MEDIUM NC#112 Blind #19 2-D Dynamic Programming
1143. Longest Common Subsequence
📖 Problem
Given two strings text1 and text2, return the length of their longest common subsequence.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •Array DP table updates
- •State transition thinking
- •Base case initialization
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Dynamic Programming reasoning pattern
💡 Approach
- → Use dp[i][j] = LCS length for text1[0:i] and text2[0:j+1]
- → If characters match: dp[i][j] = dp[i-1][j-1] + 1
- → If they don't: dp[i][j] = max(dp[i-1][j], dp[i][j-1])
- → Time: O(m * n), Space: O(m * n)
🧭 Prerequisites
🛠️ Hints & Pitfalls
Hints
- •Use dp[i][j] = LCS length for text1[0:i] and text2[0:j+1]
- •If characters match: dp[i][j] = dp[i-1][j-1] + 1
- •If they don't: dp[i][j] = max(dp[i-1][j], dp[i][j-1])
Common Pitfalls
- •Time: O(m * n), Space: O(m * n)
🧪 Test Cases
Test Case 1
Not run Input:
longestCommonSubsequence('abcde', 'ace'); Expected:
null Test Case 2
Not run Input:
longestCommonSubsequence('abc', 'abc'); Expected:
null Test Case 3
Not run Input:
longestCommonSubsequence(text1: string, text2: string); Expected:
Computed from hidden reference 📝 Code Editor
📤 Output