Create CuttingRodProblem

Author: nkatre

CuttingRodProblem

@@ -0,0 +1,73 @@
+
+/*
+ * Question: What is the best price for cutting a rod of length n
+ * 
+ * Question and Answer Source: http://www.geeksforgeeks.org/dynamic-programming-set-13-cutting-a-rod/
+ * 							   https://www.youtube.com/watch?v=U-09Gs6cbsQ&list=PL962BEE1A26238CA3&index=4
+ * 
+ */
+
+package dynamicProgramming;
+
+import java.util.Scanner;
+
+public class CuttingRodProblem {
+	 public static void main(String[] args){
+		    Scanner in = new Scanner(System.in);
+		    try{
+		    	System.out.println("Enter the number of cuts");
+		    	int n = in.nextInt();
+		    	int[] prices = new int[n];
+		    	System.out.println("Enter the prices of rod cuts");
+		    	for(int i=0;i<n;i++){
+		    		System.out.println("Enter price for "+(i+1)+" rod length");
+		    		prices[i] = in.nextInt();
+		    	}
+		    	System.out.println("The best price for rod cut of size "+n+" using Recursion is: "+byRecursion(prices, n));
+		    	System.out.println("The best price for rod cut of size "+n+" using DP is: "+byDP(prices, n));
+		    }
+		    finally{
+		    	in.close();
+		    }
+	    }
+	    public static int byRecursion(int[] prices, int rodLength){
+	         
+	         // Base Case
+	         if(rodLength<=0)
+	             return 0;     // if rod length is 0 then there is no best price
+	         
+	         // Recursive Step
+	         int maxPrice = 0;
+	         for (int i = 0; i<rodLength; i++)
+	             maxPrice = Math.max(maxPrice, prices[i]+byRecursion(prices,rodLength-1-i));
+	             
+	         
+	         // Returning Step
+	         return maxPrice;
+	           
+	    }
+	    /*
+		    * Analysis:
+		    * Time Complexity = O(2^n)  // not sure
+		    * Space Complexity = O(1) used by dp array 
+		    */
+	    public static int byDP(int[] prices, int rodLength){
+	        
+	        int[] dp = new int[rodLength+1];   // +1 because 0 rodLength will have best prices of 0
+	        dp[0] = 0;
+	        
+	        int maxPrice = -1;
+	        for(int i=1;i<=rodLength;i++){  // i represents rodLength
+	            for(int j=0;j<i;j++)        // price of cut from 0 to rodLength-1
+	                maxPrice = Math.max(maxPrice,prices[j]+dp[i-j-1]);  // here i represents rodLength
+	            dp[i] = maxPrice;
+	        }
+	        
+	        return dp[rodLength];
+	    }
+	   /*
+	    * Analysis:
+	    * Time Complexity = O(n^2)
+	    * Space Complexity = O(n) used by dp array 
+	    */
+}