Fill a small 0/1 knapsack table where each row decides whether one more item is available.

Algorithm

@steps

  1. Create a table with one extra row for using zero items.
  2. Process the four items in fixed order.
  3. For each capacity, inherit when the item is too heavy.
  4. Otherwise compare skip and take from the previous row.
  5. Print the best value and the full deterministic table. @end @complexity
  • Time: O(item_count * capacity)
  • Space: O(item_count * capacity) @end
state transition `dp[i][w]` compares skipping item `i` with taking it and reading the remaining capacity from the previous row.

Ruby DSA Implementation

basic.rb
def row_string(row)
  "[" + row.join(", ") + "]"
end
def table_string(table)
  "[" + table.map { |row| row_string(row) }.join(", ") + "]"
end

weights = [2, 3, 4, 5]
values = [3, 4, 5, 6]
capacity = 5
dp = Array.new(weights.length + 1) { Array.new(capacity + 1, 0) }
(1..weights.length).each do |item|
  weight = weights[item - 1]
  value = values[item - 1]
  (0..capacity).each do |cap|
    if weight > cap
      dp[item][cap] = dp[item - 1][cap]
    else
      skip = dp[item - 1][cap]
      take = value + dp[item - 1][cap - weight]
      dp[item][cap] = [skip, take].max
    end
  end
end
puts dp[weights.length][capacity]
puts table_string(dp)

@end @output 7 [[0, 0, 0, 0, 0, 0], [0, 0, 3, 3, 3, 3], [0, 0, 3, 4, 4, 7], [0, 0, 3, 4, 5, 7], [0, 0, 3, 4, 5, 7]] @end