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++

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// CPP progarm to find Sum of digits with even 
// number of 1’s in their binary representation
#include <bits/stdc++.h>
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;
}

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Java





// C# progarm 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


[tabby title="Python3"]

                











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# Python3 progarm 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 



















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Output:


15






          
          
          
          

            

    

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