Unbounded Knapsack - Implementation

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

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

The inner loop tries adding each item to capacity $w$. Since items are unlimited, you use dp[w-w[i]] from the same row. This lets you pick the same item multiple times.

Compare this to 0/1 knapsack where you use dp[i-1][w-w[i]] to prevent reuse. The single-row approach naturally allows repetition.