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Nearest Fibonacci Number to N

  • Last Updated : 13 Apr, 2021

Given a positive integer N, the task is to find the nearest Fibonacci number to the given integer N. If there are two Fibonacci Numbers having same difference from N, then print the smaller value.

Examples:

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Input: N = 20
Output: 21
Explanation: Nearest Fibonacci number to 20 is 21.

Input: N = 17
Output: 13



Approach: Follow the steps below to solve the problem:

  • If N is equal to 0, then print 0 as the result.
  • Initialize a variable, say ans, to store the Fibonacci Number nearest to N.
  • Initialize two variables, say First as 0, and Second as 1, to store the first and second terms of the Fibonacci Series.
  • Store the sum of First and Second in a variable, say Third.
  • Iterate until the value of Third is at most N and perform the following steps: 
    • Update the value of First to Second and Second to Third.
    • Store the sum of First and Second in the variable Third.
  • If the absolute difference of Second and N is at most the value of Third and N, then update the value of ans as Second.
  • Otherwise, update the value of ans as Third.
  • After completing the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the Fibonacci
// number which is nearest to N
void nearestFibonacci(int num)
{
    // Base Case
    if (num == 0) {
        cout << 0;
        return;
    }
 
    // Initialize the first & second
    // terms of the Fibonacci series
    int first = 0, second = 1;
 
    // Store the third term
    int third = first + second;
 
    // Iterate until the third term
    // is less than or equal to num
    while (third <= num) {
 
        // Update the first
        first = second;
 
        // Update the second
        second = third;
 
        // Update the third
        third = first + second;
    }
 
    // Store the Fibonacci number
    // having smaller difference with N
    int ans = (abs(third - num)
               >= abs(second - num))
                  ? second
                  : third;
 
    // Print the result
    cout << ans;
}
 
// Driver Code
int main()
{
    int N = 17;
    nearestFibonacci(N);
 
    return 0;
}

Java




// Java program for the above approach
class GFG{
     
// Function to find the Fibonacci
// number which is nearest to N
static void nearestFibonacci(int num)
{
     
    // Base Case
    if (num == 0)
    {
        System.out.print(0);
        return;
    }
 
    // Initialize the first & second
    // terms of the Fibonacci series
    int first = 0, second = 1;
 
    // Store the third term
    int third = first + second;
 
    // Iterate until the third term
    // is less than or equal to num
    while (third <= num)
    {
         
        // Update the first
        first = second;
 
        // Update the second
        second = third;
 
        // Update the third
        third = first + second;
    }
 
    // Store the Fibonacci number
    // having smaller difference with N
    int ans = (Math.abs(third - num) >=
               Math.abs(second - num)) ?
               second : third;
 
    // Print the result
     System.out.print(ans);
}
 
// Driver Code
public static void main (String[] args)
{
    int N = 17;
     
    nearestFibonacci(N);
}
}
 
// This code is contributed by AnkThon

Python3




# Python3 program for the above approach
 
# Function to find the Fibonacci
# number which is nearest to N
def nearestFibonacci(num):
     
    # Base Case
    if (num == 0):
        print(0)
        return
 
    # Initialize the first & second
    # terms of the Fibonacci series
    first = 0
    second = 1
 
    # Store the third term
    third = first + second
 
    # Iterate until the third term
    # is less than or equal to num
    while (third <= num):
         
        # Update the first
        first = second
 
        # Update the second
        second = third
 
        # Update the third
        third = first + second
 
    # Store the Fibonacci number
    # having smaller difference with N
    if (abs(third - num) >=
        abs(second - num)):
        ans =  second
    else:
        ans = third
 
    # Print the result
    print(ans)
 
# Driver Code
if __name__ == '__main__':
     
    N = 17
     
    nearestFibonacci(N)
 
# This code is contributed by SURENDRA_GANGWAR

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the Fibonacci
// number which is nearest to N
static void nearestFibonacci(int num)
{
     
    // Base Case
    if (num == 0)
    {
        Console.Write(0);
        return;
    }
 
    // Initialize the first & second
    // terms of the Fibonacci series
    int first = 0, second = 1;
 
    // Store the third term
    int third = first + second;
 
    // Iterate until the third term
    // is less than or equal to num
    while (third <= num)
    {
         
        // Update the first
        first = second;
 
        // Update the second
        second = third;
 
        // Update the third
        third = first + second;
    }
 
    // Store the Fibonacci number
    // having smaller difference with N
    int ans = (Math.Abs(third - num) >=
              Math.Abs(second - num)) ?
                       second : third;
 
    // Print the result
     Console.Write(ans);
}
 
// Driver Code
public static void Main(string[] args)
{
    int N = 17;
     
    nearestFibonacci(N);
}
}
 
// This code is contributed by sanjoy_62

Javascript




<script>
// Javascript program for the above approach
 
// Function to find the Fibonacci
// number which is nearest to N
function nearestFibonacci(num)
{
    // Base Case
    if (num == 0) {
        document.write(0);
        return;
    }
 
    // Initialize the first & second
    // terms of the Fibonacci series
    let first = 0, second = 1;
 
    // Store the third term
    let third = first + second;
 
    // Iterate until the third term
    // is less than or equal to num
    while (third <= num) {
 
        // Update the first
        first = second;
 
        // Update the second
        second = third;
 
        // Update the third
        third = first + second;
    }
 
    // Store the Fibonacci number
    // having smaller difference with N
    let ans = (Math.abs(third - num)
               >= Math.abs(second - num))
                  ? second
                  : third;
 
    // Print the result
    document.write(ans);
}
 
// Driver Code
    let N = 17;
    nearestFibonacci(N);
 
// This code is contributed by subhammahato348.
</script>
Output: 
13

 

Time Complexity: O(log N)
Auxiliary Space: O(1)




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