0/1 Knapsack - Implementation

Here's the complete solution:

for i from 1 to n: for w from 0 to W: if w[i] <= w: dp[i][w] = max(dp[i-1][w], dp[i-1][w-w[i]] + v[i]) else: dp[i][w] = dp[i-1][w]
return dp[n][W]

Time: $O(n \cdot W)$. Space: $O(n \cdot W)$.

You can reduce space to $O(W)$ by noticing each row only depends on the previous row. Process weights from $W$ down to $0$ to avoid overwriting values you still need. The implementation follows directly from the transition formula. Each line of code corresponds to part of the mathematical formula.