# 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.

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.

Algorithm:

1. Define a function named checkIsFibonacci that takes an array of integers and its size as input.
2. Check if the size of the array is 1 or 2. If yes, return true as an array of 1 or 2 elements can always form a Fibonacci series.
3. Sort the array in ascending order using the sort() function from the algorithm header.
4. Traverse the sorted array from index 2 to n-1.
5. Check if the current element is equal to the sum of the previous two elements of the array. If not, return false.
6. If all the elements pass the above condition, return true.
7. In the main function:

a. Define an array of integers and its size.

b. Call the checkIsFibonacci() function with the array and its size as arguments.

c. If the function returns true, print “Yes” to the console. Otherwise, print “No”.

8. End of the program.

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[0]);`   `    ``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`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(N Log N)
Auxiliary Space: O(1)

Approach 2: Using Stacks;

Here’s how the stack checks if an array can form a Fibonacci series:

We start by pushing the first two elements of the array onto the stack.
Then, for each subsequent element in the array, we check if it is equal to the sum of the two elements on top of the stack.
If it is, we push the element onto the stack.
If it isn’t, we return false, indicating that the array cannot form a Fibonacci series.
If we reach the end of the array without returning false, we return true, indicating that the array can form a Fibonacci series.

## C++

 `#include ` `#include ` `#include `   `using` `namespace` `std;`   `bool` `checkIsFibonacci(``int` `arr[], ``int` `n) {`   `    ``if` `(n == 1 || n == 2) {` `        ``return` `true``;` `    ``}`   `    ``// Sort array` `    ``sort(arr, arr+n);`   `    ``// Use stack to check if every element is equal to the sum of previous 2 elements` `    ``stack<``int``> s;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(i < 2) {` `            ``s.push(arr[i]);` `        ``} ``else` `{` `            ``if` `(s.top() + s.size() - 2 == arr[i]) {` `                ``s.push(arr[i]);` `            ``} ``else` `{` `                ``return` `false``;` `            ``}` `        ``}` `    ``}`   `    ``return` `true``;` `}`   `int` `main() {`   `    ``int` `arr[] = {8, 3, 5, 13};` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);`   `    ``if` `(checkIsFibonacci(arr, n)) {` `        ``cout << ``"No"` `<< endl;` `    ``} ``else` `{` `        ``cout << ``"Yes"` `<< endl;` `    ``}`   `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `public` `class` `CheckFibonacci {` `    ``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"``);` `        ``}` `    ``}`   `    ``public` `static` `boolean` `checkIsFibonacci(``int``[] arr, ``int` `n) {` `        ``if` `(n == ``1` `|| n == ``2``) {` `            ``return` `true``;` `        ``}`   `        ``// Sort array` `        ``Arrays.sort(arr);`   `        ``// Use stack to check if every element is equal to the sum of previous 2 elements` `        ``Stack stack = ``new` `Stack<>();` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(i < ``2``) {` `                ``stack.push(arr[i]);` `            ``} ``else` `{` `                ``if` `(stack.peek() + stack.get(stack.size() - ``2``) == arr[i]) {` `                    ``stack.push(arr[i]);` `                ``} ``else` `{` `                    ``return` `false``;` `                ``}` `            ``}` `        ``}`   `        ``return` `true``;` `    ``}` `}`

## Python3

 `# Python3 program to check if the` `# elements of a given array` `# can form a Fibonacci Series` `# using stack`   `# 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()`   `    ``# Use stack to check if every` `    ``# element is equal to the` `    ``# sum of previous 2 elements`   `    ``stack ``=` `[]` `    ``for` `i ``in` `range``(n):` `        ``if` `i < ``2``:` `            ``stack.append(arr[i])` `        ``else``:` `            ``if` `stack[``-``1``] ``+` `stack[``-``2``] ``=``=` `arr[i]:` `                ``stack.append(arr[i])` `            ``else``:` `                ``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"``)`

## C#

 `// C# code addition `   `using` `System;` `using` `System.Collections.Generic;` `using` `System.Linq;`   `class` `Program {`   `  ``// Function to check whether the given array number are fibonnaci ` `  ``static` `bool` `CheckIsFibonacci(``int``[] arr, ``int` `n) {` `    ``if` `(n == 1 || n == 2) {` `      ``return` `true``;` `    ``}`   `    ``// Sort array` `    ``Array.Sort(arr);`   `    ``// Use stack to check if every element is equal to the sum of previous 2 elements` `    ``Stack<``int``> s = ``new` `Stack<``int``>();` `    ``for` `(``int` `i = 0; i < n; i++) {` `      ``if` `(i < 2) {` `        ``s.Push(arr[i]);` `      ``} ``else` `{` `        ``if` `(s.Peek() + s.ElementAt(s.Count - 2) == arr[i]) {` `          ``s.Push(arr[i]);` `        ``} ``else` `{` `          ``return` `false``;` `        ``}` `      ``}` `    ``}`   `    ``return` `true``;` `  ``}`   `  ``// Driver code. ` `  ``static` `void` `Main(``string``[] args) {` `    ``int``[] arr = {8, 3, 5, 13};` `    ``int` `n = arr.Length;`   `    ``if` `(CheckIsFibonacci(arr, n)) {` `      ``Console.WriteLine(``"No"``);` `    ``} ``else` `{` `      ``Console.WriteLine(``"Yes"``);` `    ``}` `  ``}` `}`   `// The code is contributed by Arushi Goel. `

## Javascript

 `// Javascript program to check if the` `// elements of a given array` `// can form a Fibonacci Series` `// using stack`   `// Returns true if a permutation` `// of arr[0..n-1] can form a` `// Fibonacci Series` `function` `checkIsFibonacci(arr, n) {` `    ``if` `(n == 1 || n == 2) {` `        ``return` `true``;` `    ``}`   `    ``// Sort array` `    ``arr.sort((a, b) => a - b);`   `    ``// Use stack to check if every` `    ``// element is equal to the` `    ``// sum of previous 2 elements`   `    ``let stack = [];` `    ``for` `(let i = 0; i < n; i++) {` `        ``if` `(i < 2) {` `            ``stack.push(arr[i]);` `        ``} ``else` `{` `            ``if` `(stack[stack.length - 1] + stack[stack.length - 2] == arr[i]) {` `                ``stack.push(arr[i]);` `            ``} ``else` `{` `                ``return` `false``;` `            ``}` `        ``}` `    ``}`   `    ``return` `true``;` `}`   `// Driver Code` `let arr = [8, 3, 5, 13];` `let n = arr.length;`   `if` `(checkIsFibonacci(arr, n)) {` `    ``console.log(``"Yes"``);` `} ``else` `{` `    ``console.log(``"No"``);` `}`   `// Contributed by adityasha4x71`

Output

`Yes`

Time Complexity: O(N Log N)
Auxiliary Space: O(N)

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