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Sum of digits with even number of 1’s in their binary representation
  • Difficulty Level : Basic
  • Last Updated : 17 Mar, 2021

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 <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;
}

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




<script>
 
    // Javascript 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
    function countOne(n)
    {
        let 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
    function sumDigits(n)
    {
        let sum = 0;
        while (n != 0) {
            sum += n % 10;
            n = parseInt(n / 10, 10);
        }
 
        return sum;
    }
     
    let arr = [ 4, 9, 15 ];
       
    let n = arr.length;
    let total_sum = 0;
   
    // Iterate through the array
    for (let i = 0; i < n; i++) {
        if (countOne(arr[i]))
            total_sum += sumDigits(arr[i]);
    }
   
    document.write(total_sum);
 
</script>
Output: 
15

 

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