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Non Fibonacci Numbers

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  • Difficulty Level : Medium
  • Last Updated : 24 Jun, 2022

Given a positive integer n, the task is to print the nth non-Fibonacci number. The Fibonacci numbers are defined as: 

Fib(0) = 0
Fib(1) = 1
for n >1, Fib(n) = Fib(n-1) + Fib(n-2)

First few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ……..

Examples: 

Input : n = 2
Output : 6

Input : n = 5
Output : 10

A naive solution is to find Fibonacci numbers and then print the first n numbers not present in the found Fibonacci numbers. A better solution is to use the formula of Fibonacci numbers and keep adding of gap between two consecutive Fibonacci numbers. Value of sum of a gap is count of non-Fibonacci numbers seen so far. 

Below is the implementation of the above idea.
 

C++




// C++ program to find n'th Fibonacci number
 
#include <bits/stdc++.h>
using namespace std;
 
// Returns n'th Non-Fibonacci number
int nonFibonacci(int n)
{
    // curr is to keep track of current fibonacci
    // number, prev is previous, prevPrev is
    // previous of previous.
    int prevPrev = 1, prev = 2, curr = 3;
 
    // While count of non-fibonacci numbers
    // doesn't become negative or zero
    while (n > 0) {
        // Simple Fibonacci number logic
        prevPrev = prev;
        prev = curr;
        curr = prevPrev + prev;
 
        // (curr - prev - 1) is count of
        // non-Fibonacci numbers between curr
        // and prev.
        n = n - (curr - prev - 1);
    }
 
    // n might be negative now. Make sure it
    // becomes positive by removing last added
    // gap.
    n = n + (curr - prev - 1);
 
    // n must be now positive and less than or equal
    // to gap between current and previous, i.e.,
    // (curr - prev - 1);
 
    // Now add the positive n to previous Fibonacci
    // number to find the n'th non-fibonacci.
    return prev + n;
}
 
// Driver code
int main()
{
    cout << nonFibonacci(5);
    return 0;
}

C



// C program to find n'th Fibonacci number
 
#include<stdio.h>
 
 
// Returns n'th Non-Fibonacci number
int nonFibonacci(int n)
{
  // curr is to keep track of current fibonacci
  // number, prev is previous, prevPrev is
  // previous of previous.
  int prevPrev=1, prev=2, curr=3;
 
  // While count of non-fibonacci numbers
  // doesn't become negative or zero
  while (n > 0)
  {
    // Simple Fibonacci number logic
    prevPrev = prev;
    prev = curr;
    curr = prevPrev + prev;
 
    // (curr - prev - 1) is count of
    // non-Fibonacci numbers between curr
    // and prev.
    n = n - (curr - prev - 1);
  }
 
  // n might be negative now. Make sure it
  // becomes positive by removing last added
  // gap.
  n = n + (curr - prev - 1);
 
  // n must be now positive and less than or equal
  // to gap between current and previous, i.e.,
  // (curr - prev - 1);
 
  // Now add the positive n to previous Fibonacci
  // number to find the n'th non-fibonacci.
  return prev + n;
}
 
// Driver code
int main()
{
  printf("%d",nonFibonacci(5));
  return 0;
}
 
// This code is contributed by allwink45.
Java



// Java program to find
// n'th Fibonacci number
 
import java.io.*;
 
class GFG {
    // Returns n'th Non-
    // Fibonacci number
    static int nonFibonacci(int n)
    {
 
        // curr is to keep track of
        // current fibonacci number,
        // prev is previous, prevPrev
        // is previous of previous.
        int prevPrev = 1, prev = 2, curr = 3;
 
        // While count of non-fibonacci
        // numbers doesn't become
        // negative or zero
        while (n > 0) {
            // Simple Fibonacci number logic
            prevPrev = prev;
            prev = curr;
            curr = prevPrev + prev;
 
            // (curr - prev - 1) is count
            // of non-Fibonacci numbers
            // between curr and prev.
            n = n - (curr - prev - 1);
        }
 
        // n might be negative now. Make
        // sure it becomes positive by
        // removing last added gap.
        n = n + (curr - prev - 1);
 
        // n must be now positive and less
        // than or equal to gap between
        // current and previous, i.e.,
        // (curr - prev - 1);
 
        // Now add the positive n to
        // previous Fibonacci number
        // to find the n'th non-fibonacci.
        return prev + n;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        System.out.println(nonFibonacci(5));
    }
}
 
// This code is contributed by aj_36
Python



# Python program to find n'th
# Fibonacci number
 
# Returns n'th Non-Fibonacci
# number
 
 
def nonFibonacci(n):
 
    # curr is to keep track of
    # current fibonacci number,
    # prev is previous, prevPrev
    # is previous of previous.
    prevPrev = 1
    prev = 2
    curr = 3
 
    # While count of non-fibonacci
    # numbers doesn't become
    # negative or zero
    while n > 0:
        prevPrev = prev
        prev = curr
        curr = prevPrev + prev
 
        # (curr - prev - 1) is
        # count of non-Fibonacci
        # numbers between curr
        # and prev.
        n = n - (curr - prev - 1)
 
    # n might be negative now.
    # Make sure it becomes positive
    # by removing last added gap.
    n = n + (curr - prev - 1)
 
    # n must be now positive and
    # less than or equal to gap
    # between current and previous,
    # i.e., (curr - prev - 1)
 
    # Now add the positive n to
    # previous Fibonacci number to
    # find the n'th non-fibonacci.
    return prev + n
 
 
# Driver code
print(nonFibonacci(5))
 
# This code is contributed by anuj_67.
C#



// C# program to find
// n'th Fibonacci number
 
using System;
 
class GFG
{
    // Returns n'th Non-
    // Fibonacci number
    static int nonFibonacci (int n)
    {
         
    // curr is to keep track of
    // current fibonacci number,
    // prev is previous, prevPrev
    // is previous of previous.
    int prevPrev = 1, prev = 2, curr = 3;
 
    // While count of non-fibonacci
    // numbers doesn't become
    // negative or zero
    while (n > 0)
    {
        // Simple Fibonacci number logic
        prevPrev = prev;
        prev = curr;
        curr = prevPrev + prev;
 
        // (curr - prev - 1) is count
        // of non-Fibonacci numbers
        // between curr and prev.
        n = n - (curr - prev - 1);
    }
 
    // n might be negative now. Make
    // sure it becomes positive by
    // removing last added gap.
    n = n + (curr - prev - 1);
 
    // n must be now positive and less
    // than or equal to gap between 
    // current and previous, i.e.,
    // (curr - prev - 1);
 
    // Now add the positive n to
    // previous Fibonacci number
    // to find the n'th non-fibonacci.
    return prev + n;
    }
     
    // Driver Code
    public static void Main ()
    {
    Console.WriteLine (nonFibonacci(5));
    }
}
 
//This code is contributed by aj_36
PHP



<?php
// PHP program to find
// n'th Fibonacci number
 
// Returns n'th Non-
// Fibonacci number
function nonFibonacci($n)
{
    // curr is to keep track of
    // current fibonacci number,
    // prev is previous, prevPrev
    // is previous of previous.
    $prevPrev = 1;
    $prev = 2;
    $curr = 3;
 
    // While count of non-fibonacci
    // numbers doesn't become
    // negative or zero
    while ($n > 0)
    {
        // Simple Fibonacci
        // number logic
        $prevPrev = $prev;
        $prev = $curr;
        $curr = $prevPrev + $prev;
 
        // (curr - prev - 1) is count
        // of non-Fibonacci numbers
        // between curr and prev.
        $n = $n - ($curr - $prev - 1);
    }
 
    // n might be negative now. Make
    // sure it becomes positive by
    // removing last added gap.
    $n = $n + ($curr - $prev - 1);
 
    // n must be now positive and
    // less than or equal to gap
    // between current and previous, 
    // i.e., (curr - prev - 1);
 
    // Now add the positive n to
    // previous Fibonacci number
    // to find the n'th non-fibonacci.
    return $prev + $n;
}
 
// Driver code
echo nonFibonacci(5);
 
// This code is contributed by m_kit
?>
Javascript



<script>
 
// Javascript program to find n'th Fibonacci number
 
// Returns n'th Non-Fibonacci number
function nonFibonacci(n)
{
    // curr is to keep track of current fibonacci
    // number, prev is previous, prevPrev is
    // previous of previous.
    let prevPrev=1, prev=2, curr=3;
 
    // While count of non-fibonacci numbers
    // doesn't become negative or zero
    while (n > 0)
    {
        // Simple Fibonacci number logic
        prevPrev = prev;
        prev = curr;
        curr = prevPrev + prev;
 
        // (curr - prev - 1) is count of
        // non-Fibonacci numbers between curr
        // and prev.
        n = n - (curr - prev - 1);
    }
 
    // n might be negative now. Make sure it
    // becomes positive by removing last added
    // gap.
    n = n + (curr - prev - 1);
 
    // n must be now positive and less than or equal
    // to gap between current and previous, i.e.,
    // (curr - prev - 1);
 
    // Now add the positive n to previous Fibonacci
    // number to find the n'th non-fibonacci.
    return prev + n;
}
 
// Driver code
 
    document.write(nonFibonacci(5));
 
// This code is contributed by Mayank Tyagi
 
</script>
Output : 
10
Time Complexity : O(n) , Auxiliary Space : O(1)
Now geeks you must be wondering what if we were supposed to print Non-Fibonacci Series in a range, then the code is as follows: 
C++



#include <iostream>
using namespace std;
 
int main()
{
    int i = 0, j = 1, k, m, no, b[10];
 
    // Range is 10
    no = 10;
    b[1] = 0;
    b[2] = 1;
 
    // Check if range is less equals to 1
    if (no <= 1) {
        cout << "You have enter a wrong range";
    }
 
    // check if range is greater than 1
    // and less equals to 5
    else if (no <= 5 && no > 1) {
        cout << "\nThere is not any Non-Fibonacci series "
                "that lies between 1 to "
             << no << " term of Fibonacci Series.";
    }
 
    // If range is greater than 5
    else {
 
        // Loop to calculate fibonacci series till
        // range
        for (m = 2; m < no; m++) {
            k = i + j;
            i = j;
            j = k;
 
            // Store fibonacci series into b[]
            // array
            b[m] = k;
        }
        i = 5;
        cout << "\nThe Non-Fibonacci series that lies "
                "between 1 to "
             << no << " term of Fibonacci Series is: \n";
 
        // Loop to calculate Non-Fibonacci
        // series
        for (int ans = 4; ans < b[no - 1]; ans++) {
            if (ans != b[i])
 
                // Print Non-Fibonacci Series
                cout << ans << "  ";
            else
                i++;
        }
    }
    return 0;
}
C



#include <stdio.h>
#include <stdlib.h>
int main()
{
    int i = 0, j = 1, k, m, no, b[10];
 
    // Range is 10
    no = 10;
    b[1] = 0;
    b[2] = 1;
 
    // Check if range is less equals to 1
    if (no <= 1) {
        printf("You have enter a wrong range");
    }
 
    // check if range is greater than 1 and less equals to 5
    else if (no <= 5 && no > 1) {
        printf("\nThere is not any Non-Fibonacci series "
               "that lies between 1 to %d term of "
               "Fibonacci Series.",
               no);
    }
 
    // If range is greater than 5
    else {
 
        // Loop to calculate fibonacci series till range
        for (m = 2; m < no; m++) {
            k = i + j;
            i = j;
            j = k;
 
            // Store fibonacci series into b[] array
            b[m] = k;
        }
 
        i = 5;
        printf(
            "\nThe Non-Fibonacci series that lies between "
            "1 to %d term of Fibonacci Series is: \n",
            no);
 
        // Loop to calculate Non-Fibonacci series
        for (int ans = 4; ans < b[no - 1]; ans++) {
            if (ans != b[i])
 
                // Print Non-Fibonacci Series
                printf("%d  ", ans);
            else
                i++;
        }
    }
    return 0;
}
Java



/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    int[] holes = {21, 3, 6};
    int i = 0, j = 1, k, m, no;
    int[] b = new int[10];
 
    // Range is 10
    no = 10;
    b[1] = 0;
    b[2] = 1;
 
    // Check if range is less equals to 1
    if (no <= 1) {
        System.out.print("You have enter a wrong range");
    }
 
    // check if range is greater than 1
    // and less equals to 5
    else if (no <= 5 && no > 1) {
        System.out.print("\n" + "There is not any Non-Fibonacci series that lies between 1 to" + no +
        " term of Fibonacci Series.");
    }
 
    // If range is greater than 5
    else {
 
        // Loop to calculate fibonacci series till
        // range
        for (m = 2; m < no; m++) {
            k = i + j;
            i = j;
            j = k;
 
            // Store fibonacci series into b[]
            // array
            b[m] = k;
        }
        i = 5;
        System.out.println("\n" + "The Non-Fibonacci series that lies between 1 to "
             + no + " term of Fibonacci Series is: "+ "\n");
 
        // Loop to calculate Non-Fibonacci
        // series
        for (int ans = 4; ans < b[no - 1]; ans++) {
            if (ans != b[i])
 
                // Print Non-Fibonacci Series
                System.out.print(ans + "  ");
            else
                i++;
        }
    }
}
 
// This Solution is contributed by shinjanpatra.
Python3



i = 0
j = 1
b = []
no = 10  # Range is 10
b.append(0)
b.append(1)
if(no <= 1):  # Check if range is less equals to 1
    print("You have enter a wrong range...")
elif(no <= 5 and no > 1):  # check if range is greater than 1 and less equals to 5
    print("\nThere is not any Non-Fibonacci series that lies between 1 to ",
          no, " term of Fibonacci Series.")
else:  # If range is greater than 5
    for m in range(2, no):  # Loop to calculate fibonacci series till range
        k = i+j
        i = j
        j = k
        b.append(k)  # Store fibonacci series into list b
    i = 5
    print("\nThe Non-Fibonacci series that lies between 1 to ",
          no, " term of Fibonacci Series is:")
    for ans in range(4, b[no-1]):  # Loop to calculate Non-Fibonacci series
        if ans != b[i]:
            print(ans, end="  ")  # Print Non-Fibonacci Series
        else:
            i = i+1
Javascript



<script>
 
 
// driver code
 
let i = 0, j = 1, k, m, no, b = new Array(10);
 
// Range is 10
no = 10;
b[1] = 0;
b[2] = 1;
 
// Check if range is less equals to 1
if (no <= 1) {
    console.log("You have enter a wrong range");
}
 
// check if range is greater than 1
// and less equals to 5
else if (no <= 5 && no > 1) {
    document.write("</br>","There is not any Non-Fibonacci series that lies between 1 to " + no + " term of Fibonacci Series.");
}
 
// If range is greater than 5
else {
 
    // Loop to calculate fibonacci series till
    // range
    for (m = 2; m < no; m++) {
        k = i + j;
        i = j;
        j = k;
 
        // Store fibonacci series into b[]
        // array
        b[m] = k;
    }
    i = 5;
    document.write("</br>","The Non-Fibonacci series that lies between 1 to " + no + " term of Fibonacci Series is: ","</br>");
 
        // Loop to calculate Non-Fibonacci
        // series
    for (let ans = 4; ans < b[no - 1]; ans++) {
        if (ans != b[i])
 
            // Print Non-Fibonacci Series
            document.write(ans , "  ");
        else
            i++;
    }
}
 
// This code is contributed by shinjanpatra
 
</script>
Output
The Non-Fibonacci series that lies between 1 to 10 term of Fibonacci Series is: 
4  6  7  9  10  11  12  14  15  16  17  18  19  20  22  23  24  25  26  27  28  29  30  31  32  33  
Time Complexity : O(n) , Auxiliary Space : O(n)
The above problem and solution are contributed by Hemang Sarkar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above. 

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