Find the previous fibonacci number

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Given a Fibonacci number N, the task is to find the previous Fibonacci number.
Examples:

Input: N = 8
Output: 5
5 is the previous fibonacci number before 8.
Input: N = 5
Output: 3

Approach: The ratio of two adjacent numbers in the Fibonacci series rapidly approaches ((1 + sqrt(5)) / 2). So if N is divided by ((1 + sqrt(5)) / 2) and then rounded, the resultant number will be the previous Fibonacci number.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to return the previous
// fibonacci number
int previousFibonacci(int n)
{
    double a = n / ((1 + sqrt(5)) / 2.0);
    return round(a);
}
 
// Driver code
int main()
{
    int n = 8;
    cout << (previousFibonacci(n));
}
 
// This code is contributed by Mohit Kumar

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
         
// Function to return the previous
// fibonacci number
static int previousFibonacci(int n)
{
    double a = n / ((1 + Math.sqrt(5)) / 2.0);
    return (int)Math.round(a);
}
 
// Driver code
public static void main (String[] args) 
{
    int n = 8;
    System.out.println(previousFibonacci(n));
}
}
 
// This code is contributed by ajit. 

Python3

# Python3 implementation of the approach 
from math import *
 
# Function to return the previous 
# fibonacci number 
def previousFibonacci(n): 
    a = n/((1 + sqrt(5))/2.0)
    return round(a) 
 
# Driver code 
n = 8
print(previousFibonacci(n)) 

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the previous
// fibonacci number
static int previousFibonacci(int n)
{
    double a = n / ((1 + Math.Sqrt(5)) / 2.0);
    return (int)Math.Round(a);
}
 
// Driver code
public static void Main()
{
    int n = 8;
    Console.Write(previousFibonacci(n));
}
}
 
// This code is contributed by Akanksha_Rai

Javascript

<script>
// Javascript implementation of the approach
 
// Function to return the previous
// fibonacci number
function previousFibonacci(n)
{
    var a = n / ((1 + Math.sqrt(5)) / 2);
    return Math.round(a);
}
 
// Driver code
var n = 8;
document.write(previousFibonacci(n));
 
// This code is contributed by rutvik_56.
</script>

Output:

Time Complexity: O(1)

Auxiliary Space: O(1)

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