Given an integer N, the task is to print the first N terms of the Fibonacci series in reverse order using Recursion.
Examples:
Input: N = 5
Output: 3 2 1 1 0
Explanation: First five terms are – 0 1 1 2 3.Input: N = 10
Output: 34 21 13 8 5 3 2 1 1 0
Approach: The idea is to use recursion in a way that keeps calling the same function again till N is greater than 0 and keeps on adding the terms and after that starts printing the terms.
Follow the steps below to solve the problem:
- Define a function fibo(int N, int a, int b) where
- N is the number of terms and
- a and b are the initial terms with values 0 and 1.
- If N is greater than 0, then call the function again with values N-1, b, a+b.
- After the function call, print a as the answer.
Below is the implementation of the above approach.
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to print the fibonacci // series in reverse order. void fibo( int n, int a, int b)
{ if (n > 0) {
// Function call
fibo(n - 1, b, a + b);
// Print the result
cout << a << " " ;
}
} // Driver Code int main()
{ int N = 10;
fibo(N, 0, 1);
return 0;
} |
// Java program for the above approach import java.util.*;
public class GFG
{ // Function to print the fibonacci // series in reverse order. static void fibo( int n, int a, int b)
{ if (n > 0 ) {
// Function call
fibo(n - 1 , b, a + b);
// Print the result
System.out.print(a + " " );
}
} // Driver Code public static void main(String args[])
{ int N = 10 ;
fibo(N, 0 , 1 );
} } // This code is contributed by Samim Hossain Mondal. |
# Python program for the above approach # Function to print the fibonacci # series in reverse order. def fibo(n, a, b):
if (n > 0 ):
# Function call
fibo(n - 1 , b, a + b)
# Print the result
print (a, end = " " )
# Driver Code if __name__ = = "__main__" :
N = 10
fibo(N, 0 , 1 )
# This code is contributed by Samim Hossain Mondal.
|
// C# program for the above approach using System;
class GFG
{ // Function to print the fibonacci // series in reverse order. static void fibo( int n, int a, int b)
{ if (n > 0) {
// Function call
fibo(n - 1, b, a + b);
// Print the result
Console.Write(a + " " );
}
} // Driver Code public static void Main()
{ int N = 10;
fibo(N, 0, 1);
} } // This code is contributed by Samim Hossain Mondal. |
<script> // Javascript program for the above approach // Function to print the fibonacci // series in reverse order. function fibo(n, a, b)
{ if (n > 0) {
// Function call
fibo(n - 1, b, a + b);
// Print the result
document.write(a + " " );
}
} // Driver Code let N = 10; fibo(N, 0, 1); // This code is contributed by Samim Hossain Mondal. </script> |
34 21 13 8 5 3 2 1 1 0
Time Complexity: O(N)
Auxiliary Space: O(N)