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:
- Define a function named checkIsFibonacci that takes an array of integers and its size as input.
- 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.
- Sort the array in ascending order using the sort() function from the algorithm header.
- Traverse the sorted array from index 2 to n-1.
- Check if the current element is equal to the sum of the previous two elements of the array. If not, return false.
- If all the elements pass the above condition, return true.
- 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 <bits/stdc++.h>using namespace std;// Returns true if a permutation// of arr[0..n-1] can form a// Fibonacci Seriesbool 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 Codeint 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 codepublic 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 Seriesdef 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 Codeif __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
<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> |
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 <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;} |
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<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
# 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 Seriesdef 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 Codeif __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 Seriesfunction 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 Codelet 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
