Count Fibonacci numbers in given range in O(Log n) time and O(1) space

Given a range, count Fibonacci numbers in given range. First few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ..

Example :

Input: low = 10, high = 100
Output: 5
There are five Fibonacci numbers in given
range, the numbers are 13, 21, 34, 55 and 89.

Input: low = 10, high = 20
Output: 1
There is only one Fibonacci Number, 13.

Input: low = 0, high = 1
Output: 3
Fibonacci numbers are 0, 1 and 1

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A Brute Force Solution is to one by one find all Fibonacci Numbers and count all Fibonacci numbers in given range

An Efficient Solution is to use previous Fibonacci Number to generate next using simple Fibonacci formula that fn = fn-1 + fn-2.

C++

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// C++ program to count Fibonacci numbers in given range
#include <bits/stdc++.h>
using namespace std;
  
// Returns count of fibonacci numbers in [low, high]
int countFibs(int low, int high)
{
    // Initialize first three Fibonacci Numbers
    int f1 = 0, f2 = 1, f3 = 1;
  
    // Count fibonacci numbers in given range
    int result = 0;
  
    while (f1 <= high)
    {
        if (f1 >= low)
           result++;
        f1 = f2;
        f2 = f3;
        f3 = f1 + f2;
    }
  
    return result;
}
  
// Driver program
int main()
{
   int low = 10, high = 100;
   cout << "Count of Fibonacci Numbers is "
        << countFibs(low, high);
   return 0;
}

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Java

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// Java program to count Fibonacci
// numbers in given range
public class GFG
{     
  
    // Returns count of fibonacci 
    // numbers in [low, high]
    static int countFibs(int low, 
                         int high)
    {
          
        // Initialize first three 
        // Fibonacci Numbers
        int f1 = 0, f2 = 1, f3 = 1;
      
        // Count fibonacci numbers 
        // in given range
        int result = 0;
      
        while (f1 <= high)
        {
            if (f1 >= low)
            result++;
            f1 = f2;
            f2 = f3;
            f3 = f1 + f2;
        }
      
        return result;
    }
      
      
    // Driver Code
    public static void main(String args[])
    {
        int low = 10, high = 100;
        System.out.println("Count of Fibonacci Numbers is "
                           + countFibs(low, high));
    }
}
      
// This code is contributed by Sam007.

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Python3

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# Python3 program to count Fibonacci 
# numbers in given range
  
# Returns count of fibonacci 
# numbers in [low, high]
def countFibs(low, high):
      
    # Initialize first three 
    # Fibonacci Numbers
    f1, f2, f3 = 0, 1, 1
  
    # Count fibonacci numbers in 
    # given range
    result = 0
  
    while (f1 <= high):
        if (f1 >= low):
            result += 1
        f1 = f2
        f2 = f3
        f3 = f1 + f2
  
    return result
  
# Driver Code
low, high = 10, 100
print("Count of Fibonacci Numbers is"
                 countFibs(low, high))
  
# This code is contributed 
# by mohit kumar

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C#

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// C# program to count Fibonacci
// numbers in given range
using System;
  
public class GFG
      
    // Returns count of fibonacci 
    // numbers in [low, high]
    static int countFibs(int low, 
                        int high)
    {
          
        // Initialize first three 
        // Fibonacci Numbers
        int f1 = 0, f2 = 1, f3 = 1;
      
        // Count fibonacci numbers 
        // in given range
        int result = 0;
      
        while (f1 <= high)
        {
            if (f1 >= low)
            result++;
            f1 = f2;
            f2 = f3;
            f3 = f1 + f2;
        }
      
        return result;
    }
      
    // Driver Code
    public static void Main(String []args)
    {
        int low = 10, high = 100;
        Console.WriteLine("Count of Fibonacci Numbers is "
                        + countFibs(low, high));
    }
}
      
// This code is contributed by Sam007.

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PHP

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<?php
// PHP program to count 
// Fibonacci numbers in 
// given range
  
// Returns count of fibonacci 
// numbers in [low, high]
function countFibs($low, $high)
{
    // Initialize first 
    // three Fibonacci Numbers
    $f1 = 0; $f2 = 1; $f3 = 1;
  
    // Count fibonacci 
    // numbers in given range
    $result = 0;
  
    while ($f1 <= $high)
    {
        if ($f1 >= $low)
        $result++;
        $f1 = $f2;
        $f2 = $f3;
        $f3 = $f1 + $f2;
    }
  
    return $result;
}
  
// Driver Code
$low = 10; $high = 100;
echo "Count of Fibonacci Numbers is ",
               countFibs($low, $high);
  
// This code is contributed by nitin mittal.
?>

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Output :

Count of Fibonacci Numbers is 5

Time Complexity Analysis:
Consider the that Fibonacci Numbers can be written as below
fib(n)=\left [ \frac {1}{\sqrt{5}}\left ( \frac {1+\sqrt{5}}{2} \right )^n \right ]\sim c*1.62^n

for n\sim c*1.62^n we make O(n')comparisons,we,thus,needO(log(n))comparisons.
So the value of Fibonacci numbers grow exponentially. It means that the while loop grows exponentially till it reaches ‘high’. So the loop runs O(Log (high)) times.

One solution could be directly use above formula to find count of Fibonacci Numbers, but that is not practically feasible (See this for details).

Auxiliary Space: O(1)

This article is contributed by Sudhanshu Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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