Check if the sum of perfect squares in an array is divisible by x

Given an array arr[] and an integer x, the task is to check whether the sum of all the perfect squares from the array is divisible by x or not. If divisible then print Yes else print No.

Examples:

Input: arr[] = {2, 3, 4, 6, 9, 10}, x = 13
Output: Yes
4 and 9 are the only perfect squares from the array
sum = 4 + 9 = 13 (which is divisible by 13)

Input: arr[] = {2, 4, 25, 49, 3, 8}, x = 9
Output: No



Approach: Run a loop from i to n – 1 and check whether arr[i] is a perfect square or not. If arr[i] is a perfect square then update sum = sum + arr[i]. If in the end sum % x = 0 then print Yes else print No. To check whether an element is a perfect square or not, follow the following steps:

Let num be an integer element
float sq = sqrt(x)
if floor(sq) = ceil(sq) then num is a perfect square else not.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if the the sum of all the
// perfect squares of the given array is divisible by x
bool check(int arr[], int x, int n)
{
    long long sum = 0;
    for (int i = 0; i < n; i++) {
        double x = sqrt(arr[i]);
  
        // If arr[i] is a perfect square
        if (floor(x) == ceil(x)) {
            sum += arr[i];
        }
    }
  
    if (sum % x == 0)
        return true;
    else
        return false;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 3, 4, 9, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 13;
  
    if (check(arr, x, n)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the appraoch 
public class GFG{
  
    // Function that returns true if the the sum of all the 
    // perfect squares of the given array is divisible by x 
    static boolean check(int arr[], int x, int n) 
    
        long sum = 0
        for (int i = 0; i < n; i++) { 
            double y = Math.sqrt(arr[i]); 
      
            // If arr[i] is a perfect square 
            if (Math.floor(y) == Math.ceil(y)) { 
                sum += arr[i]; 
            
        
      
        if (sum % x == 0
            return true
        else
            return false
    
  
  
  
    // Driver Code 
    public static void main(String []args){
        int arr[] = { 2, 3, 4, 9, 10 }; 
        int n = arr.length ;
        int x = 13
  
        if (check(arr, x, n)) { 
            System.out.println("Yes"); 
        
        else
           System.out.println("No"); 
        
    }
    // This code is contributed by Ryuga
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
import math
  
# Function that returns true if the the sum of all the 
# perfect squares of the given array is divisible by x
def check (a, y):
    sum = 0
    for i in range(len(a)):
          
        x = math.sqrt(a[i])
  
        # If a[i] is a perfect square
        if (math.floor(x) == math.ceil(x)):
            sum = sum + a[i]
      
    if (sum % y == 0):
        return True
    else:
        return False
          
  
# Driver code
a = [2, 3, 4, 9, 10]
x = 13
  
if check(a, x) :
    print("Yes")
else:
    print("No")

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the appraoch 
  
using System;
public class GFG{
   
    // Function that returns true if the the sum of all the 
    // perfect squares of the given array is divisible by x 
    static bool check(int[] arr, int x, int n) 
    
        long sum = 0; 
        for (int i = 0; i < n; i++) { 
            double y = Math.Sqrt(arr[i]); 
       
            // If arr[i] is a perfect square 
            if (Math.Floor(y) == Math.Ceiling(y)) { 
                sum += arr[i]; 
            
        
       
        if (sum % x == 0) 
            return true
        else
            return false
    
   
   
   
    // Driver Code 
    public static void Main(){
        int[] arr = { 2, 3, 4, 9, 10 }; 
        int n = arr.Length ;
        int x = 13; 
   
        if (check(arr, x, n)) { 
            Console.Write("Yes"); 
        
        else
           Console.Write("No"); 
        
    }    
}

chevron_right


PHP

Output:

Yes


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.



Improved By : Ryuga, Ita_c, Sach_Code