Find the next Non-Fibonacci number
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 4Input: 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 Fibinacci Number bool isFibonacci(int N) { // N is Fibinacci if either // (5*N*N + 4), (5*N*N - 4) or both // is a perferct square return isPerfectSquare(5 * N * N + 4) || isPerfectSquare(5 * N * N - 4); } // Function to find // the next Non-Fibinacci Number int nextNonFibonacci(int N) { // Case 1 // If N<=3, then 4 will be // next Non-Fibinacci Number if (N <= 3) return 4; // Case 2 // If N+1 is Fibinacci, then N+2 // will be next Non-Fibinacci Number if (isFibonacci(N + 1)) return N + 2; // If N+1 is Non-Fibinacci, then N+2 // will be next Non-Fibinacci 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; } |
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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 Fibinacci Number static boolean isFibonacci(int N) { // N is Fibinacci if either // (5*N*N + 4), (5*N*N - 4) or both // is a perferct square return isPerfectSquare(5 * N * N + 4) || isPerfectSquare(5 * N * N - 4); } // Function to find // the next Non-Fibinacci Number static int nextNonFibonacci(int N) { // Case 1 // If N<=3, then 4 will be // next Non-Fibinacci Number if (N <= 3) return 4; // Case 2 // If N+1 is Fibinacci, then N+2 // will be next Non-Fibinacci Number if (isFibonacci(N + 1)) return N + 2; // If N+1 is Non-Fibinacci, then N+2 // will be next Non-Fibinacci 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 |
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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 Fibinacci Number def isFibonacci(N): # N is Fibinacci if either # (5*N*N + 4), (5*N*N - 4) or both # is a perferct square return isPerfectSquare(5 * N * N + 4) or \ isPerfectSquare(5 * N * N - 4) # Function to find # the next Non-Fibinacci Number def nextNonFibonacci(N): # Case 1 # If N<=3, then 4 will be # next Non-Fibinacci Number if (N <= 3): return 4 # Case 2 # If N+1 is Fibinacci, then N+2 # will be next Non-Fibinacci Number if (isFibonacci(N + 1)): return N + 2 # If N+1 is Non-Fibinacci, then N+2 # will be next Non-Fibinacci 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 |
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C#
// 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 Fibinacci Number static bool isFibonacci(int N) { // N is Fibinacci if either // (5*N*N + 4), (5*N*N - 4) or both // is a perferct square return isPerfectSquare(5 * N * N + 4) || isPerfectSquare(5 * N * N - 4); } // Function to find // the next Non-Fibinacci Number static int nextNonFibonacci(int N) { // Case 1 // If N<=3, then 4 will be // next Non-Fibinacci Number if (N <= 3) return 4; // Case 2 // If N+1 is Fibinacci, then N+2 // will be next Non-Fibinacci Number if (isFibonacci(N + 1)) return N + 2; // If N+1 is Non-Fibinacci, then N+2 // will be next Non-Fibinacci 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 |
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Output:
4 6 9
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