0/1 Knapsack - Transition

For each item $i$ and capacity $w$, you have two choices:

$1.$ Skip item $i$: Keep the best value from items $1$ to $i-1$ with the same capacity. That's $dp[i-1][w]$.

$2.$ Take item $i$: Add $v_i$ to the best value from items $1$ to $i-1$ with reduced capacity $w - w_i$. That's $dp[i-1][w-w_i] + v_i$. You can only take item $i$ if it fits: $w_i \le w$. The transition (the formula to compute each state) is: $dp[i][w] = \max(dp[i-1][w], dp[i-1][w-w_i] + v_i)$ if $w_i \le w$, else $dp[i][w] = dp[i-1][w]$.