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 <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> |
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
1. Start with an answer variable set to 0.
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.
C++
#include <iostream> #include <vector> #include <bitset> // Function to calculate the sum of digits for numbers with an even count of binary ones int sumOfDigitsWithEvenOnes(std::vector< int > arr) { int ans = 0; // Loop through each number in the vector for ( int num : arr) { // Convert the number to binary and count the number of ones in its binary representation std::bitset<32> binary(num); int countOnes = binary.count(); // Check if the count of ones is even if (countOnes % 2 == 0) { int numCopy = num; int digitSum = 0; // Calculate the sum of digits for the current number while (numCopy > 0) { digitSum += numCopy % 10; numCopy /= 10; } // Add the digit sum to the overall result ans += digitSum; } } // Return the final result return ans; } // Main function int main() { // Initialize a vector of numbers std::vector< int > arr = {4, 9, 15}; // Call the function and print the result int result = sumOfDigitsWithEvenOnes(arr); std::cout << result << std::endl; // Return 0 to indicate successful execution return 0; } |
Java
import java.util.ArrayList; public class Main { // Function to calculate the sum of digits for numbers // with an even count of binary ones static int sumOfDigitsWithEvenOnes(ArrayList<Integer> arr) { int ans = 0 ; // Loop through each number in the ArrayList for ( int num : arr) { // Convert the number to binary and count the // number of ones in its binary representation String binaryString = Integer.toBinaryString(num); int countOnes = Integer.bitCount(num); // Check if the count of ones is even if (countOnes % 2 == 0 ) { int numCopy = num; int digitSum = 0 ; // Calculate the sum of digits for the // current number while (numCopy > 0 ) { digitSum += numCopy % 10 ; numCopy /= 10 ; } // Add the digit sum to the overall result ans += digitSum; } } // Return the final result return ans; } // Main function public static void main(String[] args) { // Initialize an ArrayList of numbers ArrayList<Integer> arr = new ArrayList<>(); arr.add( 4 ); arr.add( 9 ); arr.add( 15 ); // Call the function and print the result int result = sumOfDigitsWithEvenOnes(arr); System.out.println(result); } } |
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)) |
C#
using System; using System.Collections.Generic; class Program { // Function to count set bits in an integer static int CountSetBits( int num) { int count = 0; while (num > 0) { count += num & 1; num >>= 1; } return count; } // Function to calculate the sum of digits for numbers // with an even count of binary ones static int SumOfDigitsWithEvenOnes(List< int > arr) { int ans = 0; // Loop through each number in the list foreach ( int num in arr) { // Count the number of ones in its binary // representation int countOnes = CountSetBits(num); // Check if the count of ones is even if (countOnes % 2 == 0) { int numCopy = num; int digitSum = 0; // Calculate the sum of digits for the // current number while (numCopy > 0) { digitSum += numCopy % 10; numCopy /= 10; } // Add the digit sum to the overall result ans += digitSum; } } // Return the final result return ans; } // Main function static void Main() { // Initialize a list of numbers List< int > arr = new List< int >{ 4, 9, 15 }; // Call the function and print the result int result = SumOfDigitsWithEvenOnes(arr); Console.WriteLine(result); // Pause execution to see the result Console.ReadLine(); } } |
Javascript
// Function to calculate the sum of digits for numbers with an even count of binary ones function sumOfDigitsWithEvenOnes(arr) { let ans = 0; // Loop through each number in the array for (let num of arr) { // Convert the number to binary and count the number of ones in its binary representation let binary = num.toString(2); let countOnes = (binary.match(/1/g) || []).length; // Check if the count of ones is even if (countOnes % 2 === 0) { let numCopy = num; let digitSum = 0; // Calculate the sum of digits for the current number while (numCopy > 0) { digitSum += numCopy % 10; numCopy = Math.floor(numCopy / 10); } // Add the digit sum to the overall result ans += digitSum; } } // Return the final result return ans; } // Main function function main() { // Initialize an array of numbers let arr = [4, 9, 15]; // Call the function and print the result let result = sumOfDigitsWithEvenOnes(arr); console.log(result); } // Call the main function main(); |
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).
Contact Us