Added A greedy problem: Cutting a Rod

Author: deepakumarvu

Greedy/Cutting_a_Rod/CuttingARod.java

@@ -0,0 +1,33 @@
+/**
+ * Time Complexity : O(n^2) 
+ * which is much better than the 
+ * worst case time complexity of Naive Recursive implementation.
+ */
+
+import java.util.*;
+import java.io.*;
+
+class CuttingARod {
+
+    public static int cutCutCut(int price[], int size) {
+
+        int memory[] = new int[size+1];
+        memory[0] = 0;
+        
+        for(int i=1; i<=size; i++){
+            int maximum = Integer.MIN_VALUE;
+            for(int j=0; j<i; j++) {
+                maximum = Math.max(maximum, price[j]+memory[i-j-1]);
+            }
+            memory[i] = maximum;
+        }
+        return memory[size];
+    }
+
+    public static void main(String[] args) {
+        int price[] = new int[] {1, 5, 8, 9, 10, 17, 17, 20};
+        int size = price.length;
+        System.out.println("Maximum Profit: " +
+                            cutCutCut(price, size));
+    }
+}

Greedy/Cutting_a_Rod/Problem_Statement.txt

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+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. 
+For example, if length of the rod is 8 and the values of different pieces are given as following, 
+then the maximum obtainable value is 22 (by cutting in two pieces of lengths 2 and 6)
+
+length   | 1   2   3   4   5   6   7   8  
+--------------------------------------------
+price    | 1   5   8   9  10  17  17  20