# Count of consecutive Fibonacci pairs in the given Array

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: 1
The only pair is (3, 5) which is consecutive fibonacci pair in the array

Input: arr[] = { 3, 5, 8, 11 }
Output: 2
There are two pairs (3, 5) and (5, 8) in the array, which is consecutive Fibonacci pair.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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

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).

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