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Count of total subarrays whose sum is a Fibonacci Numbers

  • Last Updated : 13 Apr, 2021
Geek Week

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:
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:
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) 
 

 

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 <bits/stdc++.h>
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[0]);
 
    // 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

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to check whether a number
// is perfect square or not
function isPerfectSquare(x)
{
    var s = parseInt(Math.sqrt(x));
    return (s * s == x);
}
 
// Function to check whether a number
// is fibonacci number or not
function 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)
           || isPerfectSquare(5 * n * n - 4);
}
 
// Function to count the subarray with
// sum fibonacci number
function fibonacciSubarrays(arr, n)
{
    var count = 0;
 
    // Traverse the array arr[] to find
    // the sum of each subarray
    for (var i = 0; i < n; ++i) {
 
        // To store the sum
        var sum = 0;
 
        for (var j = i; j < n; ++j) {
            sum += arr[j];
 
            // Check whether sum of subarray
            // between [i, j] is fibonacci
            // or not
            if (isFibonacci(sum)) {
                ++count;
            }
        }
    }
 
    document.write( count);
}
 
// Driver Code
 
var arr = [ 6, 7, 8, 9 ];
var n = arr.length;
 
// Function Call
fibonacciSubarrays(arr, n);
 
</script>
Output: 



3

 

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

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