diff --git a/0/1 Knapsack problem in CPP.cpp b/0/1 Knapsack problem in CPP.cpp
new file mode 100644
index 0000000..ed7d33d
--- /dev/null
+++ b/0/1 Knapsack problem in CPP.cpp	
@@ -0,0 +1,47 @@
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+// Structure to represent an item
+struct Item {
+    int weight;
+    int value;
+};
+
+// Function to solve the 0/1 Knapsack problem using dynamic programming
+int knapsack(vector<Item>& items, int capacity) {
+    int n = items.size();
+    // Create a 2D DP table to store the maximum value for each subproblem
+    vector<vector<int>> dp(n + 1, vector<int>(capacity + 1, 0));
+
+    // Fill the DP table using bottom-up approach
+    for (int i = 1; i <= n; ++i) {
+        for (int w = 1; w <= capacity; ++w) {
+            // If the current item's weight is greater than the current capacity,
+            // we can't include it in the knapsack
+            if (items[i - 1].weight > w) {
+                dp[i][w] = dp[i - 1][w];
+            } else {
+                // Otherwise, we have two choices: include the item or exclude it
+                // Take the maximum of these two choices
+                dp[i][w] = max(dp[i - 1][w], items[i - 1].value + dp[i - 1][w - items[i - 1].weight]);
+            }
+        }
+    }
+
+    // The maximum value is stored in dp[n][capacity]
+    return dp[n][capacity];
+}
+
+int main() {
+    // Example usage
+    int capacity = 10;
+    vector<Item> items = {{2, 6}, {2, 10}, {3, 12}, {5, 15}, {7, 22}};
+
+    int max_value = knapsack(items, capacity);
+
+    cout << "Maximum value that can be obtained: " << max_value << endl;
+
+    return 0;
+}