Count of elements in an Array whose set bits are in a multiple of K
Last Updated :
04 Oct, 2023
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 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 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 = 2
print(find_count(arr, K))
# This code is contributed by Sanjit_Prasad
// 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
<script>
// Javascript implementation of above approach
// Function to find the count of numbers
function find_count(arr, k)
{
var ans = 0;
for(var i = 0; i <= arr.length; i++)
{
// Get the set-bits count of each element
var 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;
}
function countSetBits(x)
{
var setBits = 0;
while (x != 0)
{
x = x & (x - 1);
setBits++;
}
return setBits;
}
var arr = [ 12, 345, 2, 68, 7896 ];
var K = 2;
document.write(find_count(arr, K));
// This code is contributed by SoumikMondal
</script>
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)
Using Bit Manipulation and Brute Force in python:
Approach:
- Define a function named countSetBits that takes an integer num as input.
- Initialize a variable count to 0.
- Loop through the binary representation of num using bit manipulation.
- If the last bit is 1, increment the count variable.
- Right shift the binary representation of num by 1 bit.
- Repeat steps 4-5 until the binary representation of num becomes 0.
- Return the count variable.
- Define a function named countMultiples that takes a list arr and an integer K as input.
- Initialize a variable res to 0.
- Loop through each element num in arr.
- Call the countSetBits function with num as input and store the result in a variable count.
- If count is a multiple of K, increment the res variable.
- Repeat steps 10-12 until all elements in arr have been processed.
- Return the res variable.
C++
#include <iostream>
using namespace std;
// Function to count the number of set bits in a number
int countSetBits(int num)
{
int count = 0;
while (num) {
count += num
& 1; // Count the number of set bits in num
num >>= 1; // Right shift num to check the next bit
}
return count;
}
// Function to count the numbers in the array for which the
// count of set bits is divisible by K
int countMultiples(int arr[], int n, int K)
{
int res = 0;
for (int i = 0; i < n; i++) {
if (countSetBits(arr[i]) % K
== 0) { // Check if the number of set bits is
// divisible by K
res += 1;
}
}
return res;
}
int main()
{
// Example inputs
int arr1[] = { 1, 2, 3, 4, 5 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int K1 = 2;
int arr2[] = { 10, 20, 30, 40 };
int n2 = sizeof(arr2) / sizeof(arr2[0]);
int K2 = 4;
// Example outputs
cout << countMultiples(arr1, n1, K1)
<< endl; // Output: 2
cout << countMultiples(arr2, n2, K2)
<< endl; // Output: 1
return 0;
}
public class CountSetBits {
// Function to count the number of set bits in a number
public static int countSetBits(int num)
{
int count = 0;
while (num > 0) {
count += num & 1; // Count the number of set
// bits in num
num >>= 1; // Right shift num to check the next
// bit
}
return count;
}
// Function to count the numbers in the array for which
// the count of set bits is divisible by K
public static int countMultiples(int[] arr, int K)
{
int res = 0;
for (int num : arr) {
if (countSetBits(num) % K
== 0) { // Check if the number of set bits
// is divisible by K
res += 1;
}
}
return res;
}
public static void main(String[] args)
{
// Example inputs
int[] arr1 = { 1, 2, 3, 4, 5 };
int K1 = 2;
int[] arr2 = { 10, 20, 30, 40 };
int K2 = 4;
// Example outputs
System.out.println(
countMultiples(arr1, K1)); // Output: 2
System.out.println(
countMultiples(arr2, K2)); // Output: 1
}
}
def countSetBits(num):
count = 0
while num:
count += num & 1
num >>= 1
return count
def countMultiples(arr, K):
res = 0
for num in arr:
if countSetBits(num) % K == 0:
res += 1
return res
# Example inputs
arr1 = [1, 2, 3, 4, 5]
K1 = 2
arr2 = [10, 20, 30, 40]
K2 = 4
# Example outputs
print(countMultiples(arr1, K1)) # 2
print(countMultiples(arr2, K2)) # 1
using System;
class Program {
// Function to count the number of set bits in a number
static int CountSetBits(int num)
{
int count = 0;
while (num != 0) {
count += num & 1; // Count the number of set
// bits in num
num >>= 1; // Right shift num to check the next
// bit
}
return count;
}
// Function to count the numbers in the array for which
// the count of set bits is divisible by K
static int CountMultiples(int[] arr, int n, int K)
{
int res = 0;
for (int i = 0; i < n; i++) {
if (CountSetBits(arr[i]) % K
== 0) // Check if the number of set bits is
// divisible by K
{
res += 1;
}
}
return res;
}
static void Main(string[] args)
{
// Example inputs
int[] arr1 = { 1, 2, 3, 4, 5 };
int n1 = arr1.Length;
int K1 = 2;
int[] arr2 = { 10, 20, 30, 40 };
int n2 = arr2.Length;
int K2 = 4;
// Example outputs
Console.WriteLine(
CountMultiples(arr1, n1, K1)); // Output: 2
Console.WriteLine(
CountMultiples(arr2, n2, K2)); // Output: 1
}
}
function countSetBits(num) {
let count = 0;
while (num) {
count += num & 1;
num >>= 1;
}
return count;
}
function countMultiples(arr, K) {
let res = 0;
for (let num of arr) {
if (countSetBits(num) % K === 0) {
res += 1;
}
}
return res;
}
// Example inputs
const arr1 = [1, 2, 3, 4, 5];
const K1 = 2;
const arr2 = [10, 20, 30, 40];
const K2 = 4;
// Example outputs
console.log(countMultiples(arr1, K1)); // 2
console.log(countMultiples(arr2, K2)); // 1
Time Complexity: O(n * log(max(arr)))
Space Complexity: O(1)
Similar Reads
Sum of array elements whose count of set bits are unique Given an array arr[] consisting of N positive integers, the task is to find the sum of all array elements having a distinct count of set bits in the array. Examples: Input: arr[] = {8, 3, 7, 5, 3}Output: 15Explanation:The count of set bits in each array of elements is: arr[0] = 8 = (1000)2, has 1 se
7 min read
Count of integers in an Array whose length is a multiple of K Given an array arr of N elements and an integer K, the task is to count all the elements whose length is a multiple of K.Examples: Input: arr[]={1, 12, 3444, 544, 9}, K = 2 Output: 2 Explanation: There are 2 numbers whose digit count is multiple of 2 {12, 3444}. Input: arr[]={12, 345, 2, 68, 7896},
5 min read
Count the number of elements in an array which are divisible by k Given an array of integers. The task is to calculate the count of a number of elements which are divisible by a given number k. Examples: Input: arr[] = { 2, 6, 7, 12, 14, 18 }, k = 3 Output: 3 Numbers which are divisible by k are { 6, 12, 18 } Input: arr[] = { 2, 6, 7, 12, 14, 18 }, k = 2 Output: 5
6 min read
Count of even and odd set bit with array element after XOR with K Given an array arr[] and a number K. The task is to find the count of the element having odd and even number of the set-bit after taking XOR of K with every element of the given arr[].Examples: Input: arr[] = {4, 2, 15, 9, 8, 8}, K = 3 Output: Even = 2, Odd = 4 Explanation: The binary representation
9 min read
Queries to count array elements from a given range having a single set bit Given an array arr[] consisting of N integers and a 2D array Q[][] consisting of queries of the following two types: 1 L R: Print the count of numbers from the range [L, R] having only a single set bit.2 X V: Update the array element at Xth index with V.Examples: Input: arr[] = { 12, 11, 16, 2, 32 }
15+ min read