Count integers in an Array which are multiples their bits counts
Last Updated : 24 Mar, 2023
Given an array arr[] of N elements, the task is to count all the elements which are a multiple of their set bits count.
Examples:
Input : arr[] = { 1, 2, 3, 4, 5, 6 } Output : 4 Explanation : There numbers which are multiple of their setbits count are { 1, 2, 4, 6 }. Input : arr[] = {10, 20, 30, 40} Output : 3 Explanation : There numbers which are multiple of their setbits count are { 10, 20, 40 } Approach: Loop through each array elements one by one. Count the set bits of every number in the array. Check if the current integer is a multiple of its set bits count or not. If 'yes' then increment the counter by 1, else skip that integer. Below is the implementation of above approach:
C++ #include <bits/stdc++.h> using namespace std; // Function to find the count of numbers // which are multiple of its set bits count int find_count(vector<int>& arr) { // variable to store count int ans = 0; // iterate over elements of array for (int i : arr) { // Get the set-bits count of each element int x = __builtin_popcount(i); // Check if the setbits count // divides the integer i if (i % x == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } // Driver code int main() { vector<int> arr = { 1, 2, 3, 4, 5, 6 }; cout << find_count(arr); return 0; } class GFG{ // Function to find the count of numbers // which are multiple of its set bits count static int find_count(int []arr) { // variable to store count int ans = 0; // iterate over elements of array for (int i : arr) { // Get the set-bits count of each element int x = Integer.bitCount(i); // Check if the setbits count // divides the integer i if (i % x == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } // Driver code public static void main(String[] args) { int []arr = { 1, 2, 3, 4, 5, 6 }; System.out.print(find_count(arr)); } } // This code contributed by Princi Singh # Python3 implementation of above approach # function to return set bits count def bitsoncount(x): return bin(x).count('1') # Function to find the count of numbers # which are multiple of its set bits count def find_count(arr) : # variable to store count ans = 0 # iterate over elements of array for i in arr : # Get the set-bits count of each element x = bitsoncount(i) # Check if the setbits count # divides the integer i if (i % x == 0): # Increment the count # of required numbers by 1 ans += 1 return ans # Driver code arr = [ 1, 2, 3, 4, 5, 6 ] print(find_count(arr)) # This code is contributed by Sanjit_Prasad using System; public class GFG{ // Function to find the count of numbers // which are multiple of its set bits count static int find_count(int []arr) { // Variable to store count int ans = 0; // Iterate over elements of array foreach (int i in arr) { // Get the set-bits count of each element int x = bitCount(i); // Check if the setbits count // divides the integer i if (i % x == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } static int bitCount(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 = { 1, 2, 3, 4, 5, 6 }; Console.Write(find_count(arr)); } } // This code contributed by Princi Singh <script> // Function to find the count of numbers // which are multiple of its set bits count function find_count(arr) { // Variable to store count var ans = 0; // Iterate over elements of array for (var i=0;i<=arr.length;i++) { // Get the set-bits count of each element var x = bitCount(i); // Check if the setbits count // divides the integer i if (i % x == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } function bitCount( x) { var setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits; } var arr = [ 1, 2, 3, 4, 5, 6 ]; document.write(find_count(arr)); // This code contributed by SoumikMondal </script>
Time complexity:- O(nlog(max(arr[])), where n is the size of the array,
Space complexity:- O(1)
Approach 2:
Another approach would be to precompute the set bits count of all integers up to a certain limit (e.g., the maximum value in the array), and store them in an array or a hash table. Then, we can loop through the array and look up the set bits count of each element in the table, and check if it divides the element. This approach can be faster if we have to process multiple arrays with the same set of integers.
Here's an example implementation of the second approach:
C++ #include <bits/stdc++.h> using namespace std; const int MAX_N = 1000000; // Precomputed set bits count int set_bits_count[MAX_N + 1]; void precompute_set_bits_count() { for (int i = 1; i <= MAX_N; i++) { set_bits_count[i] = set_bits_count[i >> 1] + (i & 1); } } // Function to find the count of numbers // which are multiple of its set bits count int find_count(vector<int>& arr) { // variable to store count int ans = 0; // iterate over elements of array for (int i : arr) { // Get the set-bits count of each element int x = set_bits_count[i]; // Check if the setbits count // divides the integer i if (i % x == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } // Driver code int main() { precompute_set_bits_count(); vector<int> arr = { 1, 2, 3, 4, 5, 6 }; cout << find_count(arr); return 0; } import java.util.*; public class Main { static final int MAX_N = 1000000; // Precomputed set bits count static int[] set_bits_count = new int[MAX_N + 1]; static void precompute_set_bits_count() { for (int i = 1; i <= MAX_N; i++) { set_bits_count[i] = set_bits_count[i >> 1] + (i & 1); } } // Function to find the count of numbers // which are multiple of its set bits count static int find_count(ArrayList<Integer> arr) { // variable to store count int ans = 0; // iterate over elements of array for (int i : arr) { // Get the set-bits count of each element int x = set_bits_count[i]; // Check if the setbits count // divides the integer i if (i % x == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } // Driver code public static void main(String[] args) { precompute_set_bits_count(); ArrayList<Integer> arr = new ArrayList<>(Arrays.asList(1, 2, 3, 4, 5, 6)); System.out.println(find_count(arr)); } } MAX_N = 1000000 # Precomputed set bits count set_bits_count = [0] * (MAX_N + 1) def precompute_set_bits_count(): for i in range(1, MAX_N + 1): set_bits_count[i] = set_bits_count[i >> 1] + (i & 1) # Function to find the count of numbers # which are multiple of its set bits count def find_count(arr): # variable to store count ans = 0 # iterate over elements of array for i in arr: # Get the set-bits count of each element x = set_bits_count[i] # Check if the setbits count # divides the integer i if i % x == 0: # Increment the count # of required numbers by 1 ans += 1 return ans # Driver code if __name__ == '__main__': precompute_set_bits_count() arr = [1, 2, 3, 4, 5, 6] print(find_count(arr))
using System; using System.Collections.Generic; public class MainClass { const int MAX_N = 1000000; // Precomputed set bits count static int[] set_bits_count = new int[MAX_N + 1]; static void precompute_set_bits_count() { for (int i = 1; i <= MAX_N; i++) { set_bits_count[i] = set_bits_count[i >> 1] + (i & 1); } } // Function to find the count of numbers // which are multiple of its set bits count static int find_count(List<int> arr) { // variable to store count int ans = 0; // iterate over elements of array foreach (int i in arr) { // Get the set-bits count of each element int x = set_bits_count[i]; // Check if the setbits count // divides the integer i if (i % x == 0) // Increment the count // of required numbers by 1 ans += 1; } return ans; } // Driver code public static void Main() { precompute_set_bits_count(); List<int> arr = new List<int>() { 1, 2, 3, 4, 5, 6 }; Console.WriteLine(find_count(arr)); } } const MAX_N = 1000000; // Precomputed set bits count const set_bits_count = new Array(MAX_N + 1).fill(0); function precompute_set_bits_count() { for (let i = 1; i <= MAX_N; i++) { set_bits_count[i] = set_bits_count[i >> 1] + (i & 1); } } // Function to find the count of numbers // which are multiple of its set bits count function find_count(arr) { // variable to store count let ans = 0; // iterate over elements of array for (let i of arr) { // Get the set-bits count of each element let x = set_bits_count[i]; // Check if the setbits count // divides the integer i if (i % x == 0) { // Increment the count // of required numbers by 1 ans += 1; } } return ans; } // Driver code if (require.main === module) { precompute_set_bits_count(); const arr = [1, 2, 3, 4, 5, 6]; console.log(find_count(arr)); } // Contributed by sdeadityasharma Time complexity:- O(nlog(max(arr[])), where n is the size of the array,
Auxiliary Space:- O(N)
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