Count of elements in an Array whose set bits are in a multiple of K
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++
#include <bits/stdc++.h>
using namespace std;
int find_count(vector< int > arr, int k)
{
int ans = 0;
for ( int i : arr) {
int x = __builtin_popcount(i);
if (x % k == 0)
ans += 1;
}
return ans;
}
int main()
{
vector< int > arr = { 12, 345, 2, 68, 7896 };
int K = 2;
cout << find_count(arr, K);
return 0;
}
|
Java
class GFG{
static int find_count( int []arr, int k)
{
int ans = 0 ;
for ( int i : arr) {
int x = Integer.bitCount(i);
if (x % k == 0 )
ans += 1 ;
}
return ans;
}
public static void main(String[] args)
{
int []arr = { 12 , 345 , 2 , 68 , 7896 };
int K = 2 ;
System.out.print(find_count(arr, K));
}
}
|
Python3
def bitsoncount(x):
return bin (x).count( '1' )
def find_count(arr, k) :
ans = 0
for i in arr:
x = bitsoncount(i)
if (x % k = = 0 ) :
ans + = 1
return ans
arr = [ 12 , 345 , 2 , 68 , 7896 ]
K = 2
print (find_count(arr, K))
|
C#
using System;
class GFG{
static int find_count( int []arr, int k)
{
int ans = 0;
foreach ( int i in arr)
{
int x = countSetBits(i);
if (x % k == 0)
ans += 1;
}
return ans;
}
static int countSetBits( long x)
{
int setBits = 0;
while (x != 0)
{
x = x & (x - 1);
setBits++;
}
return setBits;
}
public static void Main(String[] args)
{
int []arr = { 12, 345, 2, 68, 7896 };
int K = 2;
Console.Write(find_count(arr, K));
}
}
|
Javascript
<script>
function find_count(arr, k)
{
var ans = 0;
for ( var i = 0; i <= arr.length; i++)
{
var x = countSetBits(i);
if (x % k == 0)
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));
</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;
int countSetBits( int num)
{
int count = 0;
while (num) {
count += num
& 1;
num >>= 1;
}
return count;
}
int countMultiples( int arr[], int n, int K)
{
int res = 0;
for ( int i = 0; i < n; i++) {
if (countSetBits(arr[i]) % K
== 0) {
res += 1;
}
}
return res;
}
int main()
{
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;
cout << countMultiples(arr1, n1, K1)
<< endl;
cout << countMultiples(arr2, n2, K2)
<< endl;
return 0;
}
|
Java
public class CountSetBits {
public static int countSetBits( int num)
{
int count = 0 ;
while (num > 0 ) {
count += num & 1 ;
num >>= 1 ;
}
return count;
}
public static int countMultiples( int [] arr, int K)
{
int res = 0 ;
for ( int num : arr) {
if (countSetBits(num) % K
== 0 ) {
res += 1 ;
}
}
return res;
}
public static void main(String[] args)
{
int [] arr1 = { 1 , 2 , 3 , 4 , 5 };
int K1 = 2 ;
int [] arr2 = { 10 , 20 , 30 , 40 };
int K2 = 4 ;
System.out.println(
countMultiples(arr1, K1));
System.out.println(
countMultiples(arr2, K2));
}
}
|
Python3
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
arr1 = [ 1 , 2 , 3 , 4 , 5 ]
K1 = 2
arr2 = [ 10 , 20 , 30 , 40 ]
K2 = 4
print (countMultiples(arr1, K1))
print (countMultiples(arr2, K2))
|
C#
using System;
class Program {
static int CountSetBits( int num)
{
int count = 0;
while (num != 0) {
count += num & 1;
num >>= 1;
}
return count;
}
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)
{
res += 1;
}
}
return res;
}
static void Main( string [] args)
{
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;
Console.WriteLine(
CountMultiples(arr1, n1, K1));
Console.WriteLine(
CountMultiples(arr2, n2, K2));
}
}
|
Javascript
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;
}
const arr1 = [1, 2, 3, 4, 5];
const K1 = 2;
const arr2 = [10, 20, 30, 40];
const K2 = 4;
console.log(countMultiples(arr1, K1));
console.log(countMultiples(arr2, K2));
|
Time Complexity: O(n * log(max(arr)))
Space Complexity: O(1)
Last Updated :
04 Oct, 2023
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