LeetCode Longest Common Subsequence Problem

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Summary by CodeRabbit

• New Features

  • Added problem 1250 - Longest Common Subsequence to the Medium problems collection.
  • Total problems solved increased from 166 to 167; Medium problems from 72 to 73.

• Tests

  • Added comprehensive test suite covering edge cases and input scenarios.

• Documentation

  • Added problem description with examples and constraints.
  • Added post-mortem template for solution documentation.

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Claude finished @pertrai1's task in 1m 39s —— View job

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Code Review Complete ✅

Reviewing LeetCode #1250 - Longest Common Subsequence

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1. Correctness ✅

The solution is correct and handles all edge cases properly:

• ✅ Works for all provided examples
• ✅ Handles empty strings (test cases included)
• ✅ Handles identical strings
• ✅ Handles completely different strings
• ✅ Handles repeated characters
• ✅ Works within the constraints (1 ≤ length ≤ 1000)

The algorithm correctly builds a 2D DP table where dp[i][j] represents the length of the LCS for text1[0..i-1] and text2[0..j-1].

Test coverage is excellent - 15 test cases covering edge cases, simple cases, and complex scenarios.

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2. Complexity Analysis ✅

Time Complexity: O(m × n) ✅

• Correct explanation: The nested loops iterate through each character in text1 (m iterations) for each character in text2 (n iterations).
• Each cell computation is O(1), resulting in O(m × n) overall.

Space Complexity: O(m × n) ✅

• Correct explanation: The 2D DP array has dimensions (m + 1) × (n + 1).

Potential Optimization: Space complexity can be reduced to O(min(m, n)) by using only two rows (current and previous), since we only need the previous row to compute the current row. However, the current solution is optimal for understanding and clarity.

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3. Learning Focus - Pattern Recognition

🎯 Pattern: 2D Dynamic Programming (String Subsequence)

This is a classic 2D DP problem for finding subsequences. The pattern involves:

• Building a table where dp[i][j] stores the solution for subproblems
• Two choices at each cell: characters match or don't match
• Optimal substructure: solution depends on previously computed subproblems

💡 Key Insight

The non-obvious trick that makes this solution work:

When characters match (text1[i-1] === text2[j-1]), we extend the LCS from the diagonal cell (dp[i-1][j-1] + 1). When they don't match, we take the maximum of excluding either the current character from text1 or from text2 (i.e., max(dp[i-1][j], dp[i][j-1])).

This works because:

• If characters match, both must be included in the optimal LCS
• If they don't match, at least one must be excluded, so we try both options and take the best

The DP table naturally encodes all possible subsequence combinations without needing to generate them explicitly.

🔄 Why This Pattern Works Here

LCS is a subsequence problem (not substring), meaning:

• Order matters, but elements don't need to be consecutive
• We can't use sliding window or two pointers
• We need to consider all combinations of including/excluding characters
• DP lets us reuse solutions to overlapping subproblems

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4. Code Quality - Good Practices ✅

Strengths:

1. ✅ Clear variable names: dp, m, n are standard for DP problems
2. ✅ Excellent comments: Each section is well-explained
   • Line 9: Explains the DP table purpose
   • Lines 12-14: Describe the nested loop iteration
   • Lines 16-23: Explain both branches of the recurrence relation
3. ✅ Modern JavaScript: Uses Array.from() for 2D array creation
4. ✅ JSDoc comment: Documents time and space complexity at the top
5. ✅ Clean structure: Easy to follow logic flow

Minor Suggestions:

1. Typo in comment (line 12, 14): "iterative" should be "iterate"

   // iterative through each character in text1
   // Should be: iterate through each character in text1

2. Consider adding visual example in comments: A small DP table example could help illustrate how the algorithm works:

   // Example: text1 = "ace", text2 = "abcde"
   //     ""  a  b  c  d  e
   // ""   0  0  0  0  0  0
   // a    0  1  1  1  1  1
   // c    0  1  1  2  2  2
   // e    0  1  1  2  2  3  <- answer: 3

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5. Related Problems for Practice

To build pattern recognition for 2D DP subsequence/substring problems:

Similar LCS Variations:

1. LeetCode 1143 - Longest Common Subsequence (same problem, different number) - Practice the same pattern
2. LeetCode 583 - Delete Operation for Two Strings - Uses LCS to find minimum deletions (Medium)
3. LeetCode 712 - Minimum ASCII Delete Sum for Two Strings - LCS variant with weighted deletions (Medium)

Related 2D DP Problems:

4. LeetCode 72 - Edit Distance - Similar 2D DP but with insertions/deletions/replacements (Hard)
5. LeetCode 718 - Maximum Length of Repeated Subarray - Like LCS but for subarrays (contiguous) instead of subsequences (Medium)
6. LeetCode 1092 - Shortest Common Supersequence - Combines both strings using LCS (Hard)

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6. Alternative Approaches

Current Approach: Bottom-Up 2D DP

• Time: O(m × n)
• Space: O(m × n)
• Best for: Understanding, clarity, and interview explanation

Alternative 1: Space-Optimized DP

• Space: O(min(m, n)) - Use only 2 rows
• Trade-off: Slightly more complex code, but significant space savings for large inputs
• When to use: Production code with large inputs

Alternative 2: Top-Down DP (Memoization)

• Time: O(m × n)
• Space: O(m × n) + recursion stack
• Trade-off: More intuitive recursive logic, but overhead from recursion
• When to use: When recursive solution is easier to conceptualize

Alternative 3: Recursive (No Memoization)

• Time: O(2^(m+n)) - exponential!
• Not recommended - Only useful for understanding the problem structure before optimization

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Learning Summary 🎓

Problem Type

2D Dynamic Programming - String Subsequence Comparison

Core Concept

This problem teaches the fundamental pattern of using a 2D DP table to compare two sequences, where each cell represents the optimal solution for subproblems. The recurrence relation captures two scenarios: characters match (extend previous LCS) or don't match (take the best option from excluding one character).

Key Takeaway from Post-Mortem

Your post-mortem shows excellent self-reflection:

• ✅ Recognized the DP pattern after initially trying greedy approach
• ✅ Identified this is a classic DP subsequence problem
• ✅ Good time tracking (30 min design + 30 min coding)

Growth Tip: For future subsequence problems, ask yourself:

1. Do I need to maintain order? (Yes → consider DP)
2. Can I skip elements? (Yes → likely DP, not sliding window)
3. Do I need all combinations? (Yes → DP over greedy)

Pattern Recognition Trigger

Whenever you see "subsequence" in a problem (especially comparing two sequences), think:

• 2D DP table
• Recurrence relation based on match/no-match
• Similar to LCS, Edit Distance family

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Summary

This is a solid, well-implemented solution with excellent test coverage and clear documentation. The code follows best practices and demonstrates a strong understanding of 2D DP patterns. The only minor issue is a typo in the comments.

Overall Rating: ⭐⭐⭐⭐⭐ (5/5)

• Correctness: Perfect
• Complexity: Optimal for clarity
• Code Quality: Professional
• Learning Value: High

Great work on the post-mortem reflection - recognizing the DP pattern is a crucial skill for interview success! 🎯

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📝 Walkthrough

Walkthrough

A new LeetCode Medium difficulty problem (Longest Common Subsequence) has been added to the repository with implementation, comprehensive test suite, documentation, and post-mortem template. The main progress tracking README has been updated to reflect the addition of this problem.

Changes

Cohort / File(s)	Summary
Progress tracking README.md	Updated LeetCode problem count from 166 to 167 and Medium problems from 72 to 73. Added new problem entry for 1250 - Longest Common Subsequence.
Problem implementation leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.ts	Exported longestCommonSubsequence function with updated JSDoc formatting and trailing newline.
Problem documentation leetcode/medium/1250-longest-common-subsequence/README.md, POST_MORTEM_TEMPLATE.md	Added problem description with constraints and examples; added post-mortem template for documenting problem-solving approach and learnings.
Problem tests leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.test.ts	Added comprehensive test suite covering basic matches, empty strings, full matches, partial overlaps, repeated characters, and edge cases.
Deleted file 1250-longest-common-subsequence/README.md	Removed legacy documentation file.

Estimated code review effort

🎯 2 (Simple) | ⏱️ ~15 minutes

Poem

🐰 A subsequence quest we did complete,
With tests and docs, all tidy and neat,
One-two-five-oh joins the Medium fold,
Our LeetCode progress, a tale to be told! ✨

Pre-merge checks and finishing touches

✅ Passed checks (3 passed)

Check name	Status	Explanation
Description Check	✅ Passed	Check skipped - CodeRabbit’s high-level summary is enabled.
Title check	✅ Passed	The title accurately describes the main change: adding a new LeetCode problem (Longest Common Subsequence) with implementation, tests, and documentation.
Docstring Coverage	✅ Passed	Docstring coverage is 100.00% which is sufficient. The required threshold is 80.00%.

✨ Finishing touches

• 📝 Generate docstrings

🧪 Generate unit tests (beta)

• Create PR with unit tests
• Post copyable unit tests in a comment
• Commit unit tests in branch leetcode-12252025

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🧮 Solution Analysis

PR: LeetCode Longest Common Subsequence Problem
Files Changed: leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.test.ts leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.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

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Analysis generated for PR files only

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✅ Test Results

Tests have been executed for this pull request.

📊 Coverage Summary

Metric	Coverage
Statements	93.3%
Branches	87.6%
Functions	81.81%
Lines	93.39%

Check the workflow logs for detailed test output.

[github-actions]

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

Here's a review of the Longest Common Subsequence solution:

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1. Pattern Identification

This solution uses the Dynamic Programming (DP) algorithmic pattern. Specifically, it's a classic example of bottom-up DP for sequence alignment problems.

2. Complexity Verification

• Stated Time Complexity: O(m * n)
  • Verification: Accurate. The solution uses two nested loops. The outer loop iterates m times (where m is text1.length), and the inner loop iterates n times (where n is text2.length). Inside the loops, operations are constant time (array access, comparison, Math.max). Therefore, the total time complexity is proportional to m * n.
• Stated Space Complexity: O(m * n)
  • Verification: Accurate. A 2D array (DP table) named dp of size (m + 1) * (n + 1) is created to store intermediate results. This consumes O(m * n) space.

3. Key Insight

The key insight for solving the Longest Common Subsequence problem with dynamic programming lies in breaking it down into smaller, overlapping subproblems and building up the solution.

Let dp[i][j] represent the length of the Longest Common Subsequence of text1[0...i-1] and text2[0...j-1].

The 'aha!' moment comes from the recurrence relation:

1. If text1[i-1] (current character in text1) equals text2[j-1] (current character in text2):
   This means we found a common character. We can extend the LCS from the previous subproblem where both characters were excluded. So, dp[i][j] = 1 + dp[i-1][j-1].
2. If text1[i-1] does not equal text2[j-1]:
   We cannot include both characters in the current LCS. We must either exclude text1[i-1] and find the LCS of text1[0...i-2] and text2[0...j-1] (which is dp[i-1][j]), OR exclude text2[j-1] and find the LCS of text1[0...i-1] and text2[0...j-2] (which is dp[i][j-1]). We take the maximum of these two possibilities: dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]).

The base cases are dp[0][j] = 0 and dp[i][0] = 0 for all i, j, meaning an empty string has no common subsequence with any other string. By filling this table from top-left to bottom-right, the final answer will be in dp[m][n].

4. Edge Case Analysis

The current solution correctly handles the following edge cases:

• Empty strings: If text1 or text2 is empty, m or n will be 0. The dp table will be initialized with all zeros, and the loops won't run for i=0 or j=0, correctly returning dp[m][n], which will be 0.
• Single characters:
  • text1 = "a", text2 = "a": dp[1][1] will be 1 + dp[0][0] = 1. Correct.
  • text1 = "a", text2 = "b": dp[1][1] will be Math.max(dp[0][1], dp[1][0]) = Math.max(0, 0) = 0. Correct.
• No common characters: e.g., text1 = "abc", text2 = "def". The if condition text1Char === text2Char will never be true, so dp[i][j] will always be Math.max(dp[i-1][j], dp[i][j-1]), propagating zeros, resulting in 0. Correct.
• All characters common: e.g., text1 = "abc", text2 = "abc". The if condition will always be true, leading to dp[i][j] = 1 + dp[i-1][j-1], correctly accumulating the length.
• Constraints: 1 <= text1.length, text2.length <= 1000. The O(m*n) complexity means for 1000 * 1000 = 1,000,000 operations, which is well within typical time limits for competitive programming (usually around 10^8 operations per second). Memory for a 1001 * 1001 array of numbers is also manageable.

5. Learning Points

• Similar problems using this pattern:
  • Longest Common Substring: Similar DP structure, but the recurrence changes: if characters don't match, the current substring length resets to 0.
  • Edit Distance (Levenshtein Distance): Calculates the minimum number of operations (insertions, deletions, substitutions) required to transform one string into another. Uses a similar 2D DP table.
  • Longest Increasing Subsequence: Can be solved with DP, though a more optimized O(N log N) solution using binary search also exists.
  • Minimum Deletions/Insertions to make strings equal: Directly related to LCS. min_deletions = m - LCS_length, min_insertions = n - LCS_length.
  • Shortest Common Supersequence: The length is m + n - LCS_length.
• Common mistakes people make with this pattern:
  • Off-by-one errors: Incorrectly mapping string indices to DP table indices (e.g., using i instead of i-1 when accessing text1[i-1]). The (m+1) * (n+1) table size with dp[0][j] and dp[i][0] as base cases correctly handles this by allowing i-1 and j-1 to refer to valid characters.
  • Incorrect base cases: Not initializing dp[0][j] or dp[i][0] to 0, which would lead to incorrect accumulation.
  • Greedy approaches: Attempting to solve it with two pointers or a greedy strategy often fails because subsequences don't require contiguous characters. For example, text1 = "ab", text2 = "acb", a greedy approach might miss "ab". DP ensures all possibilities are considered.
• Variations of this problem exist:
  • Printing the actual LCS: Instead of just the length, you can reconstruct the LCS by backtracking through the dp table from dp[m][n], following the path that led to the maximum value.
  • Space-optimized LCS: Can be solved with O(min(m, n)) or O(n) space by noticing that dp[i][j] only depends on dp[i-1][j-1], dp[i-1][j], and dp[i][j-1]. This means you only need to store the current row and the previous row of the DP table.

6. Code Quality

The code quality is generally good:

• Variable Naming: m, n, dp, text1Char, text2Char are all clear and standard.
• Code Structure: The initialization of the dp table, the nested loops, and the final return statement are well-structured and easy to follow.
• Readability: The comments for time and space complexity are helpful. The logic within the loops directly reflects the recurrence relation, making it readable.
• TypeScript Best Practices: Using export for the function is good for testability. The array initialization new Array(n + 1).fill(0) is concise and correct.

7. Alternative Approaches

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

   • Approach: Define a recursive function solve(i, j) that returns the LCS length for text1[0...i-1] and text2[0...j-1]. This function would have the same base cases and recurrence relation as the iterative DP. To avoid redundant computations, a memoization table (e.g., a 2D array or a Map) would store results for (i, j) pairs.
   • Trade-offs:
     • Pros: Often more intuitive to write as it directly translates the recursive definition. Can be more space-efficient in cases where not all subproblems are visited (though for LCS, all are typically visited).
     • Cons: Can lead to stack overflow issues for very large inputs due to deep recursion (though constraints here 1000 might be acceptable). Performance might be slightly slower due to function call overhead. Same O(m*n) time and space complexity.

2. Space-Optimized Dynamic Programming:

   • Approach: Notice that dp[i][j] only depends on values from the previous row (i-1) and the current row (i) up to j-1. We can reduce the O(m*n) space to O(n) (or O(min(m, n))) by only keeping track of two rows: prev_dp and current_dp.
   • Trade-offs:
     • Pros: Significantly reduces space complexity, which can be critical for problems with larger constraints where m*n space is too much.
     • Cons: Can be slightly trickier to implement correctly, especially when managing the dp[i-1][j-1] term (which requires storing a "diagonal" value from the prev_dp row before it gets overwritten). Time complexity remains O(m*n).

[github-actions]

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

Here's a detailed review of the provided solution for the "Longest Common Subsequence" problem:

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1. Pattern Identification

This solution utilizes the Dynamic Programming (2D DP) pattern. It's a classic problem that demonstrates how to build up a solution to a larger problem by solving smaller, overlapping subproblems and storing their results in a table.

2. Complexity Verification

• Stated Time Complexity: O(m * n)
  • Verification: Accurate. The solution uses two nested loops. The outer loop iterates m times (where m is text1.length), and the inner loop iterates n times (where n is text2.length). Inside the loops, operations like character comparison, array access, and Math.max are all constant time. Therefore, the total time complexity is proportional to m * n.
• Stated Space Complexity: O(m * n)
  • Verification: Accurate. A 2D array dp of size (m + 1) x (n + 1) is created to store the results of subproblems. The memory required for this table is directly proportional to m * n.

3. Key Insight

The key insight for solving the Longest Common Subsequence problem with dynamic programming lies in defining a recurrence relation based on whether the current characters of the two strings match or not.

Let dp[i][j] represent the length of the Longest Common Subsequence of text1[0...i-1] and text2[0...j-1].

1. Base Cases: If either i or j is 0 (meaning one of the strings is empty), the LCS length is 0. This is naturally handled by initializing the first row and column of the dp table to 0.

2. Recursive Step:

   • If text1[i-1] (the i-th character of text1) is equal to text2[j-1] (the j-th character of text2):
     If the current characters match, they can be part of the LCS. So, the LCS length for these prefixes is 1 plus the LCS length of the previous prefixes (i.e., text1[0...i-2] and text2[0...j-2]).
     dp[i][j] = dp[i-1][j-1] + 1

   • If text1[i-1] is NOT equal to text2[j-1]:
     If the current characters do not match, we cannot include both in the LCS. We must consider two possibilities and take the maximum:

     1. The LCS of text1[0...i-1] and text2[0...j-2] (ignoring text2[j-1]). This is dp[i][j-1].
     2. The LCS of text1[0...i-2] and text2[0...j-1] (ignoring text1[i-1]). This is dp[i-1][j].
        dp[i][j] = Math.max(dp[i][j-1], dp[i-1][j])

By filling this dp table iteratively from top-left to bottom-right, dp[m][n] will contain the length of the LCS for the entire text1 and text2.

4. Edge Case Analysis

The provided test suite covers a good range of edge cases:

• Empty strings: text1 = "", text2 = "abc"; text1 = "abc", text2 = ""; text1 = "", text2 = ""
  • The solution correctly handles these because the dp table is initialized with zeros, and the loops start from i=1 and j=1. If m or n is 0, the loops won't run, and dp[0][0] (which is 0) will effectively be returned.
• Single characters: text1 = "a", text2 = "a"; text1 = "a", text2 = "b"
  • These are handled correctly by the recurrence relation.
• No common characters: text1 = "abc", text2 = "def"
  • The Math.max condition will always apply, leading to dp[m][n] being 0.
• Identical strings: text1 = "abc", text2 = "abc"
  • The matching condition will apply for all characters, correctly summing up to the string length.
• One string is a subsequence of another: text1 = "abcde", text2 = "abc"
  • Handled correctly.
• Repeated characters: text1 = "aaa", text2 = "aa"
  • The character-by-character comparison and dp table logic correctly account for repetitions.

The constraints 1 <= text1.length, text2.length <= 1000 indicate that the lengths are non-zero (except for the empty string edge cases you added in tests, which is good practice). The lowercase English characters constraint simplifies character comparison. The solution is robust for these.

5. Learning Points

• Similar Problems using this pattern:
  • Longest Common Substring: Similar, but requires contiguous characters. The dp[i][j] would reset to 0 if characters don't match.
  • Edit Distance (Levenshtein Distance): Calculates the minimum number of operations (insertions, deletions, substitutions) to transform one string into another. Uses a very similar 2D DP table and recurrence relation.
  • Longest Palindromic Subsequence: Find the longest subsequence that reads the same forwards and backwards. Can be solved by finding the LCS of the original string and its reverse.
  • Shortest Common Supersequence: Find the shortest string that has both given strings as subsequences. Its length is len(text1) + len(text2) - len(LCS).
  • Minimum Deletions/Insertions to Make Two Strings Equal: Directly derivable from LCS.
• Common Mistakes with this pattern:
  • Greedy Approach: A common initial mistake, as noted in your POST_MORTEM_TEMPLATE, is to try a greedy approach (e.g., matching the first available character). This fails because local optimal choices don't guarantee global optimality for subsequences. Dynamic Programming systematically explores all possibilities.
  • Off-by-One Errors: Incorrectly mapping string indices to dp table indices (e.g., using text1[i] when the dp table is (m+1)x(n+1) and i refers to the dp table row/column). Your solution correctly uses text1[i-1] and text2[j-1] for dp[i][j].
  • Incorrect Base Cases: Not properly initializing the first row/column of the dp table or handling empty string scenarios.
• Variations of this problem:
  • Printing the LCS: Instead of just returning the length, reconstruct the actual subsequence by backtracking through the dp table from dp[m][n].
  • Space Optimization: Reduce the space complexity from O(m*n) to O(min(m, n)) by observing that dp[i][j] only depends on values from the current and previous rows (or columns). This requires careful updating of two 1D arrays.

6. Code Quality

The code quality is excellent:

• Variable Naming: text1, text2, m, n, dp are standard and clear.
• Code Structure: The function is well-defined, with clear initialization and iterative loops.
• Readability: The logic for filling the dp table is straightforward and easy to follow, directly reflecting the recurrence relation.
• Comments: The @time and @space complexity comments are very helpful.
• Export: Using export function is good for modularity and testing in a TypeScript/JavaScript project.
• TypeScript: Type annotations (: string, : number, [][]) are correctly used.

7. Alternative Approaches

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

   • Approach: Instead of building the dp table iteratively from base cases, define a recursive function solve(i, j) that calculates LCS(text1[i...], text2[j...]). This function would have base cases for i or j reaching the end of the string. To avoid redundant calculations, a 2D memoization table would store the results of solve(i, j) calls.
   • Trade-offs: Can be more intuitive to write initially as it directly maps to the recursive definition. However, it might incur overhead from function calls and can lead to stack overflow errors for very large inputs in some languages/environments (though m, n <= 1000 is usually fine for JavaScript's stack). Time and space complexity remain O(m*n).

2. Space-Optimized Iterative DP (O(min(m, n)) space):

   • Approach: This optimization observes that to calculate the current row dp[i], we only need the values from the previous row dp[i-1] and the current row's already computed values dp[i][j-1]. We can reduce the O(m*n) space to O(min(m,n)) by using only two 1D arrays (one for the "previous" row and one for the "current" row). If implemented carefully, it can even be done with a single 1D array.
   • Trade-offs: Significantly reduces space complexity, which can be crucial for problems with much larger constraints. However, the implementation can be slightly more complex and less intuitive to get right, especially with the indexing for the "diagonal" dp[i-1][j-1] term. Time complexity remains O(m*n).

The current solution is the most common and clear iterative bottom-up DP approach, striking a good balance between optimality and readability.

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Actionable comments posted: 0

🧹 Nitpick comments (1)

leetcode/medium/1250-longest-common-subsequence/POST_MORTEM_TEMPLATE.md (1)

24-24: Minor grammar improvement: hyphenate compound adjective.

The phrase "greedy 2 pointer approach" would be more standard as "greedy two-pointer approach" (with hyphen and spelled-out number when used as an adjective).

🔎 Suggested improvement

-- I first attempted using a greedy 2 pointer approach, but the solution was not going to work because it would miss subsequences.
+- I first attempted using a greedy two-pointer approach, but the solution was not going to work because it would miss subsequences.

📜 Review details

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Review profile: CHILL

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Reviewing files that changed from the base of the PR and between a778934 and 93645d9.

📒 Files selected for processing (6)

• 1250-longest-common-subsequence/README.md
• README.md
• leetcode/medium/1250-longest-common-subsequence/POST_MORTEM_TEMPLATE.md
• leetcode/medium/1250-longest-common-subsequence/README.md
• leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.test.ts
• leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.ts

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📓 Path-based instructions (5)

**/*.ts

📄 CodeRabbit inference engine (AGENTS.md)

**/*.ts: Use explicit type annotations for function parameters and return types in TypeScript solutions
Follow strict TypeScript configuration with ES2020 target and CommonJS modules in TypeScript solutions
Enforce full type safety in TypeScript solutions

Files:

• leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.test.ts
• leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.ts

**/*.{js,ts}

📄 CodeRabbit inference engine (AGENTS.md)

**/*.{js,ts}: Use clear, descriptive variable names across all solutions
Add comments for non-obvious logic or algorithm steps in code
Keep functions focused and single-purpose across all solutions
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

**/*.{js,ts}: Use clear, descriptive variable names (not just i, j, k unless in simple loops) and add comments for non-obvious logic or algorithm steps
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
Use modern ES6+ syntax (arrow functions, destructuring, spread operators) where appropriate
Prefer built-in methods (.map(), .filter(), .reduce()) when appropriate and use strict equality (===) over loose equality (==)

Files:

• leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.test.ts
• leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.ts

**/*.{ts,tsx}

📄 CodeRabbit inference engine (.github/copilot-instructions.md)

**/*.{ts,tsx}: Use explicit type annotations for function parameters and return types in TypeScript
Follow strict TypeScript configuration as defined in tsconfig.json with ES2020 target and CommonJS modules

Files:

• leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.test.ts
• leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.ts

**/*.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:

• README.md
• leetcode/medium/1250-longest-common-subsequence/POST_MORTEM_TEMPLATE.md
• leetcode/medium/1250-longest-common-subsequence/README.md

**/README.md

📄 CodeRabbit inference engine (CLAUDE.md)

Each problem directory should include a README.md containing: problem title and difficulty badge, link to the problem on the respective platform, problem description, examples and constraints

Files:

• README.md
• leetcode/medium/1250-longest-common-subsequence/README.md

🧠 Learnings (2)

📚 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:

• README.md
• leetcode/medium/1250-longest-common-subsequence/README.md

📚 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]*-*/*.{js,ts} : Use standard algorithmic patterns (two pointers, sliding window, hash maps, DFS, BFS, DP) and leverage appropriate data structures (Map, Set, arrays)

Applied to files:

• README.md

🧬 Code graph analysis (1)

leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.test.ts (1)

leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.ts (1)

• longestCommonSubsequence (5-29)

🪛 LanguageTool

leetcode/medium/1250-longest-common-subsequence/POST_MORTEM_TEMPLATE.md

[grammar] ~24-~24: Use a hyphen to join words.
Context: ...s? - I first attempted using a greedy 2 pointer approach, but the solution was n...

(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)

• GitHub Check: Agent
• GitHub Check: review

🔇 Additional comments (6)

leetcode/medium/1250-longest-common-subsequence/README.md (1)

1-40: Excellent documentation structure!

The README follows all coding guidelines perfectly, including the problem title with difficulty badge, comprehensive problem description with subsequence definitions, three detailed examples with inputs/outputs/explanations, and clear constraints. Well done!

leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.ts (2)

1-4: Accurate complexity analysis!

The time and space complexity annotations are correct. The dynamic programming solution fills an (m+1) × (n+1) table, resulting in O(m*n) time and space complexity.

---

5-29: Clean and correct dynamic programming implementation!

The algorithm correctly implements the classic LCS solution:

• Proper DP table initialization with base cases
• Correct indexing (1-based DP indices mapping to 0-based string indices)
• Accurate recurrence relation for character matches and mismatches
• Clear comments explaining the logic at each step

The code follows TypeScript best practices with explicit type annotations and uses modern ES6+ syntax appropriately.

leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.test.ts (1)

1-64: Excellent comprehensive test coverage!

The test suite thoroughly covers:

• ✅ All three official examples from the problem statement
• ✅ Edge cases: empty strings (both, first, second)
• ✅ Boundary cases: single matching/non-matching characters
• ✅ Various scenarios: partial overlap, repeated characters, completely different strings, substring relationships, reversed strings
• ✅ Complex cases: longer strings with multiple possible subsequences

All test expectations are correct, and the test descriptions clearly communicate what is being validated. Great work ensuring the solution is well-tested!

README.md (1)

9-9: Accurate documentation updates!

All counts have been correctly updated to reflect the new problem:

• Total LeetCode problems: 166 → 167 ✓
• Medium problems: 72 → 73 ✓
• Section headers updated consistently ✓
• New problem entry properly placed in the Dynamic Programming section with correct formatting and badge ✓

The documentation changes accurately track the addition of problem 1250.

Also applies to: 21-21, 185-185, 243-243

leetcode/medium/1250-longest-common-subsequence/POST_MORTEM_TEMPLATE.md (1)

1-67: Valuable post-mortem documentation!

The post-mortem template provides excellent reflection on the problem-solving process, documenting the transition from an initial greedy approach to the optimal dynamic programming solution. This kind of retrospective is invaluable for learning and pattern recognition.

[Copilot]

Pull request overview

This PR adds a solution for the LeetCode "Longest Common Subsequence" problem with a TypeScript implementation using dynamic programming, comprehensive unit tests using Vitest, and supporting documentation including a README and post-mortem template.

Key changes:

• Adds a TypeScript solution with O(m*n) time and space complexity using a 2D DP table
• Includes 14 comprehensive test cases covering edge cases (empty strings, single characters, no matches, etc.)
• Provides problem documentation with examples and constraints
• Includes a detailed post-mortem reflection on the problem-solving approach
• Updates main README to increment problem count and add link to new solution
• Removes incorrectly placed directory from repository root

Reviewed changes

Copilot reviewed 6 out of 6 changed files in this pull request and generated 4 comments.

Show a summary per file

File	Description
leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.ts	Adds the main solution function with export keyword for testing; implements classic DP approach with 2D memoization table
leetcode/medium/1250-longest-common-subsequence/longest-common-subsequence.test.ts	Adds comprehensive test suite with 14 test cases covering LeetCode examples, edge cases, and various string patterns
leetcode/medium/1250-longest-common-subsequence/README.md	Provides problem statement, examples, and constraints from LeetCode
leetcode/medium/1250-longest-common-subsequence/POST_MORTEM_TEMPLATE.md	Documents the problem-solving journey, time tracking, solution exploration, and self-assessment
README.md	Updates problem counts (total and medium difficulty) and adds link to new problem in Dynamic Programming section
1250-longest-common-subsequence/README.md	Removes old incorrectly placed directory from repository root

[Copilot]

The diff shows this line changed from "71 solved" to "73 solved", which is an increment of 2. However, this PR only adds 1 new medium problem (1250 - Longest Common Subsequence). The count should change from 72 to 73, not from 71 to 73. Please verify the previous count was actually 72, not 71.

[Copilot]

Incorrect problem number. The LeetCode problem "Longest Common Subsequence" is problem number 1143, not 1250. This is confirmed by checking the LeetCode URL in the README (https://leetcode.com/problems/longest-common-subsequence/) and other references in the codebase (see docs/topics/ai-llm.md:13). The directory should be named "1143-longest-common-subsequence" and this link should reference problem 1143.

Suggested change

	- [1250 - Longest Common Subsequence](./leetcode/medium/1250-longest-common-subsequence) 
	- [1143 - Longest Common Subsequence](./leetcode/medium/1143-longest-common-subsequence) 

[Copilot]

Incorrect problem number in URL. The correct LeetCode problem number for "Longest Common Subsequence" is 1143, not 1250. The URL should be https://leetcode.com/problems/longest-common-subsequence/ which is problem 1143. The entire directory structure should use 1143 instead of 1250.

Suggested change

	# [Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence) 
	# [Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/) 

[Copilot]

Incorrect problem number in URL. The correct LeetCode problem number for "Longest Common Subsequence" is 1143, not 1250. The URL should be https://leetcode.com/problems/longest-common-subsequence/ which corresponds to problem 1143.

[github-actions]

📅 Spaced Repetition Reviews Scheduled!

Great job solving #1250 - Longest Common Subsequence! 🎉

To help you retain this knowledge long-term, I've scheduled 5 review sessions using spaced repetition:

Review	Interval	Schedule Logic
1st Review	1 day after solving	Scheduled now
2nd Review	3 days after 1st review	Auto-scheduled when 1st completes
3rd Review	7 days after 2nd review	Auto-scheduled when 2nd completes
4th Review	14 days after 3rd review	Auto-scheduled when 3rd completes
5th Review	30 days after 4th review	Auto-scheduled when 4th completes

What to expect:

• Your 1st review is scheduled for tomorrow
• Each subsequent review is scheduled automatically when you complete the previous one
• This ensures proper spacing even if you complete a review a few days late
• GitHub issues will be created automatically for each review
• Each issue will link back to your solution

🧠 Why Spaced Repetition?

Research shows that reviewing material at increasing intervals dramatically improves retention. This adaptive scheduling ensures optimal spacing based on when you actually complete each review!

Check docs/reviews/review-schedule.json to see your review schedule.