diff --git a/Rod Cutting - DP b/Rod Cutting - DP
new file mode 100644
index 0000000..914844c
--- /dev/null
+++ b/Rod Cutting - DP	
@@ -0,0 +1,70 @@
+// CPP program for above approach
+#include <iostream>
+using namespace std;
+
+// Global Array for
+// the purpose of memoization.
+int t[9][9];
+
+// A recursive program, using ,
+// memoization, to implement the
+// rod cutting problem(Top-Down).
+int un_kp(int price[], int length[],
+					int Max_len, int n)
+{
+
+	// The maximum price will be zero,
+	// when either the length of the rod
+	// is zero or price is zero.
+	if (n == 0 || Max_len == 0)
+	{
+		return 0;
+	}
+
+	// If the length of the rod is less
+	// than the maximum length, Max_lene will
+	// consider it.Now depending
+	// upon the profit,
+	// either Max_lene we will take
+	// it or discard it.
+	if (length[n - 1] <= Max_len)
+	{
+		t[n][Max_len]
+			= max(price[n - 1]
+					+ un_kp(price, length,
+						Max_len - length[n - 1], n),
+				un_kp(price, length, Max_len, n - 1));
+	}
+
+	// If the length of the rod is
+	// greater than the permitted size,
+	// Max_len we will not consider it.
+	else
+	{
+		t[n][Max_len]
+			= un_kp(price, length,
+							Max_len, n - 1);
+	}
+
+	// Max_lene Max_lenill return the maximum
+	// value obtained, Max_lenhich is present
+	// at the nth roMax_len and Max_lenth column.
+	return t[n][Max_len];
+}
+
+/* Driver program to
+test above functions */
+int main()
+{
+	int price[] = { 1, 5, 8, 9, 10, 17, 17, 20 };
+	int n = sizeof(price) / sizeof(price[0]);
+	int length[n];
+	for (int i = 0; i < n; i++) {
+		length[i] = i + 1;
+	}
+	int Max_len = n;
+
+	// Function Call
+	cout << "Maximum obtained value is "
+		<< un_kp(price, length, n, Max_len) << endl;
+}