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++ program to check if the
// elements of a given array
// can form a Fibonacci Series

#include <bits/stdc++.h>
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 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 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# 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

<script>

// Javascript 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
    function checkIsFibonacci(arr , n) 
    {
        if (n == 1 || n == 2)
            return true;

        // Sort array
        arr.sort((a, b) => a - b);

        // After sorting, check if every
        // element is equal to the sum
        // of previous 2 elements
        for (i = 2; i < n; i++) {
            if ((arr[i - 1] + arr[i - 2]) != arr[i])
                return false;
        }
        return true;
    }

    // Driver code
    
        var arr = [ 8, 3, 5, 13 ];
        var n = arr.length;

        if (checkIsFibonacci(arr, n))
            document.write("Yes");
        else
            document.write("No");

// This code contributed by umadevi9616 

</script>

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.

#include <iostream>
#include <stack>
#include <algorithm>

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;
}

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<Integer> 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 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# 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 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

Yes

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

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Article Tags :

• Mathematical
• DSA
• Arrays
• Fibonacci
• series

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Practice Tags :

• Arrays
• Fibonacci
• Mathematical
• series