# 0-1 Knapsack Problem using C++ In 0-1 Knapsack problem, we are given a set of items, each with a weight and a value and 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 a given limit and the total value is as large as possible.

Please note that the items are indivisible; we can either take an item or not (0-1 property)

### Example Code :

#include <iostream>

#include <climits>

using namespace std;

// Values (stored in array v)

// Weights (stored in array w)

// Number of distinct items (n)

// Knapsack capacity W

int knapSack(int v[], int w[], int n, int W)

{

// base case: Negative capacity

if (W < 0)

return INT_MIN;

// base case: no items left or capacity becomes 0

if (n < 0 || W == 0)

return 0;

// Case 1. include current item n in knapSack (v[n]) and recur for

// remaining items (n – 1) with decreased capacity (W – w[n])

int include = v[n] + knapSack(v, w, n – 1, W – w[n]);

// Case 2. exclude current item n from knapSack and recur for

// remaining items (n – 1)

int exclude = knapSack(v, w, n – 1, W);

// return maximum value we get by including or excluding current item

return max (include, exclude);

}

// 0-1 Knapsack problem

int main()

{

// Input: set of items each with a weight and a value

int v[] = { 20, 5, 10, 40, 15, 25 };

int w[] = {  1, 2,  3,  8,  7, 4 };

// Knapsack capacity

int W = 10;

// number of items

int n = sizeof(v) / sizeof(v);

cout << “Knapsack value is ” << knapSack(v, w, n – 1, W);

return 0;

}

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