Find the previous fibonacci number

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++ 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 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 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# 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

<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>

5

Time Complexity: O(1)

Auxiliary Space: O(1)