Given an array arr[] of N elements and an integer K, the task is to count all the elements whose number of set bits is a multiple of K.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}, K = 2
Output: 2
Explanation:
Two numbers whose setbits count is multiple of 2 are {3, 5}.Input: arr[] = {10, 20, 30, 40}, K = 4
Output: 1
Explanation:
There number whose setbits count is multiple of 4 is {30}.
Approach:
- Traverse the numbers in the array one by one.
- Count the set bits of every number in the array.
- Check if the setbits count is a multiple of K or not.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std; // Function to find the count of numbers int find_count(vector<int> arr, int k) { int ans = 0; for (int i : arr) { // Get the set-bits count of each element int x = __builtin_popcount(i); // Check if the setbits count // is divisible by K if (x % k == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } // Driver code int main() { vector<int> arr = { 12, 345, 2, 68, 7896 }; int K = 2; cout << find_count(arr, K); return 0; } |
Java
// Java implementation of above approach class GFG{ // Function to find the count of numbers static int find_count(int []arr, int k) { int ans = 0; for (int i : arr) { // Get the set-bits count of each element int x = Integer.bitCount(i); // Check if the setbits count // is divisible by K if (x % k == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } // Driver code public static void main(String[] args) { int []arr = { 12, 345, 2, 68, 7896 }; int K = 2; System.out.print(find_count(arr, K)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of above approach # Function to count total set bits def bitsoncount(x): return bin(x).count('1') # Function to find the count of numbers def find_count(arr, k) : ans = 0 for i in arr: # Get the set-bits count of each element x = bitsoncount(i) # Check if the setbits count # is divisible by K if (x % k == 0) : # Increment the count # of required numbers by 1 ans += 1 return ans # Driver code arr = [ 12, 345, 2, 68, 7896 ] K = 2print(find_count(arr, K)) # This code is contributed by Sanjit_Prasad |
C#
// C# implementation of above approach using System; class GFG{ // Function to find the count of numbers static int find_count(int []arr, int k) { int ans = 0; foreach(int i in arr) { // Get the set-bits count of each element int x = countSetBits(i); // Check if the setbits count // is divisible by K if (x % k == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } static int countSetBits(long x) { int setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits; } // Driver code public static void Main(String[] args) { int []arr = { 12, 345, 2, 68, 7896 }; int K = 2; Console.Write(find_count(arr, K)); } } // This code is contributed by sapnasingh4991 |
3
Time complexity: O(N * M), where N is the size of the array, and M is the bits count of the largest number in the array.
Auxiliary Space complexity: O(1)
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