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
- Create a table with one extra row for using zero items.
- Process the four items in fixed order.
- For each capacity, inherit when the item is too heavy.
- Otherwise compare
skipandtakefrom the previous row. - 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.
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";
}
@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