diff --git a/dynamic_programming/cut_rod.cpp b/dynamic_programming/cut_rod.cpp
index 9a905ba47..8dca5b8e7 100644
--- a/dynamic_programming/cut_rod.cpp
+++ b/dynamic_programming/cut_rod.cpp
@@ -8,6 +8,12 @@
  * the maximum profit obtainable by cutting up the rod and selling
  * the pieces.
  *
+ * ### Algorithm
+ * The idea is to break the given rod into every smaller piece as possible
+ * and then check profit for each piece, by calculating maximum profit for
+ * smaller pieces we will build the solution for larger pieces in bottom-up
+ * manner.
+ *
  * @author [Anmol](https://github.com/Anmol3299)
  * @author [Pardeep](https://github.com/Pardeep009)
  *
@@ -17,22 +23,29 @@
 #include <cassert>
 #include <climits>
 #include <iostream>
-
-template <size_t T>
-
 /**
- * This function cuts the rod in different pieces and stores the maximum profit
- * for each piece of the rod.
+ * @namespace dynamic_programming
+ * @brief Dynamic Programming algorithms
+ */
+namespace dynamic_programming {
+/**
+ * @namespace cut_rod
+ * @brief Implementation of cutting a rod problem
+ */
+namespace cut_rod {
+/**
+ * @brief cut_rod function cuts the rod in different pieces and stores the
+ * maximum profit for each piece of the rod.
  * @param n size of the rod in inches
  * @param price an array of prices that contains prices of all pieces of size<=n
  * @return maximum profit obtainable for @param n inch rod.
  */
-int cut_rod(const std::array<int, T> &price, const int n) {
-    // profit[i] will hold maximum profit for i inch rod
-    int *profit = new int[n + 1];
+template <size_t T>
+int maxProfitByCuttingRod(const std::array<int, T> &price, const int n) {
+    int *profit =
+        new int[n + 1];  // profit[i] will hold maximum profit for i inch rod
 
-    // if length of rod is zero, then no profit
-    profit[0] = 0;
+    profit[0] = 0;  // if length of rod is zero, then no profit
 
     // outer loop will select size of rod, starting from 1 inch to n inch rod.
     // inner loop will evaluate the maximum profit we can get for i inch rod by
@@ -46,8 +59,10 @@ int cut_rod(const std::array<int, T> &price, const int n) {
     }
     int ans = profit[n];
     delete[] profit;
-    return ans; /** return maximum profit obtainable for @param n inch rod */
+    return ans;  // returning maximum profit
 }
+}  // namespace cut_rod
+}  // namespace dynamic_programming
 
 /**
  * Function to test above algorithm
@@ -56,7 +71,8 @@ void test() {
     // Test 1
     const int n1 = 8;                                           // size of rod
     std::array<int, n1> price1 = {1, 5, 8, 9, 10, 17, 17, 20};  // price array
-    const int max_profit1 = cut_rod(price1, n1);
+    const int max_profit1 =
+        dynamic_programming::cut_rod::maxProfitByCuttingRod(price1, n1);
     const int expected_max_profit1 = 22;
     assert(max_profit1 == expected_max_profit1);
     std::cout << "Maximum profit with " << n1 << " inch road is " << max_profit1
@@ -68,7 +84,8 @@ void test() {
         1,  5,  8,  9,  10, 17, 17, 20, 24, 30,  // price array
         31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
         41, 42, 43, 44, 45, 46, 47, 48, 49, 50};
-    const int max_profit2 = cut_rod(price2, n2);
+    const int max_profit2 =
+        dynamic_programming::cut_rod::maxProfitByCuttingRod(price2, n2);
     const int expected_max_profit2 = 90;
     assert(max_profit2 == expected_max_profit2);
     std::cout << "Maximum profit with " << n2 << " inch road is " << max_profit2