# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function 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); ` `    ``cout << countNum(n, arr); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `     `  `// Function 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 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function 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 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function 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.. `

## PHP

 ` `

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++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function 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); ` `    ``cout << countNum(n, arr); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function 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 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function 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 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function 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 */`

## PHP

 ` `

Output:

```3
```

Time Complexity: O(N)

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