# Count array elements that divide the sum of all other elements

• Difficulty Level : Basic
• Last Updated : 28 Oct, 2021

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:

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Input: arr[] = {3, 10, 4, 6, 7}
Output:
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 the sum of all elements except the current element and if this element divides 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

 ``

## Javascript

 ``
Output:
`3`

Time Complexity: O(N2)

Efficient Approach: Run a single loop from 0 to N, calculate the 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 satisfies 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 satisfies 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 satisfies``        ``# 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 satisfies 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

 ``

## Javascript

 ``
Output:
`3`

Time Complexity: O(N)

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