# 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); ` `    ``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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