From 0d3bb4293c9dd60bdcb00d856419fba3793f3155 Mon Sep 17 00:00:00 2001
From: Reshma <101700293+reshmascode@users.noreply.github.com>
Date: Sat, 22 Oct 2022 10:52:03 +0530
Subject: [PATCH] added solution for Rod Cutting Problem
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

We are given an array price[], where the rod of length i has a value price[i-1]. The idea is simple – one by one, partition the given rod of length n into two parts: i and n-i. Recur for the rod of length n-i but don’t divide the rod of length i any further. Finally, take the maximum of all values.
This yields the following recursive relation:

rodcut(n) = max { price[i – 1] + rodCut(n – i) } where 1 <= i <= n
---
 Rod-Cutting-Problem.cpp | 43 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 43 insertions(+)
 create mode 100644 Rod-Cutting-Problem.cpp

diff --git a/Rod-Cutting-Problem.cpp b/Rod-Cutting-Problem.cpp
new file mode 100644
index 0000000..15b7b04
--- /dev/null
+++ b/Rod-Cutting-Problem.cpp
@@ -0,0 +1,43 @@
+#include <iostream>
+#include <string>
+#include <climits>
+using namespace std;
+ 
+// Function to find the best way to cut a rod of length `n`
+// where the rod of length `i` has a cost `price[i-1]`
+int rodCut(int price[], int n)
+{
+    // base case
+    if (n == 0) {
+        return 0;
+    }
+ 
+    int maxValue = INT_MIN;
+ 
+    // one by one, partition the given rod of length `n` into two parts
+    // of length (1, n-1), (2, n-2), (3, n-3), … ,(n-1, 1), (n, 0)
+    // and take maximum
+    for (int i = 1; i <= n; i++)
+    {
+        // rod of length `i` has a cost `price[i-1]`
+        int cost = price[i - 1] + rodCut(price, n - i);
+ 
+        if (cost > maxValue) {
+            maxValue = cost;
+        }
+    }
+ 
+    return maxValue;
+}
+ 
+int main()
+{
+    int price[] = { 1, 5, 8, 9, 10, 17, 17, 20 };
+ 
+    // rod length
+    int n = 4;
+ 
+    cout << "Profit is " << rodCut(price, n);
+ 
+    return 0;
+}