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; } |
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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 |
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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 |
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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 |
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Output:
15
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