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>
Output
34 21 13 8 5 3 2 1 1 0
Time Complexity: O(N)
Auxiliary Space: O(N)