diff --git a/O-1 Knapsack problem.cpp b/O-1 Knapsack problem.cpp
new file mode 100644
index 0000000..f689488
--- /dev/null
+++ b/O-1 Knapsack problem.cpp	
@@ -0,0 +1,72 @@
+#include<iostream>
+#include<vector>
+#include<algorithm>
+
+using namespace std;
+
+// Recursive approach to 0-1 Knapsack problem
+int Knapsack (int numitems, int capacity, vector<int> & weight, vector<int> & value) {
+
+    // No item can be put in the sack of capacity 0 so maximum value for sack of capcacity 0 is 0
+    if (capacity == 0)
+       return 0;
+
+    // If 0 items are put in the sack, then maximum value for sack is 0
+    if (numitems == 0)
+       return 0;
+
+    /* Note : Here the number of item is limited (unlike coin change / integer partition problem) 
+       hence the numitems -> (numitems - 1) when the item is included in the knapsack */
+    if (capacity >= weight[numitems-1])
+       return max (Knapsack (numitems-1, capacity, weight, value), // item not included in the sack
+                   Knapsack (numitems-1, capacity-weight[numitems-1], weight, value) + value[numitems-1]); // Item included.
+    else
+       return Knapsack(numitems-1, capacity, weight, value);
+}
+
+// Dynamic programming approach to 0-1 Knapsack problem
+int DPKnapsack (int capacity, vector<int> & weight, vector<int> & value) {
+
+    int numitems = weight.size();
+    int maxval[numitems+1][capacity+1];
+
+    // If 0 items are put in the sack of capacity 'cap' then maximum value for each sack is 0
+    for (int cap=0; cap<=capacity; cap++)
+        maxval[0][cap] = 0;
+
+    // No item can be put in the sack of capacity 0 so maximum value for sack of capcacity 0 is 0
+    for (int item=0; item<=numitems; item++)
+        maxval[item][0] = 0;
+
+    for (int item=1; item <= numitems; item++) {
+        for (int cap=1; cap <= capacity; cap++) {
+
+            /* Note : Here the number of item is limited (unlike coin change / integer partition problem) 
+               hence the numitems -> (numitems - 1) when the item is included in the knapsack */
+            if (cap >= weight[item-1]) {
+                maxval[item][cap] = max (maxval[item-1][cap], // Item not included in the knapsack 
+                                         maxval[item-1][cap-weight[item-1]] + value[item-1]); // Item included in the knapsack
+            } else {
+                maxval[item][cap] = maxval[item-1][cap];
+            }
+        }
+    }
+    return maxval[numitems][capacity];
+}
+
+int main() {
+
+    vector<vector<int>> weight = { {1, 3, 4, 6, 9},  {1, 2, 3, 5},     {3, 5},    {1, 2, 3, 4, 5} };
+    vector<vector<int>> value  = { {5, 10, 4, 6, 8}, {1, 19, 80, 100}, {80, 100}, {3, 5, 4, 8, 10} };
+    vector<int> capacity       = { 10, 6, 2, 5 };
+
+    int i = 0;
+    // 4 test cases
+    while ( i <= 3) {
+        cout << "\n[DP] 0-1 Knapsack max value : " << DPKnapsack ( capacity[i], weight[i], value[i] ) << endl;
+        cout << "[Recursion] 0-1 Knapsack max value : " << Knapsack ( weight[i].size(), capacity[i], weight[i], value[i] ) << endl;
+        i++;
+    }
+
+    return 0;
+}