# Count of total subarrays whose sum is a Fibonacci Numbers

Given an array arr[] of N integers, the task is to count total number of subarrays whose sum is a Fibonacci Number.

Examples:

Input: arr[] = {6, 7, 8, 9}
Output: 3
Explanation:
The subarray whose sum is fibonacci numbers are:
1. {6, 7}, sum = 13 (5 + 8)
2. {6, 7, 8}, sum = 21 (8 + 13)
3. {8}, sum = 8 (3 + 5)

Input: arr[] = {1, 1, 1, 1}
Output: 4
Explanation:
The subarray whose sum is fibonacci numbers are:
1. {4, 2, 2}, sum = 8 (3 + 5)
2. {2}, sum = 2 (1 + 1)
3. {2}, sum = 2 (1 + 1)
4. {2}, sum = 2 (1 + 1)

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

Approach: The idea is generate all possible subarray and find the sum of each subarray. Now for each sum check whether it is fibonacci or not by using the approach discussed in this article. If Yes then, count all those sum and print the total count.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check whether a number ` `// is perfect square or not ` `bool` `isPerfectSquare(``int` `x) ` `{ ` `    ``int` `s = ``sqrt``(x); ` `    ``return` `(s * s == x); ` `} ` ` `  `// Function to check whether a number ` `// is fibonacci number or not ` `bool` `isFibonacci(``int` `n) ` `{ ` `    ``// If 5*n*n + 4 or 5*n*n - 5 is a ` `    ``// perfect square, then the number ` `    ``// is Fibonacci ` `    ``return` `isPerfectSquare(5 * n * n + 4) ` `           ``|| isPerfectSquare(5 * n * n - 4); ` `} ` ` `  `// Function to count the subarray with ` `// sum fibonacci number ` `void` `fibonacciSubarrays(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// Traverse the array arr[] to find ` `    ``// the sum of each subarray ` `    ``for` `(``int` `i = 0; i < n; ++i) { ` ` `  `        ``// To store the sum ` `        ``int` `sum = 0; ` ` `  `        ``for` `(``int` `j = i; j < n; ++j) { ` `            ``sum += arr[j]; ` ` `  `            ``// Check whether sum of subarray ` `            ``// between [i, j] is fibonacci ` `            ``// or not ` `            ``if` `(isFibonacci(sum)) { ` `                ``++count; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``cout << count; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 7, 8, 9 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Function Call ` `    ``fibonacciSubarrays(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` ` `  `class` `GFG{ ` ` `  `// Function to check whether a number ` `// is perfect square or not ` `static` `boolean` `isPerfectSquare(``int` `x) ` `{ ` `    ``int` `s = (``int``) Math.sqrt(x); ` `    ``return` `(s * s == x); ` `} ` ` `  `// Function to check whether a number ` `// is fibonacci number or not ` `static` `boolean` `isFibonacci(``int` `n) ` `{ ` `    ``// If 5*n*n + 4 or 5*n*n - 5 is a ` `    ``// perfect square, then the number ` `    ``// is Fibonacci ` `    ``return` `isPerfectSquare(``5` `* n * n + ``4``) ` `        ``|| isPerfectSquare(``5` `* n * n - ``4``); ` `} ` ` `  `// Function to count the subarray  ` `// with sum fibonacci number ` `static` `void` `fibonacciSubarrays(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `count = ``0``; ` ` `  `    ``// Traverse the array arr[] to find ` `    ``// the sum of each subarray ` `    ``for` `(``int` `i = ``0``; i < n; ++i) { ` ` `  `        ``// To store the sum ` `        ``int` `sum = ``0``; ` ` `  `        ``for` `(``int` `j = i; j < n; ++j) { ` `            ``sum += arr[j]; ` ` `  `            ``// Check whether sum of subarray ` `            ``// between [i, j] is fibonacci ` `            ``// or not ` `            ``if` `(isFibonacci(sum)) { ` `                ``++count; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``System.out.print(count); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``6``, ``7``, ``8``, ``9` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``// Function Call ` `    ``fibonacciSubarrays(arr, n); ` `} ` `} ` ` `  `// This code contributed by PrinciRaj1992 `

## Python3

 `# Python3 program for the above approach ` `import` `math ` ` `  `# Function to check whether a number ` `# is perfect square or not ` `def` `isPerfectSquare(x): ` `     `  `    ``s ``=` `int``(math.sqrt(x)) ` `    ``if` `s ``*` `s ``=``=` `x: ` `        ``return` `True` `    ``return` `False` ` `  `# Function to check whether a number ` `# is fibonacci number or not ` `def` `isFibonacci(n): ` `     `  `    ``# If 5*n*n + 4 or 5*n*n - 5 is a ` `    ``# perfect square, then the number ` `    ``# is Fibonacci ` `    ``return` `(isPerfectSquare(``5` `*` `n ``*` `n ``+` `4``) ``or`  `            ``isPerfectSquare(``5` `*` `n ``*` `n ``-` `4``)) ` ` `  `# Function to count the subarray with ` `# sum fibonacci number ` `def` `fibonacciSubarrays(arr, n): ` `     `  `    ``count ``=` `0` `     `  `    ``# Traverse the array arr[] to find ` `    ``# the sum of each subarray ` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# To store the sum ` `        ``sum` `=` `0` `         `  `        ``for` `j ``in` `range``(i, n): ` `            ``sum` `+``=` `arr[j] ` `             `  `            ``# Check whether sum of subarray ` `            ``# between [i, j] is fibonacci ` `            ``# or not ` `            ``if` `(isFibonacci(``sum``)): ` `                ``count ``+``=` `1` `                 `  `    ``print``(count) ` ` `  `# Driver Code ` `arr ``=` `[ ``6``, ``7``, ``8``, ``9` `] ` `n ``=` `len``(arr) ` ` `  `# Function Call ` `fibonacciSubarrays(arr, n) ` ` `  `# This code is contributed by SHUBHAMSINGH10 `

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to check whether a number ` `// is perfect square or not ` `static` `bool` `isPerfectSquare(``int` `x) ` `{ ` `    ``int` `s = (``int``) Math.Sqrt(x); ` `    ``return` `(s * s == x); ` `} ` ` `  `// Function to check whether a number ` `// is fibonacci number or not ` `static` `bool` `isFibonacci(``int` `n) ` `{ ` `    ``// If 5*n*n + 4 or 5*n*n - 5 is a ` `    ``// perfect square, then the number ` `    ``// is Fibonacci ` `    ``return` `isPerfectSquare(5 * n * n + 4) || ` `           ``isPerfectSquare(5 * n * n - 4); ` `} ` ` `  `// Function to count the subarray  ` `// with sum fibonacci number ` `static` `void` `fibonacciSubarrays(``int` `[]arr, ``int` `n) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// Traverse the array []arr to find ` `    ``// the sum of each subarray ` `    ``for``(``int` `i = 0; i < n; ++i)  ` `    ``{ ` `        `  `       ``// To store the sum ` `       ``int` `sum = 0; ` `       ``for``(``int` `j = i; j < n; ++j) ` `       ``{ ` `          ``sum += arr[j]; ` `           `  `          ``// Check whether sum of subarray ` `          ``// between [i, j] is fibonacci ` `          ``// or not ` `          ``if` `(isFibonacci(sum)) ` `          ``{ ` `              ``++count; ` `          ``} ` `       ``} ` `    ``} ` `    ``Console.Write(count); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 6, 7, 8, 9 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``// Function Call ` `    ``fibonacciSubarrays(arr, n); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```3
```

Time Complexity: O(N2)
, where N is the number of elements.

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