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.
state transition `dp[i][w]` compares skipping item `i` with taking it and reading the remaining capacity from the previous row.

Complexity

  • Time: O(item_count * capacity)
  • Space: O(item_count * capacity)

Visual walkthrough

The pinned items have weights [2, 3, 4, 5], values [3, 4, 5, 6], and capacity 5. The diagrams show each item row, the overweight inherit branch, and the skip-versus-take comparison that produces 7.

Step 1 - Zero items baseline

With no items available, every capacity has value 0.

Row 0: the empty knapsack is worth zero at every capacity.rowcap 0cap 1cap 2cap 3cap 4cap 50 items000000

Step 2 - First item fills a row

Item 1 has weight 2 and value 3, so capacities below 2 inherit 0 and capacities 2 through 5 can take it.

Row 1 after processing item 1 (weight 2, value 3).rowcap 0cap 1cap 2cap 3cap 4cap 5item 1003333choiceinheritinherittaketaketaketake

Step 3 - Overweight capacity inherits

For item 2 (weight 3, value 4) at capacity 2, weight 3 is too large, so the cell inherits skip = 3 from the previous row.

The overweight branch does not read a negative-capacity cell.decisionitemcapskiptakeresultinherit2 (w=3, v=4)23too heavy3row 2 cap 2w > cap2row 1—inherit

Step 4 - Skip or take at capacity 5

At capacity 5 for item 2, skip = 3 while take = 4 + dp[1][2] = 7, so max(skip, take) is 7.

Item 2 improves capacity 5 by taking the item and using the previous row at capacity 2.cellitemcurrent capskip: prev row cap 5take: prev row cap - weight 2resultcompare2 (w=3, v=4)534 + 3 = 77readprevious rowcap 5dp[1][5]dp[1][2] (5 - 3)max

Step 5 - Final processed row reaches value 7

After processing item 4, the final DP row is [0, 0, 3, 4, 5, 7]; item 4 is only the last processed row, not a reconstructed selected-item set.

Complete 0/1 knapsack table for weights [2, 3, 4, 5] and values [3, 4, 5, 6]; the final row means all four items have been processed.rowcap 0cap 1cap 2cap 3cap 4cap 50 items000000item 1003333item 2003447item 3003457item 4 (final processed row)003457

C++ DSA Implementation

basic.cpp 
#include <algorithm>
#include <iostream>
#include <sstream>
#include <vector>
using namespace std;

string row_string(const vector<int>& row) {
  ostringstream out;
  out << "[";
  for (size_t i = 0; i < row.size(); i++) {
    if (i) out << ", ";
    out << row[i];
  }
  out << "]";
  return out.str();
}
string table_string(const vector<vector<int>>& table) {
  ostringstream out;
  out << "[";
  for (size_t i = 0; i < table.size(); i++) {
    if (i) out << ", ";
    out << row_string(table[i]);
  }
  out << "]";
  return out.str();
}
int main() {
  vector<int> weights = {2, 3, 4, 5};
  vector<int> values = {3, 4, 5, 6};
  int capacity = 5;
  vector<vector<int>> dp(weights.size() + 1, vector<int>(capacity + 1, 0));
  for (int item = 1; item <= (int)weights.size(); item++) {
    int weight = weights[item - 1];
    int value = values[item - 1];
    for (int cap = 0; cap <= capacity; cap++) {
      if (weight > cap) dp[item][cap] = dp[item - 1][cap];
      else dp[item][cap] = max(dp[item - 1][cap], value + dp[item - 1][cap - weight]);
    }
  }
  cout << dp[weights.size()][capacity] << "\n" << table_string(dp) << "\n";
}

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]]

Implementation notes

  • In C++, weights and values are parallel std::vector<int> objects: {2, 3, 4, 5} and {3, 4, 5, 6}.
  • The DP table is std::vector<std::vector<int>> dp(weights.size() + 1, std::vector<int>(capacity + 1, 0)), giving five independent rows and six columns for capacity 0..5.
  • The outer loop casts weights.size() to int and scans item = 1..4; weight and value are copied from item - 1, then the inner loop scans cap = 0..5.
  • Each cell writes into dp[item][cap]: either a copy from the previous row when weight > cap, or max(dp[item - 1][cap], value + dp[item - 1][cap - weight]).
  • The trace records row transitions [0, 0, 0, 0, 0, 0], [0, 0, 3, 3, 3, 3], [0, 0, 3, 4, 4, 7], [0, 0, 3, 4, 5, 7], and [0, 0, 3, 4, 5, 7].
  • table_string(const std::vector<std::vector<int>>& table) reads rows by const reference and calls row_string(table[i]); both formatting helpers use std::ostringstream before std::cout prints 7 and then the full table. Visible allocation is the weights, values, table rows, and formatting buffers; visible mutation is confined to DP cells.