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# 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

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[0]);``    ``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

 `// Java 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`

## Javascript

 ``

Output:

`15`

#### Approach#2: Using bin()

Traverse the array and check the binary representation of each element. If the count of 1’s in the binary representation of an element is even, add the sum of its digits to the answer variable. Return the answer variable.

#### Algorithm

2. Traverse the given array.
3. For each number in the array, convert it to binary using the in-built bin() function and remove the 0b prefix from the binary representation using the string slice binary[2:].
4. Count the number of ones in the binary representation of the current number using the string method count().
5. If the count of ones is even, add the sum of the digits of the current number to the answer variable using the sum() and map() functions.
6. Return the final answer variable.

## Python3

 `def` `sum_of_digits_with_even_ones(arr):``    ``ans ``=` `0``    ``for` `num ``in` `arr:``        ``binary ``=` `bin``(num)[``2``:]``        ``count_ones ``=` `binary.count(``'1'``)``        ``if` `count_ones ``%` `2` `=``=` `0``:``            ``ans ``+``=` `sum``(``map``(``int``, ``str``(num)))``    ``return` `ans``arr ``=` `[``4``, ``9``, ``15``]``print``(sum_of_digits_with_even_ones(arr))`

Output

```15
```

Time complexity: O(N*M), where N is the length of the array and M is the maximum number of bits in any element of the array.
Space complexity: O(1).