Find the Nth element of the modified Fibonacci series (original) (raw)

Last Updated : 13 Sep, 2022

Given two integers A and B which are the first two terms of the series and another integer N. The task is to find the Nth number using Fibonacci rule i.e. fib(i) = fib(i - 1) + fib(i - 2)
Example:

Input: A = 2, B = 3, N = 4
Output: 8
The series will be 2, 3, 5, 8, 13, 21, ...
And the 4th element is 8.
Input: A = 5, B = 7, N = 10
Output: 343

Approach: Initialize variable sum = 0 that stores sum of the previous two values. Now, run a loop from i = 2 to N and for each index update value of sum = A + B and A = B, B = sum. Then finally, return the sum which is the required Nth element.
Below is the implementation of the above approach:

C++ `

// C++ implementation of the approach #include using namespace std;

// Function to return the Nth number of // the modified Fibonacci series where // A and B are the first two terms int findNthNumber(int A, int B, int N) {

// To store the current element which
// is the sum of previous two
// elements of the series
int sum = 0;

// This loop will terminate when
// the Nth element is found
for (int i = 2; i < N; i++) {
    sum = A + B;

    A = B;

    B = sum;
}

// Return the Nth element
return sum;

}

// Driver code int main() { int A = 5, B = 7, N = 10;

cout << findNthNumber(A, B, N);

return 0;

}

Java

// Java implementation of the approach import java.util.*;

class GFG {

// Function to return the Nth number of
// the modified Fibonacci series where
// A and B are the first two terms
static int findNthNumber(int A, int B, int N) 
{

    // To store the current element which
    // is the sum of previous two
    // elements of the series
    int sum = 0;

    // This loop will terminate when
    // the Nth element is found
    for (int i = 2; i < N; i++)
    {
        sum = A + B;

        A = B;

        B = sum;
    }

    // Return the Nth element
    return sum;
}

// Driver code
public static void main(String[] args)
{
    int A = 5, B = 7, N = 10;

    System.out.println(findNthNumber(A, B, N));
}

}

// This code is contributed by PrinciRaj1992

Python3

Python3 implementation of the approach

Function to return the Nth number of

the modified Fibonacci series where

A and B are the first two terms

def findNthNumber(A, B, N):

# To store the current element which
# is the sum of previous two
# elements of the series
sum = 0

# This loop will terminate when
# the Nth element is found
for i in range(2, N):
    sum = A + B

    A = B

    B = sum

# Return the Nth element
return sum

Driver code

if name == 'main': A = 5 B = 7 N = 10

print(findNthNumber(A, B, N))

This code is contributed by Ashutosh450

C#

// C# implementation of the approach using System;

class GFG {

// Function to return the Nth number of
// the modified Fibonacci series where
// A and B are the first two terms
static int findNthNumber(int A, int B, int N) 
{

    // To store the current element which
    // is the sum of previous two
    // elements of the series
    int sum = 0;

    // This loop will terminate when
    // the Nth element is found
    for (int i = 2; i < N; i++)
    {
        sum = A + B;

        A = B;

        B = sum;
    }

    // Return the Nth element
    return sum;
}

// Driver code
public static void Main()
{
    int A = 5, B = 7, N = 10;

    Console.WriteLine(findNthNumber(A, B, N));
}

}

// This code is contributed by AnkitRai01

JavaScript

`

Time Complexity: O(N)

Auxiliary Space: O(1)