# Find the previous fibonacci number

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

Input:N = 8Output:5

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the

DSA Self Paced Courseat a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please referComplete Interview Preparation Course.In case you wish to attend

live classeswith experts, please referDSA Live Classes for Working ProfessionalsandCompetitive Programming Live for Students.

**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:**

5