnb's need for efficient algorithms in search and recommendation systems, I would also mention how this pattern applies to comparing user profile data or version histories, ensuring we handle large datasets effectively.
Common Mistakes to Avoid
- Confusing Subsequence with Substring by allowing contiguous characters instead of non-contiguous ones
- Failing to initialize the DP table correctly, leading to off-by-one errors in indices
- Implementing only the recursive solution without memoization, resulting in Time Limit Exceeded
- Neglecting to discuss space optimization when the interviewer hints at memory constraints
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