Edit Distance
Given two strings `word1` and `word2`, find the minimum number of operations required to convert `word1` to `word2`. Operations include insert, delete, or replace. Use DP.
Why Interviewers Ask This
LinkedIn asks this to evaluate a candidate's mastery of dynamic programming and their ability to decompose complex string manipulation problems into overlapping subproblems. It tests logical rigor in defining state transitions while assessing whether you can optimize space complexity, a critical skill for LinkedIn's high-scale data processing environments.
How to Answer This Question
1. Clarify the problem constraints and edge cases, such as empty strings or identical inputs, which is crucial for robust engineering at scale.
2. Define your DP state explicitly: explain that dp[i][j] represents the minimum edits to convert the first i characters of word1 to the first j characters of word2.
3. Derive the recurrence relation by analyzing the three operations: if characters match, carry forward the previous value; otherwise, take the minimum of insert, delete, or replace plus one.
4. Walk through a concrete example on a whiteboard, tracing the matrix fill process to demonstrate your thought flow clearly.
5. Discuss optimization strategies, specifically how to reduce O(MN) space to O(min(M,N)) using two rows, showing awareness of memory efficiency valued by top tech firms.
Key Points to Cover
- Explicitly define the DP state transition formula before writing code
- Correctly handle base cases for empty strings to prevent index errors
- Demonstrate understanding of why the recurrence relation covers all minimal paths
- Propose space optimization techniques to show advanced algorithmic thinking
- Walk through a specific example to validate the logic visually
Sample Answer
To solve the Edit Distance problem, I start by recognizing this as a classic dynamic programming challenge where we need to minimize operations between two substrings. First, I define our state: let dp[i][j] be the minimum operations required to transform the prefix of word1 of length i into the prefix of word2 of length j. The base cases are straightforward: converting an empty string to a string of length k requires k insertions, so dp[0][j] = j, and similarly dp[i][0] = i for deletions.
For the recursive step, I compare character word1[i-1] with word2[j-1]. If they are identical, no new operation is needed, so dp[i][j] equals dp[i-1][j-1]. If they differ, I consider three possibilities: deleting from word1 (dp[i-1][j] + 1), inserting into word1 (dp[i][j-1] + 1), or replacing (dp[i-1][j-1] + 1). We take the minimum of these three values.
After building the table, the answer resides in dp[m][n]. For example, converting 'horse' to 'ros' involves filling a 6x4 matrix. I would trace the logic to show how 'h' matches nothing, leading to deletions, until we align 'r', 'o', and 's'. Finally, I'd mention optimizing space to O(N) since we only need the previous row, demonstrating efficiency suitable for large datasets.
Common Mistakes to Avoid
- Confusing the order of operations, such as treating insertion and deletion as symmetric without adjusting indices correctly
- Failing to initialize the first row and column of the DP table, leading to incorrect base case calculations
- Overlooking the possibility that characters might match, forcing unnecessary replacements instead of carrying over previous values
- Ignoring space complexity optimization, resulting in O(MN) memory usage when O(min(M,N)) is achievable
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