From 777bc0374dc9825933d440fb3416f17503e20655 Mon Sep 17 00:00:00 2001
From: Pardeep Bhatt <pardeepbhatt5254@gmail.com>
Date: Fri, 25 Sep 2020 16:14:02 +0530
Subject: [PATCH] added requested changes in the code

---
 dynamic_programming/cut_rod.cpp | 70 +++++++++++++++++++++++----------
 1 file changed, 49 insertions(+), 21 deletions(-)

diff --git a/dynamic_programming/cut_rod.cpp b/dynamic_programming/cut_rod.cpp
index 0d566cc8c..38b02926a 100644
--- a/dynamic_programming/cut_rod.cpp
+++ b/dynamic_programming/cut_rod.cpp
@@ -1,35 +1,63 @@
-/*Given a rod of length n inches and an array of prices that
-contains prices of all pieces of size smaller than n. Determine
-the maximum value obtainable by cutting up the rod and selling
-the pieces.*/
+/**
+ * @file cut_rod.cpp
+ * @brief Implementation of cutting a rod problem
+ *
+ * @details
+ * Given a rod of length n inches and an array of prices that
+ * contains prices of all pieces of size smaller than n. Determine
+ * the maximum profit obtainable by cutting up the rod and selling
+ * the pieces.
+ *
+ */
 
 #include <array>
 #include <climits>
 #include <iostream>
 
-using std::cout;
-
 template <size_t T>
 
-int cutrod(const std::array<int, T> &p, const int n) {
-    int *r = new int[n + 1];
-    r[0] = 0;
-    for (int j = 0; j < n; j++) {
+/**
+ * This 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];
+
+    // if length of rod is zero, then no profit
+    profit[0] = 0;
+
+    // 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
+    // making every possible cut on it and will store it in profit[i].
+    for (size_t i = 1; i <= n; i++) {
         int q = INT_MIN;
-        for (int i = 0; i <= j; i++) {
-            q = std::max(q, p[i] + r[j - i]);
+        for (size_t j = 1; j <= i; j++) {
+            q = std::max(q, price[j - 1] + profit[i - j]);
         }
-        r[j + 1] = q;
+        profit[i] = q;
     }
-    int ans = r[n];
-    delete[] r;
-    return ans;
+    int ans = profit[n];
+    delete[] profit;
+    return ans; /** return maximum profit obtainable for @param n inch rod */
 }
+
+/**
+ * @brief Main function
+ * @returns 0 on exit
+ */
 int main() {
-    const int n = 30;
-    std::array<int, n> price = {1,  5,  8,  9,  10, 17, 17, 20, 24, 30,
-                                31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
-                                41, 42, 43, 44, 45, 46, 47, 48, 49, 50};
-    cout << cutrod(price, n);
+    const int n = 30;  // size of rod
+    std::array<int, n> price = {
+        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};
+
+    // maximum profit
+    std::cout << cut_rod(price, n);
+
     return 0;
 }