LeetCode Edit Distance Problem

[pertrai1]

Summary by CodeRabbit

• New Features

  • Added LeetCode problem LeetCode Word Search II Problem #72 (Edit Distance) to the Medium collection and updated progress count.

• Documentation

  • Added a comprehensive README for the Edit Distance problem (statement, operations, examples, constraints).
  • Added a POST_MORTEM_TEMPLATE for tracking solution exploration, mistakes, and retrospective.

_{✏️ Tip: You can customize this high-level summary in your review settings.}

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Walkthrough

Adds problem documentation and a post-mortem template for LeetCode 72 (Edit Distance), updates the root README progress count and entry, and makes the solution function exportable by adding an export to the minDistance implementation.

Changes

Cohort / File(s)	Change Summary
Problem README leetcode/medium/0072-edit-distance/README.md	Adds a comprehensive README for Edit Distance (problem statement, allowed operations, examples with step-by-step transformations, constraints).
Post‑mortem template leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md	Adds a post-mortem template file for tracking time, exploration, mistakes, retrospective, and self-rating.
Code export leetcode/medium/0072-edit-distance/edit-distance.ts	Adds export to minDistance function declaration; minor formatting adjustments only.
Root progress update README.md	Increments medium problem count (71 → 72) and adds new entry/badge for 0072 - Edit Distance.
Removed/modified previous README content 0072-edit-distance/README.md	Prior README content for Edit Distance was removed (documentation-only deletion).

Estimated code review effort

🎯 2 (Simple) | ⏱️ ~10 minutes

• Focus review on edit-distance.ts export to ensure module boundaries and build/export behavior are correct.
• Quick skim of added README and POST_MORTEM_TEMPLATE for accuracy and formatting.

Poem

🐰 I hopped through edits, letters in a line,
Counting swaps and gaps, making steps align.
A README to guide, a template to chart,
minDistance now exported — a small coding art! 🥕✨

Pre-merge checks and finishing touches

❌ Failed checks (1 inconclusive)

Check name	Status	Explanation	Resolution
Title check	❓ Inconclusive	The title 'LeetCode Edit Distance Problem' is vague and generic, using non-descriptive terms that don't convey the specific nature of the changes (adding solution, documentation, and exports).	Consider a more specific title that describes the main change, such as 'Add LeetCode 72 Edit Distance solution with documentation and export' or 'Implement LeetCode Edit Distance problem with post-mortem template'.

✅ Passed checks (2 passed)

Check name	Status	Explanation
Description Check	✅ Passed	Check skipped - CodeRabbit’s high-level summary is enabled.
Docstring Coverage	✅ Passed	Docstring coverage is 100.00% which is sufficient. The required threshold is 80.00%.

✨ Finishing touches

🧪 Generate unit tests (beta)

• Create PR with unit tests
• Post copyable unit tests in a comment
• Commit unit tests in branch leetcode-12222025

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📒 Files selected for processing (1)

• leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md

🧰 Additional context used

📓 Path-based instructions (1)

**/*.md

📄 CodeRabbit inference engine (AGENTS.md)

**/*.md: Create comprehensive documentation guides for arrays, trees, graphs, heaps, and other data structures
Document reusable code templates for common algorithmic patterns
Maintain Big O complexity cheatsheets and references

Files:

• leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md

🧠 Learnings (1)

📚 Learning: 2025-11-26T14:46:46.097Z

Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: .github/copilot-instructions.md:0-0
Timestamp: 2025-11-26T14:46:46.097Z
Learning: Applies to [0-9]*-*/*.md : Problem README files should include: problem title with difficulty badge, problem description from LeetCode, examples with inputs/outputs, and constraints

Applied to files:

• leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md

🪛 LanguageTool

leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md

[style] ~36-~36: You have already used this phrasing in nearby sentences. Shortening it will avoid wordiness.
Context: ...able and dynamic programming allows for making use of previous operations and calculations #...

(REP_MAKE_USE_OF)

---

[grammar] ~50-~50: Use a hyphen to join words.
Context: ... ### Bugs to add to the Bug List - Off by one iterations that go out of bounds ...

(QB_NEW_EN_HYPHEN)

---

[grammar] ~50-~50: Use a hyphen to join words.
Context: ...## Bugs to add to the Bug List - Off by one iterations that go out of bounds ##...

(QB_NEW_EN_HYPHEN)

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Claude finished @pertrai1's task in 1m 45s —— View job

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Code Review Complete ✅

Overview

Excellent work on implementing the Edit Distance (Levenshtein Distance) problem! This is a classic dynamic programming problem and your solution is correct and optimal. The reorganization into the proper directory structure is also good.

---

1. Correctness ✓

The solution correctly handles all test cases:

• Edge cases: Empty strings (handled by base case initialization)
• Character matching: Properly checks if characters match before adding cost
• All operations: Insert, delete, and replace are all represented in the recurrence relation

Mental trace with Example 1 ("horse" → "ros"):

• The DP table correctly builds up from base cases
• Final answer dp[5][3] = 3 matches expected output

Verdict: Algorithm is correct for all cases within constraints.

---

2. Complexity Analysis ✓

Your complexity annotations are accurate:

Time Complexity: O(m × n)

• Two nested loops iterate through all (m+1) × (n+1) cells
• Each cell performs constant-time operations (comparison + min of 3 values)
• Total: O(m × n) where m = word1.length, n = word2.length

Space Complexity: O(m × n)

• 2D DP table of size (m+1) × (n+1)
• No other significant space usage

Note: This can be optimized to O(min(m, n)) space by only keeping two rows at a time, but the current approach prioritizes clarity and is the standard solution.

---

3. Learning Focus: Dynamic Programming Pattern

Pattern Classification

This problem uses the 2D Dynamic Programming pattern, specifically a Bottom-Up DP Table approach.

Key Insight 💡

The breakthrough insight is recognizing that Edit Distance has optimal substructure:

The minimum distance to transform word1[0...i] to word2[0...j] can be computed from the solutions to smaller subproblems.

At each cell dp[i][j], we have 3 choices:

1. Replace: Use dp[i-1][j-1] + (0 if chars match, 1 if replace needed)
2. Delete: Use dp[i-1][j] + 1 (delete from word1)
3. Insert: Use dp[i][j-1] + 1 (insert into word1)

Taking the minimum of these three gives the optimal solution at each step!

Why This Pattern Works

• Overlapping subproblems: Computing edit distance for substrings repeatedly
• Optimal substructure: Optimal solution contains optimal solutions to subproblems
• Base cases: Empty string transformations are straightforward (all inserts or all deletes)

---

4. Code Quality Assessment

Strengths ✓

• Clear variable names (m, n, dp)
• Proper TypeScript typing
• Clean structure with separated initialization and computation
• Helpful comments explaining each section

Minor Suggestions

Comment Enhancement (edit-distance.ts:23-27):
The core recurrence relation could benefit from explaining what each option represents:

// Choose minimum of:
// 1. Replace (or match): dp[i-1][j-1] + cost
// 2. Delete from word1: dp[i-1][j] + 1
// 3. Insert into word1: dp[i][j-1] + 1
dp[i][j] = Math.min(
  dp[i - 1][j - 1] + (word1[i - 1] === word2[j - 1] ? 0 : 1),
  dp[i - 1][j] + 1,
  dp[i][j - 1] + 1
);

Base Case Clarity (edit-distance.ts:12-18):
Consider adding a comment explaining the significance:

// Base cases:
// dp[i][0] = i: Delete all i characters from word1
// dp[0][j] = j: Insert all j characters from word2

---

5. Alternative Approaches

Space-Optimized Version

Trade-off: O(min(m,n)) space vs. current O(m×n)

• Only keep current and previous row
• Same time complexity but 2-row array instead of full table
• More complex to implement and debug

Recursive + Memoization

Trade-off: Top-down vs. bottom-up approach

• Same complexity but uses call stack
• Often more intuitive to reason about
• Slightly more overhead due to recursion

When to use each:

• Your solution: Best for interviews - clear, optimal, standard approach
• Space-optimized: When memory is constrained
• Recursive: When you find top-down thinking easier

---

6. Related Problems for Practice

To master this pattern, try these similar problems:

1. Longest Common Subsequence (LeetCode 1143 - Medium)

   • Similar 2D DP structure
   • Same "compare characters" logic
   • Great for reinforcing the pattern

2. Delete Operation for Two Strings (LeetCode 583 - Medium)

   • Edit Distance variant (only delete operations)
   • Can be solved by finding LCS
   • Helps understand the relationship between these problems

3. Minimum ASCII Delete Sum for Two Strings (LeetCode 712 - Medium)

   • Weighted version of edit distance
   • Same DP structure with different cost calculation

---

Learning Summary

Core Concepts Demonstrated

1. ✅ 2D Dynamic Programming - Building a table to store subproblem solutions
2. ✅ Optimal Substructure - Combining optimal subproblem solutions
3. ✅ Base Case Initialization - Handling edge cases in DP
4. ✅ Recurrence Relations - Defining state transitions mathematically

Pattern Recognition Trigger

When you see:

• "Minimum operations" or "minimum cost" between two sequences
• Operations that transform one string to another
• Need to compare characters at different positions

Think: 2D DP with one dimension per string!

Your Post-Mortem Insights

Your self-reflection shows strong growth mindset:

• ✓ Recognized the need to trace examples first
• ✓ Identified dynamic programming as the optimal approach
• ✓ Learned that Edit Distance = Levenshtein Distance

Recommendation: Continue using that POST_MORTEM_TEMPLATE.md - it's an excellent tool for deliberate practice and pattern recognition!

---

Final Verdict

Overall Assessment: Strong implementation of a classic DP problem! 🎯

• ✅ Correct algorithm
• ✅ Optimal complexity
• ✅ Clean code
• ✅ Proper documentation
• ✅ Good self-reflection in post-mortem

The solution demonstrates solid understanding of dynamic programming fundamentals. Keep building on this pattern recognition - Edit Distance is a foundational problem that appears in many variations!

[github-actions]

✅ Test Results

Tests have been executed for this pull request.

📊 Coverage Summary

Metric	Coverage
Statements	93.63%
Branches	87.61%
Functions	82.14%
Lines	93.39%

Check the workflow logs for detailed test output.

[github-actions]

🧮 Solution Analysis

PR: LeetCode Edit Distance Problem
Files Changed: leetcode/medium/0072-edit-distance/edit-distance.ts

📊 Code Quality Results

ESLint Analysis

• Errors: 0
• Warnings: 0

🔍 Complexity Analysis

💡 Analysis Tips

• Time Complexity: Focus on the dominant operation in loops
• Space Complexity: Consider auxiliary data structures
• Optimization: Look for opportunities to improve efficiency
• Edge Cases: Ensure your solution handles empty inputs, null/undefined, edge conditions

📁 Analysis Files Created

Detailed analysis files have been automatically generated and committed to docs/analysis/ for future reference.

🎯 Next Steps

1. Review any ESLint errors/warnings above
2. Consider the complexity analysis - is this the optimal approach?
3. Test with edge cases mentioned in the problem description
4. Add comments explaining complex logic
5. Check the docs/analysis/ directory for detailed complexity analysis files

---

Analysis generated for PR files only

[github-actions]

---

🤖 Gemini Code Review (Model: gemini-2.5-flash)

Here's a detailed review of the provided minDistance solution for the Edit Distance problem:

---

1. Pattern Identification

This solution uses the Dynamic Programming (DP) algorithmic pattern.

Why this pattern is suitable:

• Optimal Substructure: The problem can be broken down into smaller subproblems, and the optimal solution to the overall problem can be constructed from the optimal solutions to these subproblems. For example, the minimum edit distance between word1[0...i] and word2[0...j] depends on the minimum edit distances of their prefixes (e.g., word1[0...i-1] and word2[0...j-1]).
• Overlapping Subproblems: The same subproblems are solved multiple times. A DP table (or memoization) is used to store the results of these subproblems, so they are computed only once.

This problem is a classic example of string DP, often referred to as the Levenshtein Distance algorithm.

---

2. Complexity Verification

• Stated Time Complexity: O(m * n)

  • Accuracy: Accurate.
  • Reasoning: The solution initializes a 2D DP table of size (m+1) x (n+1). It then iterates through this table using nested loops, where the outer loop runs m times and the inner loop runs n times. Inside the inner loop, each operation (array access, comparison, Math.min) takes constant time, O(1). Therefore, the total time complexity is O((m+1) * (n+1)), which simplifies to O(m * n).

• Stated Space Complexity: O(m * n)

  • Accuracy: Accurate.
  • Reasoning: The solution uses a 2D array dp of size (m+1) x (n+1) to store the results of subproblems. This table directly consumes O(m * n) space.

---

3. Key Insight

The key insight for solving the Edit Distance problem with dynamic programming lies in defining the state dp[i][j] and understanding the recurrence relation.

State Definition:
dp[i][j] represents the minimum number of operations required to convert the first i characters of word1 (i.e., word1.substring(0, i)) to the first j characters of word2 (i.e., word2.substring(0, j)).

Recurrence Relation (The "Aha!" Moment):

When considering dp[i][j], we look at the characters word1[i-1] and word2[j-1]:

1. If word1[i-1] === word2[j-1] (characters match):
   No operation is needed for these characters. The cost is simply the cost of converting word1[0...i-2] to word2[0...j-2].
   dp[i][j] = dp[i-1][j-1]

2. If word1[i-1] !== word2[j-1] (characters don't match):
   We must perform one of the three allowed operations, and we choose the one that results in the minimum total cost:

   • Replace: Replace word1[i-1] with word2[j-1]. The cost is 1 (for the replacement) plus the cost of converting word1[0...i-2] to word2[0...j-2].
     cost = dp[i-1][j-1] + 1
   • Delete: Delete word1[i-1]. The cost is 1 (for the deletion) plus the cost of converting word1[0...i-2] to word2[0...j-1].
     cost = dp[i-1][j] + 1
   • Insert: Insert word2[j-1] into word1. This is equivalent to converting word1[0...i-1] to word2[0...j-2] and then inserting word2[j-1]. The cost is 1 (for the insertion) plus the cost of converting word1[0...i-1] to word2[0...j-2].
     cost = dp[i][j-1] + 1

   Therefore, dp[i][j] = 1 + min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]).

The base cases are crucial for initializing the DP table:

• dp[i][0] = i: To convert word1[0...i-1] to an empty string, we need i deletions.
• dp[0][j] = j: To convert an empty string to word2[0...j-1], we need j insertions.

By building up the dp table from these base cases, dp[m][n] will contain the minimum edit distance for the entire word1 and word2.

---

4. Edge Case Analysis

The current solution correctly handles the standard edge cases due to the robust initialization of the DP table and the recurrence relation:

• Empty strings:

  • word1 = "", word2 = "": m=0, n=0. The loops won't run. dp[0][0] is initialized to 0. Correct.
  • word1 = "abc", word2 = "": m=3, n=0. dp[i][0] will correctly be i. The final result dp[3][0] will be 3 (3 deletions). Correct.
  • word1 = "", word2 = "xyz": m=0, n=3. dp[0][j] will correctly be j. The final result dp[0][3] will be 3 (3 insertions). Correct.

• Strings are identical:

  • word1 = "abc", word2 = "abc": The if (word1[i - 1] === word2[j - 1]) condition will frequently be met, leading to dp[i][j] = dp[i-1][j-1]. The result dp[3][3] will be 0. Correct.

• Strings with maximum length constraint:

  • word1.length, word2.length <= 500. The O(m*n) complexity means 500 * 500 = 250,000 operations in the worst case, which is perfectly fine within typical time limits (usually around 10^8 operations per second).

---

5. Learning Points

• Similar problems that use this same pattern (Dynamic Programming on Strings):

  • Longest Common Subsequence (LCS): Find the length of the longest subsequence common to two sequences. The recurrence relation is similar, but instead of minimizing operations, it maximizes the common part.
  • Minimum Deletions/Insertions to Make Strings Equal: Directly related to LCS and Edit Distance. The number of deletions needed for word1 and insertions needed for word2 to make them identical is m + n - 2 * LCS_length.
  • Regular Expression Matching / Wildcard Matching: These problems often involve 2D DP where dp[i][j] signifies whether s[0...i-1] matches p[0...j-1].
  • Distinct Subsequences: Count the number of distinct subsequences of s that equal t.
  • Palindrome Partitioning II: Find the minimum cuts needed to partition a string such that every partition is a palindrome.

• Common mistakes people make with this pattern:

  • Incorrect Base Cases: Misinitializing dp[i][0] or dp[0][j] can lead to incorrect results.
  • Off-by-one errors: Using i instead of i-1 for string indexing or loop bounds can be tricky, especially when mapping dp[i][j] to word1.substring(0, i). The current solution correctly uses i-1 for character access within the loops, matching the dp table's 1-based indexing for string lengths.
  • Forgetting +1: When operations like insert, delete, or replace are performed, forgetting to add 1 to the previous subproblem's cost is a common error.
  • Mixing 0-based and 1-based indexing: The DP table is often 1-indexed (meaning dp[i][j] refers to the first i characters), while strings are 0-indexed. Careful mapping (word1[i-1]) is essential.

• Variations of this problem:

  • Weighted Edit Distance: Each operation (insert, delete, replace) has a different cost. The +1 in the recurrence relation would be replaced by the specific cost of that operation.
  • Finding the Actual Operations: Instead of just returning the minimum distance, reconstruct the sequence of operations. This can be done by backtracking through the dp table from dp[m][n] to dp[0][0], always choosing the path that led to the minimum value.
  • Restricted Operations: Some problems might only allow a subset of operations (e.g., no replacements, only insert/delete).

---

6. Code Quality

The code quality is generally good and adheres to standard practices for LeetCode solutions.

• Variable Naming:

  • word1, word2: Clear.
  • m, n: Standard abbreviations for lengths, acceptable.
  • dp: Standard for dynamic programming table.
  • Overall, names are descriptive and concise.

• Code Structure:

  • The initialization of the dp table is clear.
  • The nested loops correctly iterate through the table.
  • The if/else logic for matching/non-matching characters is well-structured.

• Readability:

  • The code is quite readable for anyone familiar with DP.
  • The comments at the top clearly state time and space complexity.
  • The use of export for the function is a good practice for modularity, especially in a repository like this.

• Minor points:

  • The original file had a missing semicolon at the end of the function definition (};). The new file correctly removes this, which is good as semicolons are optional in JavaScript, and omitting them consistently can improve readability.

---

7. Alternative Approaches

1. Recursion with Memoization (Top-Down Dynamic Programming):

   • Approach: Define a recursive function solve(i, j) that returns the minimum edit distance for word1[0...i-1] and word2[0...j-1]. This function would have base cases for i=0 or j=0. For other cases, it would apply the same recurrence relation as the iterative DP. To avoid redundant computations, a memoization table (e.g., a 2D array or a Map) would store and retrieve results for (i, j) pairs.
   • Trade-offs:
     • Pros: Often more intuitive to write as it directly follows the recursive definition of the problem. Can sometimes be more efficient if only a subset of subproblems is actually needed (though for Edit Distance, most subproblems are typically visited).
     • Cons: Can incur overhead due to recursive function calls, potentially leading to stack overflow issues for very large inputs (though usually not for constraints like 500).

2. Space-Optimized Dynamic Programming (using only two rows):

   • Approach: Notice that to calculate dp[i][j], we only need values from the previous row (dp[i-1][...]) and the current row's previous column (dp[i][j-1]). This means we don't need to store the entire m x n table. We can reduce the space complexity to O(min(m, n)) by using only two rows (current and previous) or even one row if handled carefully. For instance, if n is smaller, we can use an O(n) array, updating it in place.
   • Trade-offs:
     • Pros: Significantly reduces space complexity from O(m*n) to O(min(m,n)). This is a common optimization for DP problems where the current state only depends on a limited number of previous states.
     • Cons: The logic for managing indices and updating the single/two rows can be slightly more complex and prone to off-by-one errors compared to the full 2D table approach. It might slightly reduce readability for someone new to the pattern.

The current O(m*n) time and space solution is the most straightforward and common way to implement Edit Distance and is perfectly acceptable given the problem constraints.

[github-actions]

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🤖 Gemini Code Review (Model: gemini-2.5-flash)

This is an excellent problem to solidify understanding of Dynamic Programming! Let's break down your solution.

---

1. Pattern Identification

This solution uses the Dynamic Programming (DP) algorithmic pattern. Specifically, it's a classic application of DP to solve the Levenshtein Distance problem, which is another name for Edit Distance.

• Why DP?
  • Optimal Substructure: The minimum edit distance between two strings can be found by combining the minimum edit distances of their prefixes. For example, to find the distance between word1[0...i-1] and word2[0...j-1], we look at the distances of smaller prefixes.
  • Overlapping Subproblems: When calculating the edit distance for various prefixes, the same subproblems (e.g., distance between word1[0...x] and word2[0...y]) are encountered and solved multiple times. DP stores these results to avoid redundant computations.

2. Complexity Verification

• Stated Time Complexity: O(m * n)

  • Verification: This is accurate.
  • Explanation: The solution uses a nested loop structure. The outer loop iterates m times (from i = 1 to m), and the inner loop iterates n times (from j = 1 to n). Inside the loops, operations like array access, comparison, and Math.min are all constant time operations. Therefore, the total time complexity is proportional to m * n.

• Stated Space Complexity: O(m * n)

  • Verification: This is accurate.
  • Explanation: A 2D array, dp, of size (m + 1) x (n + 1) is created to store the results of subproblems. The space required for this table is directly proportional to m * n.

3. Key Insight

The key insight for solving Edit Distance with Dynamic Programming lies in defining the recurrence relation and base cases for the dp table.

Let dp[i][j] represent the minimum number of operations required to convert the first i characters of word1 (i.e., word1[0...i-1]) to the first j characters of word2 (i.e., word2[0...j-1]).

1. Base Cases:

   • If word1 is empty (i = 0), to convert it to word2[0...j-1], we need j insertions. So, dp[0][j] = j.
   • If word2 is empty (j = 0), to convert word1[0...i-1] to an empty string, we need i deletions. So, dp[i][0] = i.

2. Recurrence Relation (for i > 0 and j > 0):

   • If word1[i-1] equals word2[j-1] (characters match): No operation is needed for these characters. The cost is simply the edit distance of the preceding substrings: dp[i][j] = dp[i-1][j-1].
   • If word1[i-1] does not equal word2[j-1] (characters mismatch): We have three choices, each incurring a cost of 1 operation:
     • Replace: Replace word1[i-1] with word2[j-1]. The cost is 1 + dp[i-1][j-1] (1 for replacement, plus the distance of the prefixes before these characters).
     • Delete: Delete word1[i-1]. The cost is 1 + dp[i-1][j] (1 for deletion, plus the distance of word1[0...i-2] to word2[0...j-1]).
     • Insert: Insert word2[j-1] into word1. The cost is 1 + dp[i][j-1] (1 for insertion, plus the distance of word1[0...i-1] to word2[0...j-2]).
   • We take the minimum of these three options:
     dp[i][j] = 1 + min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]).

The solution cleverly combines these two cases: dp[i-1][j-1] + (word1[i-1] === word2[j-1] ? 0 : 1) effectively covers the replacement cost (0 if match, 1 if mismatch). This is then compared with the deletion and insertion costs.

4. Edge Case Analysis

The current solution correctly handles the following edge cases:

• Empty strings: word1 = "", word2 = "". m=0, n=0. The loops won't run, and dp[0][0] will be initialized to 0 and returned. Correct.
• One empty string:
  • word1 = "abc", word2 = "". m=3, n=0. The dp[i][0] initialization correctly sets dp[3][0] to 3. Correct (3 deletions needed).
  • word1 = "", word2 = "xyz". m=0, n=3. The dp[0][j] initialization correctly sets dp[0][3] to 3. Correct (3 insertions needed).
• Identical strings: word1 = "abc", word2 = "abc". The word1[i-1] === word2[j-1] condition will always be true, leading to dp[i][j] = dp[i-1][j-1]. Eventually dp[m][n] will be dp[0][0] = 0. Correct.
• Strings requiring only one type of operation: E.g., word1 = "abc", word2 = "ab". The solution will correctly find 1 deletion.

The initialization of the first row and column of the dp table as i and j respectively is crucial for handling these base cases correctly within the general recurrence.

5. Learning Points

• Similar problems using this pattern:
  • Longest Common Subsequence (LCS): Another fundamental DP problem involving two strings. It calculates the length of the longest subsequence common to both strings. The DP table structure and recurrence relation are very similar, but the operations differ (no cost for matching, only for skipping).
  • Minimum Deletions/Insertions to Make Two Strings Equal: This problem is directly related to LCS and Edit Distance. The answer is (m + n - 2 * LCS_length).
  • Distinct Subsequences: Counts the number of distinct subsequences of word1 that are equal to word2.
  • Wildcard Matching / Regular Expression Matching: More complex string matching problems that can often be tackled with DP.
• Common mistakes people make with this pattern:
  • Off-by-one errors: Incorrectly mapping string indices (0 to length-1) to dp table indices (1 to length). Your solution correctly uses word1[i-1] and word2[j-1] when working with dp[i][j], which is a common and robust convention.
  • Incorrect base cases: Failing to initialize the first row/column of the DP table, or initializing them incorrectly.
  • Misunderstanding the recurrence: Confusing the meaning of dp[i-1][j], dp[i][j-1], and dp[i-1][j-1] in relation to insert, delete, and replace operations.
  • Top-down vs. Bottom-up: While top-down (recursion with memoization) can be more intuitive for some, bottom-up (iterative DP as you've done) often avoids recursion depth limits and can be more performant due to less function call overhead.
• Variations of this problem:
  • Weighted Edit Distance: Assign different costs to insertion, deletion, and replacement operations.
  • Restricted Edit Distance: Only allow a subset of the three operations (e.g., only insert and delete).
  • Edit Distance with Transpositions: Consider "transposition" (swapping two adjacent characters) as a fourth operation, often with a cost. This leads to the Damerau-Levenshtein distance.

6. Code Quality

Your code quality is very good:

• Variable Naming: m, n for lengths, dp for the dynamic programming table, word1, word2 are all standard, clear, and descriptive.
• Code Structure and Readability: The code is well-structured, easy to follow, and clearly implements the bottom-up DP approach. The initialization and nested loops are standard and clean.
• TypeScript Best Practices:
  • Using export for the function is good for modularity.
  • const is used for m and n, correctly indicating they are not reassigned.
  • The Array(m + 1).fill(0).map(() => Array(n + 1).fill(0)) is an idiomatic and correct way to initialize a 2D array in JavaScript/TypeScript.
  • The removal of the extraneous semicolon after the function declaration (};) and adding a newline at the end of the file (as seen in the diff) are minor but good stylistic improvements.

7. Alternative Approaches

1. Recursion with Memoization (Top-Down Dynamic Programming):

   • Approach: Define a recursive function solve(i, j) that computes the edit distance for word1[0...i-1] and word2[0...j-1]. Store the results of solve(i, j) in a memoization table (e.g., a 2D array or a Map) to avoid recomputing subproblems.
   • Trade-offs:
     • Pros: Can be more intuitive to write as it directly mirrors the recursive definition of the problem. Only computes the subproblems that are actually needed (lazy evaluation).
     • Cons: Can incur function call overhead, potentially slower than iterative DP for large inputs. May hit recursion depth limits in some languages/environments if not optimized (though less common in JS/TS). Space complexity remains O(m * n) for the recursion stack and memoization table.

2. Space-Optimized Dynamic Programming (Using two rows):

   • Approach: Observe that to compute dp[i][j], we only need values from the current row (dp[i][j-1]) and the previous row (dp[i-1][j-1], dp[i-1][j]). This means we don't need the entire m x n table. We can reduce the space complexity by using only two rows (current and previous) or even one row if managed carefully.
   • Trade-offs:
     • Pros: Reduces space complexity to O(min(m, n)) (if you optimize for the shorter string) or O(n) (if n is the length of the second string). This is a significant improvement for very long strings where memory might be a constraint.
     • Cons: The logic for updating the prev and curr rows, especially handling the dp[i-1][j-1] term, can be slightly more intricate and prone to off-by-one errors if not implemented carefully. Readability might decrease slightly compared to the full 2D table.

---

Overall, your solution is a clean, correct, and optimal implementation of the standard bottom-up Dynamic Programming approach for the Edit Distance problem. Well done!

[coderabbitai]

Actionable comments posted: 1

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• 0072-edit-distance/README.md
• README.md
• leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md
• leetcode/medium/0072-edit-distance/README.md
• leetcode/medium/0072-edit-distance/edit-distance.ts

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**/*.md

📄 CodeRabbit inference engine (AGENTS.md)

**/*.md: Create comprehensive documentation guides for arrays, trees, graphs, heaps, and other data structures
Document reusable code templates for common algorithmic patterns
Maintain Big O complexity cheatsheets and references

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• leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md

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• leetcode/medium/0072-edit-distance/README.md
• README.md

**/*.ts

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Follow strict TypeScript configuration with ES2020 target and CommonJS modules in TypeScript solutions
Enforce full type safety in TypeScript solutions

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**/*.{js,ts}: Use clear, descriptive variable names across all solutions
Add comments for non-obvious logic or algorithm steps in code
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Prefer readability over cleverness in code implementation

**/*.{js,ts}: Use descriptive variable names; single letters like i, j, k are acceptable for loop iterators and standard mathematical contexts
Add comments for complex logic, but prefer self-documenting code with clear logic
Minimize external dependencies and keep solutions self-contained

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Keep functions focused and single-purpose, preferring readability over cleverness
Use const by default, let only when reassignment needed; avoid var except for LeetCode solution function definitions
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Prefer built-in methods (.map(), .filter(), .reduce()) when appropriate and use strict equality (===) over loose equality (==)

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🧠 Learnings (3)

📓 Common learnings

Learnt from: CR
Repo: pertrai1/coding-challenges PR: 0
File: .github/copilot-instructions.md:0-0
Timestamp: 2025-11-26T14:46:46.097Z
Learning: Applies to [0-9]*-*/*.md : Problem README files should include: problem title with difficulty badge, problem description from LeetCode, examples with inputs/outputs, and constraints

📚 Learning: 2025-11-26T14:46:46.097Z

Learnt from: CR
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File: .github/copilot-instructions.md:0-0
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📚 Learning: 2025-12-02T17:56:01.747Z

Learnt from: CR
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Timestamp: 2025-12-02T17:56:01.747Z
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Applied to files:

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🪛 LanguageTool

leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md

[style] ~36-~36: You have already used this phrasing in nearby sentences. Shortening it will avoid wordiness.
Context: ...able and dynamic programming allows for making use of previous operations and calculations #...

(REP_MAKE_USE_OF)

---

[grammar] ~40-~40: Ensure spelling is correct
Context: ...atterns did I find or miss? - I missed dyanmic programming being the most efficient wa...

(QB_NEW_EN_ORTHOGRAPHY_ERROR_IDS_1)

---

[grammar] ~50-~50: Use a hyphen to join words.
Context: ... ### Bugs to add to the Bug List - Off by one iterations that go out of bounds ...

(QB_NEW_EN_HYPHEN)

---

[grammar] ~50-~50: Use a hyphen to join words.
Context: ...## Bugs to add to the Bug List - Off by one iterations that go out of bounds ##...

(QB_NEW_EN_HYPHEN)

⏰ Context from checks skipped due to timeout of 90000ms. You can increase the timeout in your CodeRabbit configuration to a maximum of 15 minutes (900000ms). (2)

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🔇 Additional comments (3)

README.md (1)

21-21: LGTM! Progress tracking updated correctly.

The Medium problems count has been incremented appropriately, and the new Edit Distance entry follows the established format with the correct path and difficulty badge.

Also applies to: 237-237

leetcode/medium/0072-edit-distance/README.md (1)

1-38: LGTM! Comprehensive problem documentation.

The README follows all required guidelines with a clear problem statement, detailed examples with step-by-step transformations, and complete constraints. The documentation structure aligns with the repository's standards.

Based on learnings, this meets all requirements for problem README files.

leetcode/medium/0072-edit-distance/edit-distance.ts (1)

5-32: LGTM! Clean edit distance implementation with proper export.

The dynamic programming solution correctly implements the classic edit distance algorithm:

• Proper initialization of the DP table with base cases
• Correct recurrence relation considering all three operations (insert, delete, replace)
• Accurate complexity annotations (O(m*n) time and space)

The export keyword appropriately makes the function available for external use, and the type annotations follow TypeScript best practices.

[coderabbitai]

⚠️ Potential issue | 🟡 Minor

Fix spelling and grammar issues in the template.

The post-mortem template contains a few text quality issues:

• Line 36: Consider simplifying "allows for making use of" to "allows using" for conciseness
• Line 40: Typo "dyanmic" should be "dynamic"
• Line 50: "Off by one" should be hyphenated as "Off-by-one"

🔎 Proposed fixes

-- Using dynamic programming allows for remembering the distances at each of the characters for both words
+- Using dynamic programming allows remembering the distances at each of the characters for both words
 - Making sure to populate the table initially before iterating over
 - Comparing the minimum of the current character in the table with previous characters
 - Returning the correct index of the table

-- I missed dyanmic programming being the most efficient way of solving this distance problem
+- I missed dynamic programming being the most efficient way of solving this distance problem

-- Off by one iterations that go out of bounds
+- Off-by-one iterations that go out of bounds

Also applies to: 40-40, 50-50

🧰 Tools

🪛 LanguageTool

[style] ~36-~36: You have already used this phrasing in nearby sentences. Shortening it will avoid wordiness.
Context: ...able and dynamic programming allows for making use of previous operations and calculations #...

(REP_MAKE_USE_OF)

🤖 Prompt for AI Agents

In leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md around lines 36,
40, and 50, fix the text issues: on line 36 replace "allows for making use of"
with the concise phrase "allows using"; on line 40 correct the typo "dyanmic" to
"dynamic"; and on line 50 hyphenate "Off by one" to "Off-by-one". Ensure
punctuation and surrounding sentence flow remain correct after each change.

[Copilot]

Pull request overview

This PR adds a new solution for LeetCode problem 0072 (Edit Distance), a classic dynamic programming problem also known as the Levenshtein Distance. The solution is implemented in TypeScript with proper export declarations consistent with other TypeScript solutions in the repository.

Key Changes:

• Implements a standard DP solution with O(m*n) time and space complexity
• Adds comprehensive problem documentation and a detailed post-mortem reflection
• Updates main README with the new problem entry in the Dynamic Programming section

Reviewed changes

Copilot reviewed 5 out of 5 changed files in this pull request and generated 2 comments.

Show a summary per file

File	Description
leetcode/medium/0072-edit-distance/edit-distance.ts	TypeScript implementation using 2D DP table to compute minimum edit distance; properly exported function following repository conventions
leetcode/medium/0072-edit-distance/README.md	Problem description with examples and constraints in clean markdown format
leetcode/medium/0072-edit-distance/POST_MORTEM_TEMPLATE.md	Detailed post-mortem log documenting the problem-solving process, time tracking, and self-reflection
README.md	Updated medium problem count from 71 to 72 and added link to new problem in Dynamic Programming section
0072-edit-distance/README.md	Removed old file from incorrect location (moved to proper directory structure)

[github-actions]

@codex review

[github-actions]

Claude finished @pertrai1's task in 2m 17s —— View job

---

Code Review Complete ✅

Todo List

• Read solution implementation and problem description
• Verify correctness and edge case handling
• Analyze time/space complexity
• Identify algorithmic patterns and provide learning insights
• Review code quality
• Suggest alternative approaches
• Post comprehensive review

---

Overview

Excellent work on implementing the Edit Distance (Levenshtein Distance) problem! This is a classic dynamic programming problem and your solution is correct and optimal. Your post-mortem reflection shows great self-awareness about the problem-solving process.

---

1. Correctness ✅

Verdict: The solution is correct and handles all cases properly.

Your implementation correctly:

• Handles all three operations (insert, delete, replace)
• Uses proper base case initialization
• Implements the correct recurrence relation

Edge Cases Verified:

• ✅ Empty strings: ("", "") → 0 operations
• ✅ One empty string: ("abc", "") → 3 deletions, ("", "xyz") → 3 insertions
• ✅ Identical strings: ("abc", "abc") → 0 operations
• ✅ No common characters: Works correctly
• ✅ Max constraint (500 chars): Handles within limits

Mental Trace (Example 1):
For word1 = "horse", word2 = "ros":

    ""  r  o  s
""   0  1  2  3
h    1  1  2  3
o    2  2  1  2
r    3  2  2  2
s    4  3  3  2
e    5  4  4  3

Final answer: dp[5][3] = 3 ✓

---

2. Complexity Analysis ✅

Your complexity annotations are accurate:

Time Complexity: O(m × n) ✓

• Two nested loops iterate through all (m+1) × (n+1) cells
• Each cell performs constant-time operations (comparison + min of 3 values)
• Why: Every subproblem (i, j) is computed exactly once

Space Complexity: O(m × n) ✓

• 2D DP table of size (m+1) × (n+1)
• Optimization possible: Can be reduced to O(min(m, n)) using rolling array (see Alternative Approaches)

---

3. Pattern Identification & Learning Focus

Algorithmic Pattern: 2D Dynamic Programming (String DP)

This problem is a textbook example of the 2D DP pattern for string comparison problems.

Why This Pattern Works:

1. Optimal Substructure: The minimum edit distance for full strings can be built from solutions to smaller substrings
2. Overlapping Subproblems: Computing distance for word1[0...i] to word2[0...j] requires solving many overlapping substring pairs
3. Bottom-up Construction: We build from base cases (empty strings) up to the full solution

The Key Insight 💡

The breakthrough insight is the recurrence relation at dp[i][j]:

If word1[i-1] == word2[j-1]:
    dp[i][j] = dp[i-1][j-1]  // No operation needed
Else:
    dp[i][j] = 1 + min(
        dp[i-1][j-1],  // Replace word1[i-1] with word2[j-1]
        dp[i-1][j],    // Delete word1[i-1]
        dp[i][j-1]     // Insert word2[j-1]
    )

Why this works:

• Each cell represents the optimal solution for a substring pair
• We consider all three operations and take the minimum
• The +1 represents the cost of the current operation
• Your code elegantly combines the match/replace cases: dp[i-1][j-1] + (word1[i-1] === word2[j-1] ? 0 : 1)

Pattern Recognition Trigger 🎯

When you see:

• "Minimum operations" to transform one sequence to another
• Operations that modify sequences (insert/delete/replace/swap)
• Comparing two strings/arrays at different positions

Think: 2D Dynamic Programming!

---

4. Related Problems for Practice

Master this pattern with these similar problems:

Same Pattern (String DP):

1. Longest Common Subsequence (LeetCode 1143 - Medium)

   • Same 2D DP structure
   • Instead of minimizing operations, maximize common characters
   • Connection: minOperations = m + n - 2*LCS

2. Delete Operation for Two Strings (LeetCode 583 - Medium)

   • Edit Distance with only delete operations
   • Same DP approach with modified recurrence

3. Minimum ASCII Delete Sum (LeetCode 712 - Medium)

   • Weighted version of edit distance
   • Same structure, different cost calculation

Advanced Variations:

4. Regular Expression Matching (LeetCode 10 - Hard)

   • More complex 2D DP with pattern matching

5. Wildcard Matching (LeetCode 44 - Hard)

   • Similar to regex matching but different rules

---

5. Code Quality Review

Strengths ✓

• Clear structure: Base case initialization separated from computation
• Good naming: m, n, dp are standard and clear
• Proper TypeScript: Export keyword, type annotations
• Complexity documentation: Accurate JSDoc comments
• Modern JS: Array.from() with proper initialization

Minor Enhancement Suggestions

1. Enhanced Comments (edit-distance.ts:20-28)

Consider adding operation descriptions to the recurrence:

// Fill the table using the recurrence relation
for (let i = 1; i <= m; i++) {
  for (let j = 1; j <= n; j++) {
    dp[i][j] = Math.min(
      // Replace (or match if characters are same)
      dp[i - 1][j - 1] + (word1[i - 1] === word2[j - 1] ? 0 : 1),
      // Delete from word1
      dp[i - 1][j] + 1,
      // Insert into word1
      dp[i][j - 1] + 1
    );
  }
}

2. Base Case Explanation (edit-distance.ts:12-18)

Add a comment explaining the base case meaning:

// Base cases:
// dp[i][0] = i: Need i deletions to convert word1[0...i-1] to empty string
// dp[0][j] = j: Need j insertions to convert empty string to word2[0...j-1]

JavaScript/TypeScript Best Practices ✓

• ✅ Uses const for non-reassigned variables
• ✅ Proper array initialization (avoiding .fill() reference bug)
• ✅ Strict equality (===)
• ✅ Clean, functional style

---

6. Alternative Approaches

Approach 1: Space-Optimized DP

Space: O(min(m, n)) instead of O(m × n)

Since we only need the previous row to compute the current row, we can use a rolling array:

function minDistanceOptimized(word1: string, word2: string): number {
  let [m, n] = [word1.length, word2.length];
  
  // Ensure word2 is the shorter string for space optimization
  if (m < n) {
    [word1, word2, m, n] = [word2, word1, n, m];
  }
  
  // Only keep previous and current row
  let prev = Array(n + 1).fill(0).map((_, i) => i);
  
  for (let i = 1; i <= m; i++) {
    const curr = [i];
    for (let j = 1; j <= n; j++) {
      curr[j] = Math.min(
        prev[j - 1] + (word1[i - 1] === word2[j - 1] ? 0 : 1),
        prev[j] + 1,
        curr[j - 1] + 1
      );
    }
    prev = curr;
  }
  
  return prev[n];
}

Trade-offs:

• ✅ Reduces space from O(m × n) to O(min(m, n))
• ❌ Slightly more complex logic
• ❌ Can't reconstruct the actual edit sequence

When to use: Memory-constrained environments or very long strings

---

Approach 2: Recursive + Memoization (Top-Down DP)

Same complexity, different style

function minDistanceTopDown(word1: string, word2: string): number {
  const memo = new Map<string, number>();
  
  function dp(i: number, j: number): number {
    // Base cases
    if (i === 0) return j;
    if (j === 0) return i;
    
    const key = `${i},${j}`;
    if (memo.has(key)) return memo.get(key)!;
    
    let result: number;
    if (word1[i - 1] === word2[j - 1]) {
      result = dp(i - 1, j - 1);
    } else {
      result = 1 + Math.min(
        dp(i - 1, j - 1),  // Replace
        dp(i - 1, j),      // Delete
        dp(i, j - 1)       // Insert
      );
    }
    
    memo.set(key, result);
    return result;
  }
  
  return dp(word1.length, word2.length);
}

Trade-offs:

• ✅ More intuitive (follows recursive definition)
• ✅ Only computes needed subproblems (lazy evaluation)
• ❌ Recursion overhead
• ❌ Potential stack overflow on very large inputs (though unlikely with 500 limit)

When to use: When the recursive formulation is more natural or when not all subproblems are needed

---

7. Common Mistakes with This Pattern

Be aware of these pitfalls:

1. Off-by-one errors: Mixing 0-based string indexing with 1-based DP indexing

   • ✅ Your code correctly uses word1[i-1] when working with dp[i][j]

2. Incorrect base cases: Forgetting to initialize first row/column

   • ✅ Your initialization is correct

3. Forgetting the +1: Missing the operation cost in the recurrence

   • ✅ Your code has proper +1 for operations

4. Confusing operations: Mix up what dp[i-1][j], dp[i][j-1], dp[i-1][j-1] represent

   • ✅ Your logic is correct

---

8. Learning Summary

Core Concepts Demonstrated:

✅ 2D Dynamic Programming - Building solutions from smaller subproblems
✅ Optimal Substructure - Combining optimal subsolutions
✅ State Definition - dp[i][j] represents edit distance for prefixes
✅ Recurrence Relations - Mathematical formulation of state transitions

Your Growth (from Post-Mortem):

Your post-mortem shows excellent self-reflection:

• ✅ Recognized learning: Discovered Edit Distance = Levenshtein Distance
• ✅ Pattern identification: Realized DP was the optimal approach
• ✅ Self-awareness: Noted the importance of tracing examples first
• ✅ Bug tracking: Identified off-by-one as a recurring issue

Recommendation: Your practice of using post-mortems is excellent for deliberate practice and pattern recognition. Keep it up!

Problem-Solving Insights:

1. Always trace examples first - You correctly identified this in your retrospective
2. Recognize DP triggers - "Minimum operations" + "string transformation" → 2D DP
3. Build from base cases - Empty string cases lead naturally to the full solution

---

Final Assessment

Category	Rating	Notes
Correctness	⭐⭐⭐⭐⭐	Perfect implementation, handles all edge cases
Complexity	⭐⭐⭐⭐⭐	Optimal time, accurate analysis
Code Quality	⭐⭐⭐⭐⭐	Clean, well-documented, follows best practices
Learning Value	⭐⭐⭐⭐⭐	Excellent post-mortem reflection

Overall: This is a textbook-perfect implementation of Edit Distance. Your solution demonstrates solid understanding of dynamic programming fundamentals. The combination of correct code + thoughtful post-mortem analysis shows you're building strong problem-solving skills.

Keep up the great work! 🎯

---

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[github-actions]

---

🤖 Gemini Code Review (Model: gemini-2.5-flash)

Here's a review of the minDistance solution for the "Edit Distance" problem.

---

1. Pattern Identification

This solution uses Dynamic Programming. Specifically, it's a classic example of a 2D DP problem where the state dp[i][j] represents the minimum number of operations required to convert the prefix word1[0...i-1] to word2[0...j-1]. This pattern is often used for string-related problems involving transformations or finding optimal substructure solutions.

2. Complexity Verification

• Time Complexity: O(m * n)

  • Stated: O(m * n)
  • Verification: Accurate. The solution initializes a 2D DP table of size (m+1) x (n+1). It then iterates through this table using nested loops, where i goes from 1 to m and j goes from 1 to n. Inside the inner loop, each dp[i][j] state is calculated in constant time O(1) by looking up three previously computed states. Therefore, the total time complexity is O(m * n).

• Space Complexity: O(m * n)

  • Stated: O(m * n)
  • Verification: Accurate. The solution uses a 2D array dp of size (m+1) x (n+1) to store the intermediate results. This table directly dictates the space usage, leading to O(m * n) space complexity.

3. Key Insight

The key insight for solving the Edit Distance problem (also known as Levenshtein Distance) with dynamic programming lies in breaking down the problem into smaller, overlapping subproblems and establishing a recurrence relation.

Consider dp[i][j] as the minimum operations to convert word1[0...i-1] to word2[0...j-1].
When comparing word1[i-1] and word2[j-1] (the last characters of the current prefixes):

1. If word1[i-1] === word2[j-1]: The characters match. No operation is needed for these characters. The cost is simply the cost of converting word1[0...i-2] to word2[0...j-2].

   • dp[i][j] = dp[i-1][j-1]

2. If word1[i-1] !== word2[j-1]: We have three choices, each costing 1 operation:

   • Replace: Replace word1[i-1] with word2[j-1]. The cost is 1 (for the replacement) plus the cost of converting word1[0...i-2] to word2[0...j-2].
     • cost_replace = dp[i-1][j-1] + 1
   • Delete: Delete word1[i-1]. The cost is 1 (for the deletion) plus the cost of converting word1[0...i-2] to word2[0...j-1] (since word1[i-1] is gone, we now need to match word1[0...i-2] with the full word2 prefix).
     • cost_delete = dp[i-1][j] + 1
   • Insert: Insert word2[j-1] into word1. The cost is 1 (for the insertion) plus the cost of converting word1[0...i-1] to word2[0...j-2] (since word2[j-1] is now "matched" by the insertion, we need to match the full word1 prefix with the remaining word2 prefix).
     • cost_insert = dp[i][j-1] + 1

The dp[i][j] will be the minimum of these three costs when characters don't match.
The base cases are crucial:

• dp[i][0] = i: To convert word1[0...i-1] to an empty string "", we need i deletions.
• dp[0][j] = j: To convert "" to word2[0...j-1], we need j insertions.

This recurrence relation and base cases allow us to build up the solution from smallest prefixes to the full strings.

4. Edge Case Analysis

The current solution correctly handles the following edge cases due to the robust initialization of the dp table:

• Both strings are empty: word1 = "", word2 = ""
  • m=0, n=0. The loops for (let i = 1; i <= m; i++) and for (let j = 1; j <= n; j++) will not run. dp[0][0] will be 0, which is correctly returned.
• One string is empty:
  • word1 = "abc", word2 = "": m=3, n=0. dp[i][0] will be i. dp[3][0] will be 3 (3 deletions).
  • word1 = "", word2 = "xyz": m=0, n=3. dp[0][j] will be j. dp[0][3] will be 3 (3 insertions).
• Identical strings: word1 = "abc", word2 = "abc"
  • The word1[i-1] === word2[j-1] condition will always be met, leading to dp[i][j] = dp[i-1][j-1]. The final result will be 0.
• Strings with maximum length (500): The O(m*n) complexity means 500*500 = 250,000 operations, which is well within typical time limits for LeetCode (usually around 10^8 operations per second). Memory for 501*501 numbers (approx 250k numbers) is also fine.

The solution is robust for these common edge cases.

5. Learning Points

• Similar problems using this pattern:

  • Longest Common Subsequence (LCS): Finding the longest subsequence common to two strings (often solved with DP, very similar table structure).
  • Minimum Deletions/Insertions to make two strings equal: Directly related to LCS and Edit Distance.
  • Minimum Deletions to Make a String Palindrome: Can be framed as finding LCS between the string and its reverse.
  • Distinct Subsequences: Counting unique subsequences.
  • Wildcard Matching / Regular Expression Matching: More complex DP problems.
  • Knapsack Problem (0/1 Knapsack): Another classic DP problem with a 2D table.

• Common mistakes with this pattern:

  • Incorrect Base Cases: Not properly initializing the first row/column of the DP table, or handling empty strings. This is critical as all subsequent calculations depend on them.
  • Off-by-One Errors: Mismatching array indices with string lengths (e.g., using word1[i] when dp[i][j] refers to word1[0...i-1]). The solution's use of i-1 and j-1 for string access is correct for dp[i][j] representing prefixes of length i and j.
  • Misinterpreting Recurrence Relation: Confusing the costs of insert, delete, and replace, or not realizing which previous states correspond to which operation.
  • Forgetting Math.min: Not taking the minimum of all possible operations when characters don't match.

• Variations of this problem:

  • Weighted Edit Distance: Different costs assigned to insert, delete, and replace operations.
  • Restricted Operations: Only allowing a subset of the three operations (e.g., no replacements).
  • Bounded Edit Distance: Finding if the edit distance is within a certain threshold k. This can sometimes be optimized.
  • Approximate String Matching: Finding substrings that are "close" to a pattern (e.g., using a sliding window with edit distance).

6. Code Quality

The code is generally well-structured and readable.

• Variable Naming: m, n for lengths, dp for the dynamic programming table, word1, word2 are all standard and clear.
• Code Structure: The initialization of the dp table, followed by nested loops for computation, is the standard and expected structure for this DP problem.
• Readability: The logic is straightforward. The Math.min call clearly shows the three options being considered.
• Improvements:
  • Semicolons: The original code was missing a semicolon after dp[i][j] = Math.min(...) and a newline at the end of the file. The diff shows these have been corrected, which is good for consistent style.
  • export keyword: The addition of export to minDistance is a good practice if this function is intended to be imported into other modules (e.g., for testing).

7. Alternative Approaches

1. Space-Optimized Dynamic Programming:

   • Approach: Notice that to compute dp[i][j], we only need values from the previous row (dp[i-1]) and the current row (dp[i][j-1]). This means we don't need to store the entire m x n table. We can optimize space to O(min(m, n)).
   • Trade-offs: Reduces space complexity significantly (e.g., from O(500*500) to O(500)), which is useful for very long strings. The time complexity remains O(m * n). The code becomes slightly more complex to manage the two rows (current and previous) or even a single row with temporary variables.

2. Recursion with Memoization (Top-Down DP):

   • Approach: Define a recursive function solve(i, j) that calculates dp[i][j]. Store the results of solve(i, j) in a memoization table (e.g., a 2D array or map) to avoid redundant computations. If solve(i, j) has been computed before, return the stored result.
   • Trade-offs: Conceptually often more intuitive to write for some developers as it mirrors the mathematical recurrence directly. However, it can incur overhead from function calls and potentially use more stack space for deep recursions. For this problem, the iterative (bottom-up) DP is generally preferred in competitive programming for performance reasons, but memoization is equally valid in terms of asymptotic time complexity.

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