Explain Longest Common Subsequence (LCS) for interviews.

Short answer: 2-D DP: if s[i]==t[j], dp[i][j] = dp[i-1][j-1]+1 else max(skip either char). Classic O(m*n) table; can compress to two rows for space.

Common follow-ups

• Print the LCS string
• Longest Common Substring (different recurrence)
• Edit Distance

Contrast subsequence (not contiguous) vs substring (contiguous) immediately.