Find the next Non-Fibonacci number

Last Updated : 26 May, 2022

Given a number N, the task is to find the next Non-Fibonacci number.
Examples:

Input: N = 4
Output: 6
6 is the next non-fibonacci number after 4
Input: N = 6
Output: 7

Approach: As the fibonacci series is given as

0, 1, 1, 2, 3, 5, 8, 13, 21, 34....

It can be observed that there does not exists any 2 consecutive fibonacci numbers. Therefore, inorder to find the next Non-Fibonacci number, the following cases arise:

If N <= 3, then the next Non-Fibonacci number will be 4
If N > 3, then we will check if (N + 1) is fibonacci number or not.
- If (N + 1) is a fibonacci number then (N + 2) will be the next Non-Fibonacci number.
- Else (N + 1) will be the answer

Below is the implementation of the above approach:

C++

// C++ implementation of the approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if a
// number is perfect square
bool isPerfectSquare(int x)
{
    int s = sqrt(x);
    return (s * s == x);
}

// Function to check if a
// number is Fibonacci Number
bool isFibonacci(int N)
{
    // N is Fibonacci if either
    // (5*N*N + 4), (5*N*N - 4) or both
    // is a perfect square
    return isPerfectSquare(5 * N * N + 4)
           || isPerfectSquare(5 * N * N - 4);
}

// Function to find
// the next Non-Fibonacci Number
int nextNonFibonacci(int N)
{

    // Case 1
    // If N<=3, then 4 will be
    // next Non-Fibonacci Number
    if (N <= 3)
        return 4;

    // Case 2
    // If N+1 is Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    if (isFibonacci(N + 1))
        return N + 2;

    // If N+1 is Non-Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    else
        return N + 1;
}

// Driver code
int main()
{
    int N = 3;
    cout << nextNonFibonacci(N)
         << endl;

    N = 5;
    cout << nextNonFibonacci(N)
         << endl;

    N = 7;
    cout << nextNonFibonacci(N)
         << endl;
}

// C implementation of the approach

#include <stdio.h>
#include<stdbool.h>
#include<math.h>


// Function to check if a
// number is perfect square
bool isPerfectSquare(int x)
{
    int s = sqrt(x);
    return (s * s == x);
}

// Function to check if a
// number is Fibonacci Number
bool isFibonacci(int N)
{
    // N is Fibonacci if either
    // (5*N*N + 4), (5*N*N - 4) or both
    // is a perfect square
    return isPerfectSquare(5 * N * N + 4)
        || isPerfectSquare(5 * N * N - 4);
}

// Function to find
// the next Non-Fibonacci Number
int nextNonFibonacci(int N)
{

    // Case 1
    // If N<=3, then 4 will be
    // next Non-Fibonacci Number
    if (N <= 3)
        return 4;

    // Case 2
    // If N+1 is Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    if (isFibonacci(N + 1))
        return N + 2;

    // If N+1 is Non-Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    else
        return N + 1;
}

// Driver code
int main()
{
    int N = 3;
    printf("%d\n",nextNonFibonacci(N));

    N = 5;
    printf("%d\n",nextNonFibonacci(N));

    N = 7;
    printf("%d",nextNonFibonacci(N));
    
}

// This code is contributed by allwink45.

Java

// Java implementation of the approach
import java.util.*;

class GFG{
 
// Function to check if a
// number is perfect square
static boolean isPerfectSquare(int x)
{
    int s = (int) Math.sqrt(x);
    return (s * s == x);
}
 
// Function to check if a
// number is Fibonacci Number
static boolean isFibonacci(int N)
{
    // N is Fibonacci if either
    // (5*N*N + 4), (5*N*N - 4) or both
    // is a perfect square
    return isPerfectSquare(5 * N * N + 4)
           || isPerfectSquare(5 * N * N - 4);
}
 
// Function to find
// the next Non-Fibonacci Number
static int nextNonFibonacci(int N)
{
 
    // Case 1
    // If N<=3, then 4 will be
    // next Non-Fibonacci Number
    if (N <= 3)
        return 4;
 
    // Case 2
    // If N+1 is Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    if (isFibonacci(N + 1))
        return N + 2;
 
    // If N+1 is Non-Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    else
        return N + 1;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
    System.out.print(nextNonFibonacci(N)
         +"\n");
 
    N = 5;
    System.out.print(nextNonFibonacci(N)
         +"\n");
 
    N = 7;
    System.out.print(nextNonFibonacci(N)
         +"\n");
}
}

// This code is contributed by 29AjayKumar

Python 3

# Python 3 implementation of the approach
from math import sqrt

# Function to check if a
# number is perfect square
def isPerfectSquare(x):
    s = sqrt(x)
    return (s * s == x)

# Function to check if a
# number is Fibonacci Number
def isFibonacci(N):

    # N is Fibonacci if either
    # (5*N*N + 4), (5*N*N - 4) or both
    # is a perfect square
    return isPerfectSquare(5 * N * N + 4) or \
            isPerfectSquare(5 * N * N - 4)

# Function to find
# the next Non-Fibonacci Number
def nextNonFibonacci(N):
    
    # Case 1
    # If N<=3, then 4 will be
    # next Non-Fibonacci Number
    if (N <= 3):
        return 4

    # Case 2
    # If N+1 is Fibonacci, then N+2
    # will be next Non-Fibonacci Number
    if (isFibonacci(N + 1)):
        return N + 2

    # If N+1 is Non-Fibonacci, then N+2
    # will be next Non-Fibonacci Number
    else:
        return N 

# Driver code
if __name__ == '__main__':
    N = 3
    print(nextNonFibonacci(N))
    N = 4
    print(nextNonFibonacci(N))

    N = 7
    print(nextNonFibonacci(N))
    
# This code is contributed by Surendra_Gangwar

// C# implementation of the approach
using System;

class GFG{
  
// Function to check if a
// number is perfect square
static bool isPerfectSquare(int x)
{
    int s = (int) Math.Sqrt(x);
    return (s * s == x);
}
  
// Function to check if a
// number is Fibonacci Number
static bool isFibonacci(int N)
{
    // N is Fibonacci if either
    // (5*N*N + 4), (5*N*N - 4) or both
    // is a perfect square
    return isPerfectSquare(5 * N * N + 4)
           || isPerfectSquare(5 * N * N - 4);
}
  
// Function to find
// the next Non-Fibonacci Number
static int nextNonFibonacci(int N)
{
  
    // Case 1
    // If N<=3, then 4 will be
    // next Non-Fibonacci Number
    if (N <= 3)
        return 4;
  
    // Case 2
    // If N+1 is Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    if (isFibonacci(N + 1))
        return N + 2;
  
    // If N+1 is Non-Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    else
        return N + 1;
}
  
// Driver code
public static void Main(String[] args)
{
    int N = 3;
    Console.Write(nextNonFibonacci(N)
         +"\n");
  
    N = 5;
    Console.Write(nextNonFibonacci(N)
         +"\n");
  
    N = 7;
    Console.Write(nextNonFibonacci(N)
         +"\n");
}
}

// This code is contributed by Princi Singh

JavaScript

<script>

// Javascript implementation of the approach

// Function to check if a
// number is perfect square
function isPerfectSquare(x)
{
    var s = parseInt(Math.sqrt(x));
    return (s * s == x);
}

// Function to check if a
// number is Fibonacci Number
function isFibonacci(N)
{
    // N is Fibonacci if either
    // (5*N*N + 4), (5*N*N - 4) or both
    // is a perfect square
    return isPerfectSquare(5 * N * N + 4)
           || isPerfectSquare(5 * N * N - 4);
}

// Function to find
// the next Non-Fibonacci Number
function nextNonFibonacci(N)
{

    // Case 1
    // If N<=3, then 4 will be
    // next Non-Fibonacci Number
    if (N <= 3)
        return 4;

    // Case 2
    // If N+1 is Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    if (isFibonacci(N + 1))
        return N + 2;

    // If N+1 is Non-Fibonacci, then N+2
    // will be next Non-Fibonacci Number
    else
        return N + 1;
}

// Driver code
var N = 3;
document.write(nextNonFibonacci(N)+"<br>");
N = 5;
document.write(nextNonFibonacci(N)+"<br>");
N = 7;
document.write(nextNonFibonacci(N)+"<br>");

// This code is contributed by rutvik_56.
</script>

Output:

4
6
9

Time Complexity: O(n^1/2)

Auxiliary Space: O(1)

Find First n Fibonacci Numbers

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Find the next Non-Fibonacci number

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