chore(CPlusPlus): add rod cutting problem (#985)

* Rod cutting in cpp completed

* Update rod-cutting.cpp

algorithms/CPlusPlus/Dynamic-Programming/rod-cutting.cpp

@@ -0,0 +1,76 @@
+// Rod Cutting Problem
+// Given a rod of length n and a list of rod prices of length i, where 1 <= i <= n, find the optimal way to cut the rod into smaller rods to maximize profit.
+
+// Rod Cutting Optimal Approach
+//     We will solve this problem in a bottom-up manner. (iteratively)  
+//     In the bottom-up approach, we solve smaller subproblems first, then move on to larger subproblems.  
+//     The following bottom-up approach computes dp[i], which stores maximum profit achieved from the rod of length i from 1 to len. 
+//     It uses the value of smaller values i already computed.
+
+// Space complexity: O(n)
+// Time complexity: O(n^n)
+
+
+// Solution
+#include <iostream>
+#include <vector>
+#include <climits>
+
+using namespace std;
+
+// Function to find the maximum revenue from cutting a rod of length (len)
+// where the rod of length (i) has cost (prices[i - 1])
+int RodCutting(vector<int> &prices, int len)
+{
+    // (dp) stores the maximum revenue achieved from cutting a rod of length (from 1 to len)
+    vector<int> dp(len + 1, 0);
+    // If the rod length is negative (invalid) or zero there's no revenue from it
+    if (len <= 0)
+    {
+        return 0;
+    }
+    // Cut a rod of length (i)
+    for (int i = 1; i <= len; i++)
+    {
+        // divide the rod of length (i) into two rods of lengths (j) and (i - j)
+        // and store the maximum revenue
+        for (int j = 0; j < i; j++)
+        {
+            // (dp[i]) stores the maximum revenue achieved from cutting a rod of length (i)
+            dp[i] = max(dp[i], prices[j] + dp[i - j - 1]);
+        }
+    }
+    // (dp[len]) contains the maximum revenue from cutting a rod of length (len)
+    return dp[len];
+}
+
+int main()
+{
+    int len;
+    cout << "Enter the rod length :";
+    cin >> len;
+
+    vector<int> prices(len);
+    for (int i = 1; i <= len; i++)
+    {
+        cout << "Enter the price of the rod of length " << i << " :";
+        cin >> prices[i - 1];
+    }
+
+    cout << "Maximum revenue = " << RodCutting(prices, len);
+    return 0;
+}
+
+
+// Input and output:
+// 1. prices[] = [1, 5, 8, 9, 10, 17, 17, 20]  
+//        rod length = 4  
+//        Best: Cut the rod into two pieces of length 2 each to gain revenue of 5 + 5 = 10
+       
+//     2. prices[] = [1, 5, 8, 9, 10, 17, 17, 20]
+//        rod length = 8
+//        Best: Cut the rod into two pieces of length 2 and 6 to gain revenue of 5 + 17 = 22
+       
+//     3. prices[] = [3, 5, 8, 9, 10, 17, 17, 20]
+//        rod length = 8
+//        Best: Cut the rod into eight pieces of length 1 to gain revenue of 8 * 3 = 24

algorithms/CPlusPlus/README.md

@@ -42,6 +42,7 @@
 - [Matrix chain Multiplication](Dynamic-Programming/matrix-chain-multiplication.cpp)
 - [Edit Distance](Dynamic-Programming/edit-distance.cpp)
 - [Fibonacci](Dynamic-Programming/fibonacci.cpp)
+- [Rod Cutting](Dynamic-Programming/rod-cutting.cpp)
 
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