chore(Java): add nth geek-onacci number (#608)

algorithms/Java/Maths/nth-geek-onacci-number.java

@@ -0,0 +1,89 @@
+/** Author : Suraj Kumar
+ * Github : https://github.com/skmodi649
+ */
+
+
+
+/** PROBLEM DESCRIPTION :
+ * Geek created a random series and given a name geek-onacci series
+ * Given four integers a, b, b, n
+ * a, b, c represents the first three numbers of geek-onacci series
+ * Find the Nth number of the series.
+ * The nth number of geek-onacci series is a sum of the last three numbers (summation of N-1th, N-2th, and N-3th geek-onacci numbers)
+ */
+
+
+/** ALGORITHM :
+ * Simply use the concept of Recursion
+ * Define the base cases of the Recursion for n equal to 1 , 2 and 3
+ * Then simply use recurrence in the same way as we used it in Fibonacci series
+ */
+
+
+
+import java.util.*;
+import java.lang.*;
+
+class GFG {
+    public static int rec(int a,int b,int c,int n)
+    {
+        if(n==1)
+            return a;
+        else if(n==2)
+            return b;
+        else if(n==3)
+            return c;
+        else
+            return rec(a,b,c,n-1)+rec(a,b,c,n-2)+rec(a,b,c,n-3);
+    }
+    public static void main (String[] args) {
+        Scanner sc = new Scanner(System.in);
+            int a,b,c,n;
+            System.out.print("Enter the first number of Geek-onacci series : ");
+            a = sc.nextInt();
+        System.out.print("Enter the second number of Geek-onacci series : ");
+            b = sc.nextInt();
+        System.out.print("Enter the third number of Geek-onacci series : ");
+            c = sc.nextInt();
+            System.out.print("Enter the values of n : ");
+            n = sc.nextInt();
+            int val = rec(a,b,c,n);
+            System.out.println("Nth Geek-onacci number : "+val);
+        }
+    }
+
+
+
+/** TEST CASES :
+ * Test Case 1 :
+ * Input : a = 1 , b = 3 , c = 2 , n = 4
+ * Output : 6
+ *
+ * Test Case 2 :
+ * Input : a = 1 , b = 3 , c = 2 , n = 5
+ * Output : 11
+ *
+ * Test Case 3 :
+ * Input : a = 1 , b = 3 , c = 2 , n = 6
+ * Output : 19
+  */
+
+
+/** Explanation for a = 1 , b = 3 , c = 2 , n = 5
+ * n = 5 then val = rec(1,3,2,4)+rec(1,3,2,3)+rec(1,3,2,2)
+ * rec(1,3,2,2) gives 3 and rec(1,3,2,3) gives 2 hence both totally give 3 + 2 = 5
+ * rec(1,3,2,4) is actually equal to rec(1,3,2,3)+rec(1,3,2,2)+rec(1,3,2,1) hence gives 2 + 3 + 1 = 6
+ * Now val = 6 + 5 = 11
+ * 11 will be returned as output
+ */
+
+
+/** Time Complexity : O(n)
+ * Auxiliary Space Complexity : O(1)
+ * Constraints : 1 <= A, B, C <= 100
+ *               N >= 4
+ */
+
+
+
+

algorithms/Java/README.md

@@ -26,6 +26,7 @@
 ## Maths
 - [Factorial](Maths/factorial_using_big_integer.java)
 - [Catalan Numbers](Maths/catalan-numbers.java)
+- [Nth Geek Onacci Number](Maths/nth-geek-onacci-number.java)
 
 ## Queues
 - [Circular Queue using Linked List](queues/circular-queue-linked-list.java)