Recursion and Dynamic Programming

0/1 Knapsack (Small)

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.

TypeScript DSA Implementation

basic.ts 
function rowString(row: number[]): string { return `[${row.join(", ")}]`; }
function tableString(table: number[][]): string { return `[${table.map(rowString).join(", ")}]`; }
const weights = [2, 3, 4, 5];
const values = [3, 4, 5, 6];
const capacity = 5;
const dp = Array.from({ length: weights.length + 1 }, () => Array(capacity + 1).fill(0));
for (let item = 1; item <= weights.length; item++) {
  const weight = weights[item - 1];
  const value = values[item - 1];
  for (let cap = 0; cap <= capacity; cap++) {
    if (weight > cap) dp[item][cap] = dp[item - 1][cap];
    else {
      const skip = dp[item - 1][cap];
      const take = value + dp[item - 1][cap - weight];
      dp[item][cap] = Math.max(skip, take);
    }
  }
}
console.log(dp[weights.length][capacity]);
console.log(tableString(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