Step 1 — Subproblem: Define dp[w] = maximum value we can get with capacity w using the first i items (we update the same array for each i). For each item we either skip it (dp unchanged) or take it (value + best for capacity w - weight).

Step 2 — Initialization: dp[0..W] = 0. No items used, so max value is 0 for any capacity.

Step 3 — Loop: For each item i (weight, value), update dp from right to left (from W down to 0) so we don't use the same item twice: for w from W down to weight, dp[w] = max(dp[w], dp[w - weight] + value). Going backwards ensures we only use one copy of the current item.

Step 4 — Result: After processing all items, dp[capacity] is the maximum value we can get. To recover which items were chosen, backtrack: if dp[w] changed when we considered item i, we took it. Time O(n × W), space O(W).