# C++ Program to Solve Knapsack Problem Using Dynamic Programming

This is a C++ program to solve 0-1 knapsack problem using dynamic programming. In 0-1 knapsack problem, a set of items are given, each with a weight and a value. We need to determine the number of each item to include in a collection so that the total weight is less than or equal to the given limit and the total value is large as possible.

## Algorithm

Begin
Input set of items each with a weight and a value
Set knapsack capacity
Create a function that returns maximum of two integers.
Create a function which returns the maximum value that can be put in a knapsack of capacity W
int knapSack(int W, int w[], int v[], int n)
int i, wt;
int K[n + 1][W + 1]
for i = 0 to n
for wt = 0 to W
if (i == 0 or wt == 0)
Do K[i][wt] = 0
else if (w[i - 1] <= wt)
Compute: K[i][wt] = max(v[i - 1] + K[i - 1][wt - w[i - 1]], K[i -1][wt])
else
K[i][wt] = K[i - 1][wt]
return K[n][W]
Call the function and print.
End

## Example Code

#include <iostream>
using namespace std;
int max(int x, int y) {
return (x > y) ? x : y;
}
int knapSack(int W, int w[], int v[], int n) {
int i, wt;
int K[n + 1][W + 1];
for (i = 0; i <= n; i++) {
for (wt = 0; wt <= W; wt++) {
if (i == 0 || wt == 0)
K[i][wt] = 0;
else if (w[i - 1] <= wt)
K[i][wt] = max(v[i - 1] + K[i - 1][wt - w[i - 1]], K[i - 1][wt]);
else
K[i][wt] = K[i - 1][wt];
}
}
return K[n][W];
}
int main() {
cout << "Enter the number of items in a Knapsack:";
int n, W;
cin >> n;
int v[n], w[n];
for (int i = 0; i < n; i++) {
cout << "Enter value and weight for item " << i << ":";
cin >> v[i];
cin >> w[i];
}
cout << "Enter the capacity of knapsack";
cin >> W;
cout << knapSack(W, w, v, n);
return 0;
}

## Output

Enter the number of items in a Knapsack:4
Enter value and weight for item 0:10
50
Enter value and weight for item 1:20
60
Enter value and weight for item 2:30
70
Enter value and weight for item 3:40
90
Enter the capacity of knapsack100
40