Skip to content
Related Articles

Related Articles

Improve Article
Count of consecutive Fibonacci pairs in the given Array
  • Last Updated : 08 Apr, 2021

Given an array arr[], the task is to count the number of consecutive Fibonacci pairs in this array.
Examples: 
 

Input: arr[] = { 3, 5, 6, 11 } 
Output:
The only pair is (3, 5) which is consecutive fibonacci pair in the array
Input: arr[] = { 3, 5, 8, 11 } 
Output:
There are two pairs (3, 5) and (5, 8) in the array, which is consecutive Fibonacci pair. 
 

 

Approach: 
 

  • Find out all the pairs of the array.
  • For each pair, 
    • Find the minimum and maximum of the pair
    • Find the next consective fibonacci number after minimum_element and check that it is equal to the maximum of the pair.
    • If the next consecutive fibonacci number is equal to the maximum element of the pair, then increment the count by 1.
  • Return the total count as the required number of pairs.

Below is the implementation of the above approach:
 



C++




// C++ implementation to count the
// consecutive fibonacci pairs in the array
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the previous
// fibonacci for the number N
int previousFibonacci(int n)
{
    double a = n / ((1 + sqrt(5)) / 2.0);
    return round(a);
}
 
// Function to find the next
// fibonacci number for the number N
int nextFibonacci(int n)
{
    double a = n * (1 + sqrt(5)) / 2.0;
    return round(a);
}
 
// Function to check that a Number
// is a perfect square or not
bool isPerfectSquare(int x)
{
    int s = sqrt(x);
    return (s * s == x);
}
 
// Function to check that a number
// is fibonacci number or not
bool isFibonacci(int n)
{
    // N is Fibinacci if one of
    // (5*n*n + 4) or (5*n*n - 4)
    // is a perferct square
    return (isPerfectSquare(5 * n * n + 4)
            || isPerfectSquare(5 * n * n - 4));
}
 
// Function to count the fibonacci
// pairs in the array
int countFibonacciPairs(int arr[], int n)
{
    int res = 0;
 
    // Loop to iterate over the array
    // to choose all pairs of the array
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
 
            // Condition to check if both
            // the number of pair is a
            // fibonacci number
            if (isFibonacci(arr[i])
                && isFibonacci(arr[j])) {
 
                int prevFib = previousFibonacci(arr[i]);
                int nextFib = nextFibonacci(arr[i]);
 
                // Condition to check if both
                // the number form consecutive
                // fibonacci numbers
                if (prevFib == arr[j]
                    || nextFib == arr[j]) {
                    res++;
                }
            }
 
    return res;
}
 
// Driver Code
int main()
{
    int a[] = { 3, 5, 8, 11 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << countFibonacciPairs(a, n);
    return 0;
}

Java




// Java implementation to count the
// consecutive fibonacci pairs in the array
import java.util.*;
import java.lang.*;
 
class GFG
{
  
// Function to find the previous
// fibonacci for the number N
static int previousFibonacci(int n)
{
    double a = n / ((1 + (int)Math.sqrt(5)) / 2.0);
    return (int)Math.round(a);
}
  
// Function to find the next
// fibonacci number for the number N
static int nextFibonacci(int n)
{
    double a = n * (1 + (int)Math.sqrt(5)) / 2.0;
    return (int)Math.round(a);
}
  
// Function to check that a Number
// is a perfect square or not
static boolean isPerfectSquare(int x)
{
    int s = (int)Math.sqrt(x);
    return (s * s == x);
}
  
// Function to check that a number
// is fibonacci number or not
static boolean isFibonacci(int n)
{
    // N is Fibinacci if one of
    // (5*n*n + 4) or (5*n*n - 4)
    // is a perferct square
    return (isPerfectSquare(5 * n * n + 4)
            || isPerfectSquare(5 * n * n - 4));
}
  
// Function to count the fibonacci
// pairs in the array
static int countFibonacciPairs(int arr[], int n)
{
    int res = 0;
  
    // Loop to iterate over the array
    // to choose all pairs of the array
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
  
            // Condition to check if both
            // the number of pair is a
            // fibonacci number
            if (isFibonacci(arr[i])
                && isFibonacci(arr[j])) {
  
                int prevFib = previousFibonacci(arr[i]);
                int nextFib = nextFibonacci(arr[i]);
  
                // Condition to check if both
                // the number form consecutive
                // fibonacci numbers
                if (prevFib == arr[j]
                    || nextFib == arr[j]) {
                    res++;
                }
            }
  
    return res;
}
  
// Driver Code
public static void main(String []args)
{
    int []a = { 3, 5, 8, 11 };
    int n = a.length;
    System.out.print(countFibonacciPairs(a, n));   
}
}
 
// This code is contributed by chitranayal

Python3




# Python3 implementation to count the
# consecutive fibonacci pairs in the array
from math import sqrt
 
# Function to find the previous
# fibonacci for the number N
def previousFibonacci(n):
 
    a = n / ((1 + sqrt(5)) / 2.0)
    return round(a)
 
# Function to find the next
# fibonacci number for the number N
def nextFibonacci(n):
 
    a = n * (1 + sqrt(5)) / 2.0
    return round(a)
 
# Function to check that a Number
# is a perfect square or not
def isPerfectSquare(x):
 
    s = sqrt(x)
    return (s * s == x)
 
# Function to check that a number
# is fibonacci number or not
def isFibonacci(n):
 
    # N is Fibinacci if one of
    # (5*n*n + 4) or (5*n*n - 4)
    # is a perferct square
    return (isPerfectSquare(5 * n * n + 4)
            or isPerfectSquare(5 * n * n - 4))
 
# Function to count the fibonacci
# pairs in the array
def countFibonacciPairs(arr, n):
 
    res = 0
 
    # Loop to iterate over the array
    # to choose all pairs of the array
    for i in range(n):
        for j in range(i+1,n):
 
            # Condition to check if both
            # the number of pair is a
            # fibonacci number
            if (isFibonacci(arr[i])
                and isFibonacci(arr[j])):
 
                prevFib = previousFibonacci(arr[i])
                nextFib = nextFibonacci(arr[i])
 
                # Condition to check if both
                # the number form consecutive
                # fibonacci numbers
                if (prevFib == arr[j]
                    or nextFib == arr[j]):
                    res += 1
 
    return res
 
# Driver Code
a = [3, 5, 8, 11]
n = len(a)
print(countFibonacciPairs(a, n))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation to count the
// consecutive fibonacci pairs in the array
using System;
 
class GFG
{
   
// Function to find the previous
// fibonacci for the number N
static int previousFibonacci(int n)
{
    double a = n / ((1 + (int)Math.Sqrt(5)) / 2.0);
    return (int)Math.Round(a);
}
   
// Function to find the next
// fibonacci number for the number N
static int nextFibonacci(int n)
{
    double a = n * (1 + Math.Sqrt(5)) / 2.0;
    return (int)Math.Round(a);
}
   
// Function to check that a Number
// is a perfect square or not
static bool isPerfectSquare(int x)
{
    int s = (int)Math.Sqrt(x);
    return (s * s == x);
}
   
// Function to check that a number
// is fibonacci number or not
static bool isFibonacci(int n)
{
    // N is Fibinacci if one of
    // (5*n*n + 4) or (5*n*n - 4)
    // is a perferct square
    return (isPerfectSquare(5 * n * n + 4)
            || isPerfectSquare(5 * n * n - 4));
}
   
// Function to count the fibonacci
// pairs in the array
static int countFibonacciPairs(int []arr, int n)
{
    int res = 0;
   
    // Loop to iterate over the array
    // to choose all pairs of the array
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
   
            // Condition to check if both
            // the number of pair is a
            // fibonacci number
            if (isFibonacci(arr[i])
                && isFibonacci(arr[j])) {
   
                int prevFib = previousFibonacci(arr[i]);
                int nextFib = nextFibonacci(arr[i]);
   
                // Condition to check if both
                // the number form consecutive
                // fibonacci numbers
                if (prevFib == arr[j]
                    || nextFib == arr[j]) {
                    res++;
                }
            }
   
    return res;
}
   
// Driver Code
public static void Main(String []args)
{
    int []a = { 3, 5, 8, 11 };
    int n = a.Length;
    Console.Write(countFibonacciPairs(a, n));   
}
}
 
// This code is contributed by sapnasingh4991

Javascript




<script>
 
// Javascript implementation to count the
// consecutive fibonacci pairs in the array
 
// Function to find the previous
// fibonacci for the number N
function previousFibonacci(n)
{
    var a = n / ((1 + Math.sqrt(5)) / 2.0);
    return Math.round(a);
}
 
// Function to find the next
// fibonacci number for the number N
function nextFibonacci(n)
{
    var a = n * (1 + Math.sqrt(5)) / 2.0;
    return Math.round(a);
}
 
// Function to check that a Number
// is a perfect square or not
function isPerfectSquare(x)
{
    var s = Math.sqrt(x);
    return (s * s == x);
}
 
// Function to check that a number
// is fibonacci number or not
function isFibonacci(n)
{
    // N is Fibinacci if one of
    // (5*n*n + 4) or (5*n*n - 4)
    // is a perferct square
    return (isPerfectSquare(5 * n * n + 4)
            || isPerfectSquare(5 * n * n - 4));
}
 
// Function to count the fibonacci
// pairs in the array
function countFibonacciPairs(arr, n)
{
    var res = 0;
 
    // Loop to iterate over the array
    // to choose all pairs of the array
    for (var i = 0; i < n; i++)
        for (var j = i + 1; j < n; j++)
 
            // Condition to check if both
            // the number of pair is a
            // fibonacci number
            if (isFibonacci(arr[i])
                && isFibonacci(arr[j])) {
 
                var prevFib = previousFibonacci(arr[i]);
                var nextFib = nextFibonacci(arr[i]);
 
                // Condition to check if both
                // the number form consecutive
                // fibonacci numbers
                if (prevFib == arr[j]
                    || nextFib == arr[j]) {
                    res++;
                }
            }
 
    return res;
}
 
// Driver Code
var a = [ 3, 5, 8, 11 ];
var n = a.length;
document.write(countFibonacciPairs(a, n));
 
</script>
Output: 
2

 

Performance Analysis: 
 

  • Time Complexity: As in the above approach, there are two nested loops which takes O(N2) time, Hence the Time Complexity will be O(N2).
  • Space Complexity: As in the above approach, there is no extra space used, Hence the space complexity will be O(1).

 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :