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 numberint previousFibonacci(int n){ double a = n / ((1 + sqrt(5)) / 2.0); return round(a);}// Driver codeint main(){ int n = 8; cout << (previousFibonacci(n));}// This code is contributed by Mohit Kumar |
Java
// Java implementation of the approachimport java.io.*;class GFG{ // Function to return the previous// fibonacci numberstatic int previousFibonacci(int n){ double a = n / ((1 + Math.sqrt(5)) / 2.0); return (int)Math.round(a);}// Driver codepublic 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 = 8print(previousFibonacci(n)) |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the previous// fibonacci numberstatic int previousFibonacci(int n){ double a = n / ((1 + Math.Sqrt(5)) / 2.0); return (int)Math.Round(a);}// Driver codepublic 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 numberfunction previousFibonacci(n){ var a = n / ((1 + Math.sqrt(5)) / 2); return Math.round(a);}// Driver codevar n = 8;document.write(previousFibonacci(n));// This code is contributed by rutvik_56.</script> |
Output:
5
Time Complexity: O(1)
Auxiliary Space: O(1)
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