From 34f5c5938718139bd2601b8d3ac26eaf0cce24e9 Mon Sep 17 00:00:00 2001
From: Himanshu Shekhar <himanshushekharb16@gmail.com>
Date: Sun, 1 Oct 2017 23:20:57 +0530
Subject: [PATCH] add rod cutting problem

---
 dynamic-programming/rod-cutting.go | 60 ++++++++++++++++++++++++++++++
 1 file changed, 60 insertions(+)
 create mode 100644 dynamic-programming/rod-cutting.go

diff --git a/dynamic-programming/rod-cutting.go b/dynamic-programming/rod-cutting.go
new file mode 100644
index 000000000..8abc4e441
--- /dev/null
+++ b/dynamic-programming/rod-cutting.go
@@ -0,0 +1,60 @@
+// Solution to Rod cutting problem
+// https://en.wikipedia.org/wiki/Cutting_stock_problem
+// http://www.geeksforgeeks.org/dynamic-programming-set-13-cutting-a-rod/
+
+package dpRodCutting
+
+// package main
+
+import "fmt"
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	} else {
+		return b
+	}
+
+}
+
+// solve the problem recursively: initial approach
+func cutRodRec(price []int, length int) int {
+	if length == 0 {
+		return 0
+	}
+
+	q := -1
+	for i := 1; i <= length; i++ {
+		q = max(q, price[i]+cutRodRec(price, length-i))
+	}
+	return q
+}
+
+// solve the same problem using dynamic programming
+func cutRodDp(price []int, length int) int {
+	r := make([]int, length+1) // a.k.a the memoization array
+	r[0] = 0                   // cost of 0 length rod is 0
+
+	for j := 1; j <= length; j++ { // for each length (subproblem)
+		q := -1
+		for i := 1; i <= j; i++ {
+			q = max(q, price[i]+r[j-i]) // avoiding recursive call
+		}
+		r[j] = q
+	}
+
+	return r[length]
+}
+
+/*
+func main() {
+	length := 10
+	price := []int{0, 1, 5, 8, 9, 17, 17, 17, 20, 24, 30}
+	// price := []int{0, 10, 5, 8, 9, 17, 17, 17, 20, 24, 30}
+
+	// fmt.Print(price[5]+price[length-5], "\n")
+
+	fmt.Print(cutRodRec(price, length), "\n")
+	fmt.Print(cutRodDp(price, length), "\n")
+}
+*/