0/1 Knapsack

/*
 * Problem : Maximum value in the bag.
 * Input   : Three lines: n and W, weights, values.
 * Output  : The maximum value.
 * Approach: Take or leave each item.
 * Time    : O(n * W)
 * Space   : O(W)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static int solve(String[] lines) {
        String[] header = lines[0].split(" ");
        int n = Integer.parseInt(header[0]);
        int capacity = Integer.parseInt(header[1]);
        String[] weightText = lines[1].split(" ");
        String[] valueText = lines[2].split(" ");

        // Step 1: best[c] = the best value for a bag with the capacity c. At the start the bag is empty, so all values are 0.
        int[] best = new int[capacity + 1];

        for (int i = 0; i < n; i++) {
            int weight = Integer.parseInt(weightText[i]);
            int value = Integer.parseInt(valueText[i]);

            // Step 2: Go from the largest capacity to the smallest. This way, every item is used at most once.
            for (int c = capacity; c >= weight; c--) {
                // Step 3: Take the item (value plus the best for the remaining capacity) or leave it.
                best[c] = Math.max(best[c], best[c - weight] + value);
            }
        }
        return best[capacity];
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        java.util.List<String> all = new java.util.ArrayList<>();
        while (scanner.hasNextLine()) {
            String line = scanner.nextLine().trim();
            if (!line.isEmpty()) {
                all.add(line);
            }
        }
        String[] lines = all.toArray(new String[0]);
        System.out.println(solve(lines));
    }
}