Count array elements that divide the sum of all other elements

Given an array arr[], the task is to count the number of elements from the array which divide the sum of all other elements.

Examples:

Input: arr[] = {3, 10, 4, 6, 7}
Output: 3
3 divides (10 + 4 + 6 + 7) i.e. 27
10 divides (3 + 4 + 6 + 7) i.e. 20
6 divides (3 + 10 + 4 + 7) i.e. 24

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Output: 2



Naive Approach: Run two loop from 0 to N, calculate sum of all elements except the current element and if this element divide that sum then increment the count.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Fuction to return the count
// of the required numbers
int countNum(int N, int arr[])
{
    // To store the count of required numbers
    int count = 0;
    for (int i = 0; i < N; i++) {
  
        // Initialize sum to 0
        int sum = 0;
        for (int j = 0; j < N; j++) {
  
            // If current element and the
            // chosen element are same
            if (i == j)
                continue;
  
            // Add all other numbers of array
            else
                sum += arr[j];
        }
  
        // If sum is divisible by the chosen element
        if (sum % arr[i] == 0)
            count++;
    }
  
    // Return the count
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 3, 10, 4, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countNum(n, arr);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
      
// Fuction to return the count
// of the required numbers
static int countNum(int N, int arr[])
{
    // To store the count of required numbers
    int count = 0;
    for (int i = 0; i < N; i++) 
    {
  
        // Initialize sum to 0
        int sum = 0;
        for (int j = 0; j < N; j++)
        {
  
            // If current element and the
            // chosen element are same
            if (i == j)
                continue;
  
            // Add all other numbers of array
            else
                sum += arr[j];
        }
  
        // If sum is divisible by the chosen element
        if (sum % arr[i] == 0)
            count++;
    }
  
    // Return the count
    return count;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 10, 4, 6, 7 };
    int n = arr.length;
    System.out.println(countNum(n, arr));
}
}
  
// This code is contributed by Code_Mech

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Python3

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# Python3 implementation of the approach
  
# Fuction to return the count
# of the required numbers
def countNum(N, arr):
  
    # To store the count of 
    # required numbers
    count = 0
  
    for i in range(N):
  
        # Initialize sum to 0
        Sum = 0
        for j in range(N):
  
            # If current element and the
            # chosen element are same
            if (i == j):
                continue
  
            # Add all other numbers of array
            else:
                Sum += arr[j]
  
        # If Sum is divisible by the 
        # chosen element
        if (Sum % arr[i] == 0):
            count += 1
      
    # Return the count
    return count
  
# Driver code
arr = [3, 10, 4, 6, 7]
n = len(arr)
print(countNum(n, arr))
  
# This code is contributed
# by Mohit Kumar

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
      
// Fuction to return the count
// of the required numbers
static int countNum(int N, int []arr)
{
    // To store the count of required numbers
    int count = 0;
    for (int i = 0; i < N; i++) 
    {
  
        // Initialize sum to 0
        int sum = 0;
        for (int j = 0; j < N; j++)
        {
  
            // If current element and the
            // chosen element are same
            if (i == j)
                continue;
  
            // Add all other numbers of array
            else
                sum += arr[j];
        }
  
        // If sum is divisible by the chosen element
        if (sum % arr[i] == 0)
            count++;
    }
  
    // Return the count
    return count;
}
  
// Driver code
public static void Main()
{
    int []arr = { 3, 10, 4, 6, 7 };
    int n = arr.Length;
    Console.WriteLine(countNum(n, arr));
}
}
  
// This code is contributed by inder_verma..

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PHP

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<?php
// Php implementation of the approach
  
// Fuction to return the count
// of the required numbers
function countNum($N, $arr)
{
// To store the count of
// required numbers
$count = 0;
  
for ($i=0;$i<$N;$i++) {
// Initialize sum to 0
    $Sum = 0;
    for ($j = 0; $j < $N; $j++) {
        // If current element and the
        // chosen element are same
        if ($i == $j)
            continue;
  
        // Add all other numbers of array
        else
            $Sum += $arr[$j];
    }
    // If Sum is divisible by the
    // chosen element
    if ($Sum % $arr[$i] == 0)
        $count += 1;
}
// Return the count
    return $count;
}
// Driver code
$arr = array(3, 10, 4, 6, 7);
$n = count($arr);
echo countNum($n, $arr);
  
// This code is contributed
// by Srathore
?>

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Output:

3

Time Complexity: O(N2)

Efficient Approach: Run a single loop from 0 to N, calculate sum of all the elements. Now run another loop from 0 to N and if (sum – arr[i]) % arr[i] = 0 then increment the count.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Fuction to return the count
// of the required numbers
int countNum(int N, int arr[])
{
    // Initialize sum and count to 0
    int sum = 0, count = 0;
  
    // Calculate sum of all
    // the array elements
    for (int i = 0; i < N; i++)
        sum += arr[i];
  
    for (int i = 0; i < N; i++)
  
        // If current element satifies the condition
        if ((sum - arr[i]) % arr[i] == 0)
            count++;
  
    // Return the count of required elements
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 3, 10, 4, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countNum(n, arr);
  
    return 0;
}

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Java

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// Java implementation of the approach
  
class GFG 
{
  
    // Fuction to return the count
    // of the required numbers
    static int countNum(int N, int arr[]) 
    {
        // Initialize sum and count to 0
        int sum = 0, count = 0;
  
        // Calculate sum of all
        // the array elements
        for (int i = 0; i < N; i++) 
        {
            sum += arr[i];
        }
          
        // If current element satifies the condition
        for (int i = 0; i < N; i++) 
        {
            if ((sum - arr[i]) % arr[i] == 0
            {
                count++;
            }
        }
  
        // Return the count of required elements
        return count;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int arr[] = {3, 10, 4, 6, 7};
        int n = arr.length;
        System.out.println(countNum(n, arr));
    }
}
  
// This code has been contributed by 29AjayKumar

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Python3

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# Python3 implementation of the approach
  
# Fuction to return the count
# of the required numbers
def countNum(N, arr):
      
    # Initialize Sum and count to 0
    Sum, count = 0, 0
  
    # Calculate Sum of all the 
    # array elements
    for i in range(N):
        Sum += arr[i]
  
    for i in range(N):
  
        # If current element satifies 
        # the condition
        if ((Sum - arr[i]) % arr[i] == 0):
            count += 1
  
    # Return the count of required 
    # elements
    return count
  
# Driver code
arr = [ 3, 10, 4, 6, 7 ]
n = len(arr)
print(countNum(n, arr))
  
# This code is contributed 
# by Mohit Kumar

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C#

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// C# implementation of the approach 
using System;
  
class GFG 
{
  
    // Fuction to return the count
    // of the required numbers
    static int countNum(int N, int []arr) 
    {
        // Initialize sum and count to 0
        int sum = 0, count = 0;
  
        // Calculate sum of all
        // the array elements
        for (int i = 0; i < N; i++) 
        {
            sum += arr[i];
        }
          
        // If current element satifies the condition
        for (int i = 0; i < N; i++) 
        {
            if ((sum - arr[i]) % arr[i] == 0) 
            {
                count++;
            }
        }
  
        // Return the count of required elements
        return count;
    }
  
    // Driver code
    public static void Main() 
    {
        int []arr = {3, 10, 4, 6, 7};
        int n = arr.Length;
        Console.WriteLine(countNum(n, arr));
    }
}
  
/* This code contributed by PrinciRaj1992 */

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PHP

Output:

3

Time Complexity: O(N)



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