Recursive solution for 0/1 knapsackProblem

KnapsackRecursion.cpp

@@ -0,0 +1,40 @@
+// 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;
+}