diff --git a/src/algorithm_practice/Dynamic Programming/knapsack.cpp b/src/algorithm_practice/Dynamic Programming/knapsack.cpp
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+
+/*
+This is a implementation of the general (0,1) knapsack problem
+
+Problem statement:Given a set of items, each with a weight and a value, 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.
+
+Code:-
+To consider all subsets of items, there can be two cases for every item: (1) the item is included in the optimal subset, (2) not included in the optimal set.
+Therefore, the maximum value that can be obtained from n items is max of following two values.
+1) Maximum value obtained by n-1 items and W weight (excluding nth item).
+2) Value of nth item plus maximum value obtained by n-1 items and W minus weight of the nth item (including nth item).
+
+We exploit this property for knapsack problem and build a 2-D array for storing value corresponding to certain weight and no of items
+
+Time Complexity: O(nW) where n is the number of items and W is the capacity of knapsack.
+
+
+
+*/
+
+
+
+
+
+
+
+
+#include<bits/stdc++.h>
+
+using namespace std;
+
+//returns maximum of two numbers
+int maxm(int a, int b) {
+    return (a > b)? a : b; 
+}
+
+
+//a DP implementation of the famous knapsack problem 
+/*
+Each element of the array k stores the optimal value for the corresponding weight size and number of items.
+We see that the relation comes out to be (from the initial discussion):
+
+k[i][j]=maxm(k[i-1][j-w[i-1]]+v[i-1],k[i-1][j]);
+
+i.e >If you take the last weight find and optimal solution as if the weight didnt exist and If you dont the optimal
+solution is just k[i-1][j].
+*/
+int knapsack(int v[],int w[],int n,int wmax){
+    
+    int k[n+1][wmax+1];
+    
+
+    for(int i=0;i<=n;i++)
+        {
+            for(int j=0;j<=wmax;j++)
+                {
+                if(i==0||j==0)
+                    k[i][j]=0;
+                    else if(w[i-1]<=j){
+                           k[i][j]=maxm(k[i-1][j-w[i-1]]+v[i-1],k[i-1][j]);
+                        }
+                        else{
+                        k[i][j]=k[i-1][j];
+                        }
+
+                }
+        }
+
+return k[n][wmax];
+
+}
+
+int main(){
+
+    int n;
+    //Enter size of the input array
+    cout<<"Enter the size:";
+    cin>>n;
+    int v[n];
+    int w[n];
+    int wmax;
+    //  Enter the values
+    for(int i =0;i<n;i++)
+        cin>>v[i];
+
+    //Enter the weights for each of the value
+    for(int i =0;i<n;i++)
+        cin>>w[i];
+
+    cin>>wmax;
+
+        //Knapsack solution
+    cout<<knapsack(v,w,n,wmax)<<endl;
+
+    }