dsa1 min read
Longest Common Subsequence — 2D DP Preview
LCS of two strings: dp[i][j] = LCS of first i chars of s1 and first j chars of s2. Full 2D DP treatment in the 2D DP section.
Read →
Lcs
6 articles
dsa4 min read
2D Dynamic Programming — Complete Guide
Master 2D DP: LCS, Edit Distance, grid path counting, matrix chain, interval DP, and stock patterns with 5-language templates.
Read →
dsa1 min read
Longest Common Subsequence — LCS 2D DP
Classic LCS: dp[i][j] = LCS length of s1[:i] and s2[:j]. If chars match add 1 to diagonal; else take max of skip either string.
Read →
dsa1 min read
Shortest Common Supersequence — LCS + Reconstruct
Find shortest string containing both strings as subsequences. Length = len(s1)+len(s2)-LCS. Reconstruct by tracing DP table.
Read →
dsa2 min read
Longest Common Substring — DP and Rolling Hash Approaches
Find the longest common substring between two strings. DP approach O(nm), binary search + rolling hash O((n+m) log(n+m)).
Read →
dsa2 min read
String DP — Edit Distance, LCS, and Interleaving
Core string DP patterns: edit distance (Wagner-Fischer), longest common subsequence, interleaving strings, and regular expression matching.
Read →