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++

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// 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

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Java

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// 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. 

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Python3

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# 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)) 

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C#

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// 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

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Output:

5


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