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Geek-onacci Number

  • Difficulty Level : Medium
  • Last Updated : 30 Apr, 2021

Given four integers A, B, C and N, where A, B, C represents the first three numbers of the geekonacci series, the task is to find the Nth term of the geekonacci series.

The Nth term of the geekonacci series is the sum of its previous three terms in the series i.e., sum of (N – 1)th, (N – 2)th, and (N – 3)th geekonacci numbers.

Examples:

Input: A = 1, B = 3, C = 2, N = 4
Output: 6
Explanation: The geekonacci series is 1, 3, 2, 6, 11, 19, ……
Therefore, the 4th geekonacci number is 6.

Input: A = 1, B = 3, C = 2, N = 6
Output: 19

Approach: The given problem can be solved by generating the geekonacci series upto N terms and print the Nth term of the series obtained. Follow the steps below to solve this problem:

  • Initialize an array arr[] of size N and initialize arr[0] = A, arr[1] = B, and arr[2] = C.
  • Iterate over the range [3, N – 1] and update the value of each every ith element, i.e. arr[i] as (arr[i – 1] + arr[i – 2] + arr[i – 3]) to get the ith term of the geekonacci series.
  • After completing the above steps, print the value of arr[N – 1] as the resultant Nth number of the geekonacci series.

Below is the implementation of the above approach:

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the
// N-th Geek-onacci Number
int find(int A, int B,
                int C, int N)
{
   
    // Stores the geekonacci series
    int arr[N];
 
    // Store the first three
    // terms of the series
    arr[0] = A;
    arr[1] = B;
    arr[2] = C;
 
    // Iterate over the range [3, N]
    for (int i = 3; i < N; i++) {
 
        // Update the value of arr[i]
        // as the sum of previous 3
        // terms in the series
        arr[i] = arr[i - 1]
                 + arr[i - 2]
                 + arr[i - 3];
    }
 
    // Return the last element
    // of arr[] as the N-th term
    return arr[N - 1];
}
 
// Driver Code
int main()
{
  int A = 1, B = 3, C = 2, N = 4;
  cout<<(find(A, B, C, N));
 
  return 0;
}
 
// This code is contributed by mohit kumar 29.

Java




// Java program for the above approach
 
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
    // Function to calculate the
    // N-th Geek-onacci Number
    static int find(int A, int B,
                    int C, int N)
    {
        // Stores the geekonacci series
        int[] arr = new int[N];
 
        // Store the first three
        // terms of the series
        arr[0] = A;
        arr[1] = B;
        arr[2] = C;
 
        // Iterate over the range [3, N]
        for (int i = 3; i < N; i++) {
 
            // Update the value of arr[i]
            // as the sum of previous 3
            // terms in the series
            arr[i] = arr[i - 1]
                     + arr[i - 2]
                     + arr[i - 3];
        }
 
        // Return the last element
        // of arr[] as the N-th term
        return arr[N - 1];
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int A = 1, B = 3, C = 2, N = 4;
        System.out.print(find(A, B, C, N));
    }
}

Python3




# Python3 program for the above approach
 
# Function to calculate the
# N-th Geek-onacci Number
def find(A, B, C, N) :
   
    # Stores the geekonacci series
    arr = [0] * N
 
    # Store the first three
    # terms of the series
    arr[0] = A
    arr[1] = B
    arr[2] = C
 
    # Iterate over the range [3, N]
    for i in range(3, N):
 
        # Update the value of arr[i]
        # as the sum of previous 3
        # terms in the series
        arr[i] = (arr[i - 1]
                 + arr[i - 2]
                 + arr[i - 3])
     
    # Return the last element
    # of arr[] as the N-th term
    return arr[N - 1]
 
# Driver Code
A = 1
B = 3
C = 2
N = 4
 
print(find(A, B, C, N))
 
# This code is contributed by sanjoy_62.

C#




// C# program for the above approach
using System;
 
class GFG{
 
  // Function to calculate the
  // N-th Geek-onacci Number
  static int find(int A, int B,
                  int C, int N)
  {
    // Stores the geekonacci series
    int[] arr = new int[N];
 
    // Store the first three
    // terms of the series
    arr[0] = A;
    arr[1] = B;
    arr[2] = C;
 
    // Iterate over the range [3, N]
    for (int i = 3; i < N; i++) {
 
      // Update the value of arr[i]
      // as the sum of previous 3
      // terms in the series
      arr[i] = arr[i - 1]
        + arr[i - 2]
        + arr[i - 3];
    }
 
    // Return the last element
    // of arr[] as the N-th term
    return arr[N - 1];
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    int A = 1, B = 3, C = 2, N = 4;
    Console.Write(find(A, B, C, N));
  }
}
 
// This code is contributed by code_hunt.

Javascript




<script>
 
// Javascript program for the above approach
 
    // Function to calculate the
    // N-th Geek-onacci Number
    function find(A, B, C, N)
    {
        // Stores the geekonacci series
        let arr = new Array(N).fill(0);
 
        // Store the first three
        // terms of the series
        arr[0] = A;
        arr[1] = B;
        arr[2] = C;
 
        // Iterate over the range [3, N]
        for (let i = 3; i < N; i++) {
 
            // Update the value of arr[i]
            // as the sum of previous 3
            // terms in the series
            arr[i] = arr[i - 1]
                     + arr[i - 2]
                     + arr[i - 3];
        }
 
        // Return the last element
        // of arr[] as the N-th term
        return arr[N - 1];
    }
 
// Driver code
     
    let A = 1, B = 3, C = 2, N = 4;
    document.write(find(A, B, C, N));
     
</script>
Output: 
6

 

Time Complexity: O(N)
Auxiliary Space: O(N)


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