KnapsackRecursion.cpp

// 0/1 knapsack problem 
//I/P : wt[]: 1 3 4 5 and val[]: 1 4 5 7 and W: 7
// output: maximum profit
// recursive apporach
/* A Naive recursive implementation of 0-1 Knapsack problem */
#include <bits/stdc++.h>
using namespace std;

// A utility function that returns maximum of two integers
// int max(int a, int b) { return (a > b) ? a : b; }

// Returns the maximum value that can be put in a knapsack of capacity W
int knapSack(int W, int wt[], int val[], int n)
{

	// Base Case
	if (n == 0 || W == 0)
		return 0;

	// If weight of the nth item is more than Knapsack capacity W, then this item cannot be included in the optimal solution
	if (wt[n - 1] > W)
		return knapSack(W, wt, val, n - 1);

	// Return the maximum of two cases:
	// (1) nth item included
	// (2) not included
	else
		return max(val[n - 1]+ knapSack(W - wt[n - 1],wt, val, n - 1),knapSack(W, wt, val, n - 1));
}

// Driver code
int main()
{
	int val[] = { 60, 100, 120 };
	int wt[] = { 10, 20, 30 };
	int W = 50;
	int n = sizeof(val) / sizeof(val[0]);
	cout << knapSack(W, wt, val, n);
	return 0;
}