# Check whether Array represents a Fibonacci Series or not

Given an array arr[] consisting of N integers, the task is to check whether a Fibonacci series can be formed using all the array elements or not. If possible, print “Yes”. Otherwise, print “No”.

Examples:

Input: arr[] = { 8, 3, 5, 13 }
Output: Yes
Explanation:
Rearrange given array as {3, 5, 8, 13} and these numbers form Fibonacci series.

Input: arr[] = { 2, 3, 5, 11 }
Output: No
Explanation:
The given array elements do not form a Fibonacci series.

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

Approach:
In order to solve the problem mentioned above, the main idea is to sort the given array. After sorting, check if every element is equal to the sum of the previous 2 elements. If so, then the array elements form a Fibonacci series.

Below is the implementation of the above approach:

## C++

 `// C++ program to check if the ` `// elements of a given array ` `// can form a Fibonacci Series ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Returns true if a permutation ` `// of arr[0..n-1] can form a ` `// Fibonacci Series ` `bool` `checkIsFibonacci(``int` `arr[], ``int` `n) ` `{ ` `    ``if` `(n == 1 || n == 2) ` `        ``return` `true``; ` ` `  `    ``// Sort array ` `    ``sort(arr, arr + n); ` ` `  `    ``// After sorting, check if every ` `    ``// element is equal to the ` `    ``// sum of previous 2 elements ` ` `  `    ``for` `(``int` `i = 2; i < n; i++) ` `        ``if` `((arr[i - 1] + arr[i - 2]) ` `            ``!= arr[i]) ` `            ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 8, 3, 5, 13 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``if` `(checkIsFibonacci(arr, n)) ` `        ``cout << ``"Yes"` `<< endl; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to check if the elements of  ` `// a given array can form a Fibonacci Series  ` `import` `java. util. Arrays; ` ` `  `class` `GFG{ ` `     `  `// Returns true if a permutation  ` `// of arr[0..n-1] can form a  ` `// Fibonacci Series  ` `public` `static` `boolean` `checkIsFibonacci(``int` `arr[],  ` `                                       ``int` `n)  ` `{  ` `    ``if` `(n == ``1` `|| n == ``2``)  ` `        ``return` `true``;  ` `     `  `    ``// Sort array  ` `    ``Arrays.sort(arr); ` `     `  `    ``// After sorting, check if every  ` `    ``// element is equal to the sum  ` `    ``// of previous 2 elements  ` `    ``for``(``int` `i = ``2``; i < n; i++) ` `    ``{ ` `       ``if` `((arr[i - ``1``] + arr[i - ``2``]) != arr[i])  ` `           ``return` `false``;  ` `    ``}  ` `    ``return` `true``;  ` `}  ` `     `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = { ``8``, ``3``, ``5``, ``13` `};  ` `    ``int` `n = arr.length;  ` `     `  `    ``if` `(checkIsFibonacci(arr, n))  ` `        ``System.out.println(``"Yes"``); ` `    ``else` `        ``System.out.println(``"No"``);  ` `} ` `} ` ` `  `// This code is contributed by divyeshrabadiya07 `

## Python3

 `# Python3 program to check if the ` `# elements of a given array ` `# can form a Fibonacci Series ` ` `  `# Returns true if a permutation ` `# of arr[0..n-1] can form a ` `# Fibonacci Series ` `def` `checkIsFibonacci(arr, n) : ` ` `  `    ``if` `(n ``=``=` `1` `or` `n ``=``=` `2``) : ` `        ``return` `True``; ` ` `  `    ``# Sort array ` `    ``arr.sort() ` ` `  `    ``# After sorting, check if every ` `    ``# element is equal to the ` `    ``# sum of previous 2 elements ` ` `  `    ``for` `i ``in` `range``(``2``, n) : ` `        ``if` `((arr[i ``-` `1``] ``+`  `             ``arr[i ``-` `2``])!``=` `arr[i]) : ` `            ``return` `False``; ` ` `  `    ``return` `True``; ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` `  `    ``arr ``=` `[ ``8``, ``3``, ``5``, ``13` `]; ` `    ``n ``=` `len``(arr); ` ` `  `    ``if` `(checkIsFibonacci(arr, n)) : ` `        ``print``(``"Yes"``); ` `    ``else` `: ` `        ``print``(``"No"``); ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# program to check if the elements of  ` `// a given array can form a fibonacci series  ` `using` `System; ` ` `  `class` `GFG{  ` `     `  `// Returns true if a permutation  ` `// of arr[0..n-1] can form a  ` `// fibonacci series  ` `public` `static` `bool` `checkIsFibonacci(``int` `[]arr,  ` `                                    ``int` `n)  ` `{  ` `    ``if` `(n == 1 || n == 2)  ` `        ``return` `true``;  ` `         `  `    ``// Sort array  ` `    ``Array.Sort(arr);  ` `         `  `    ``// After sorting, check if every  ` `    ``// element is equal to the sum  ` `    ``// of previous 2 elements  ` `    ``for``(``int` `i = 2; i < n; i++)  ` `    ``{  ` `       ``if` `((arr[i - 1] + arr[i - 2]) != arr[i])  ` `           ``return` `false``;  ` `    ``}  ` `    ``return` `true``;  ` `}  ` `         `  `// Driver code  ` `public` `static` `void` `Main(``string``[] args)  ` `{  ` `    ``int` `[]arr = { 8, 3, 5, 13 };  ` `    ``int` `n = arr.Length;  ` `         `  `    ``if` `(checkIsFibonacci(arr, n))  ` `        ``Console.WriteLine(``"Yes"``);  ` `    ``else` `        ``Console.WriteLine(``"No"``);  ` `}  ` `}  ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```Yes
```

Time Complexity: O(N Log N)

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.