LeetCode 1143 Longest Common Subsequence - Transition

At position $(i, j)$, compare A[i-1] and B[j-1]. If they match: $dp[i][j] = dp[i-1][j-1] + 1$. Found a common character, extend by 1. If they don't match: $dp[i][j] = \max(dp[i-1][j], dp[i][j-1])$. Skip a character from A or B, keep the better result. The final answer is $dp[m][n]$ where $m$ and $n$ are the lengths.

This formula tells you how to compute dp[i][j] from earlier dp values. Getting this right is the core of DP (dynamic programming).