# Sum of digits with even number of 1’s in their binary representation

Given an array arr[] of size N. The task is to find the sum of the digits of all array elements which contains even number of 1’s in it’s their binary representation.

Examples:

Input : arr[] = {4, 9, 15}
Output : 15
4 = 10, it contains odd number of 1’s
9 = 1001, it contains even number of 1’s
15 = 1111, it contains even number of 1’s
Total Sum = Sum of digits of 9 and 15 = 9 + 1 + 5 = 15

Input : arr[] = {7, 23, 5}
Output :10

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

Approach :
The number of 1’s in the binary representation of each array element is counted and if it is even then the sum of its digits is calculated.

Below is the implementation of the above approach:

## C++

 `// CPP program to find Sum of digits with even  ` `// number of 1’s in their binary representation ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count and check the  ` `// number of 1's is even or odd ` `int` `countOne(``int` `n) ` `{ ` `    ``int` `count = 0; ` `    ``while` `(n) { ` `        ``n = n & (n - 1); ` `        ``count++; ` `    ``} ` ` `  `    ``if` `(count % 2 == 0) ` `        ``return` `1; ` `    ``else` `        ``return` `0; ` `} ` ` `  `// Function to calculate the sum  ` `// of the digits of a number ` `int` `sumDigits(``int` `n) ` `{ ` `    ``int` `sum = 0; ` `    ``while` `(n != 0) { ` `        ``sum += n % 10; ` `        ``n /= 10; ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 4, 9, 15 }; ` `     `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `total_sum = 0; ` ` `  `    ``// Iterate through the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(countOne(arr[i])) ` `            ``total_sum += sumDigits(arr[i]); ` `    ``} ` ` `  `    ``cout << total_sum << ``'\n'``; ` `     `  `    ``return` `0; ` `} `

## Java

 `// C# program to find Sum of digits with even  ` `// number of 1's in their binary representation ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to count and check the  ` `// number of 1's is even or odd ` `static` `int` `countOne(``int` `n) ` `{ ` `    ``int` `count = ``0``; ` `    ``while` `(n > ``0``)  ` `    ``{ ` `        ``n = n & (n - ``1``); ` `        ``count++; ` `    ``} ` ` `  `    ``if` `(count % ``2` `== ``0``) ` `        ``return` `1``; ` `    ``else` `        ``return` `0``; ` `} ` ` `  `// Function to calculate the sum  ` `// of the digits of a number ` `static` `int` `sumDigits(``int` `n) ` `{ ` `    ``int` `sum = ``0``; ` `    ``while` `(n != ``0``)  ` `    ``{ ` `        ``sum += n % ``10``; ` `        ``n /= ``10``; ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = { ``4``, ``9``, ``15` `}; ` `     `  `    ``int` `n = arr.length; ` `    ``int` `total_sum = ``0``; ` ` `  `    ``// Iterate through the array ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``if` `(countOne(arr[i]) == ``1``) ` `            ``total_sum += sumDigits(arr[i]); ` `    ``} ` `    ``System.out.println(total_sum); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 program to find Sum of digits with even ` `# number of 1’s in their binary representation ` ` `  `# Function to count and check the ` `# number of 1's is even or odd ` `def` `countOne(n): ` `    ``count ``=` `0` `    ``while` `(n): ` `        ``n ``=` `n & (n ``-` `1``) ` `        ``count ``+``=` `1` ` `  `    ``if` `(count ``%` `2` `=``=` `0``): ` `        ``return` `1` `    ``else``: ` `        ``return` `0` ` `  `# Function to calculate the summ ` `# of the digits of a number ` `def` `summDigits(n): ` `    ``summ ``=` `0` `    ``while` `(n !``=` `0``): ` `        ``summ ``+``=` `n ``%` `10` `        ``n ``/``/``=` `10` ` `  `    ``return` `summ ` ` `  `# Driver Code ` `arr ``=` `[``4``, ``9``, ``15``] ` ` `  `n ``=` `len``(arr) ` `total_summ ``=` `0` ` `  `# Iterate through the array ` `for` `i ``in` `range``(n): ` `    ``if` `(countOne(arr[i])): ` `        ``total_summ ``+``=` `summDigits(arr[i]) ` ` `  `print``(total_summ ) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# program to find Sum of digits with even  ` `// number of 1's in their binary representation ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to count and check the  ` `// number of 1's is even or odd ` `static` `int` `countOne(``int` `n) ` `{ ` `    ``int` `count = 0; ` `    ``while` `(n > 0)  ` `    ``{ ` `        ``n = n & (n - 1); ` `        ``count++; ` `    ``} ` ` `  `    ``if` `(count % 2 == 0) ` `        ``return` `1; ` `    ``else` `        ``return` `0; ` `} ` ` `  `// Function to calculate the sum  ` `// of the digits of a number ` `static` `int` `sumDigits(``int` `n) ` `{ ` `    ``int` `sum = 0; ` `    ``while` `(n != 0)  ` `    ``{ ` `        ``sum += n % 10; ` `        ``n /= 10; ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main()  ` `{ ` `    ``int``[] arr = { 4, 9, 15 }; ` `     `  `    ``int` `n = arr.Length; ` `    ``int` `total_sum = 0; ` ` `  `    ``// Iterate through the array ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(countOne(arr[i]) == 1) ` `            ``total_sum += sumDigits(arr[i]); ` `    ``} ` `    ``Console.WriteLine(total_sum); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```15
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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