Prime Number of Set Bits in Binary Representation | Set 1
Given two integers ‘L’ and ‘R’, write a program to find the total numbers that are having prime number of set bits in their binary representation in the range [L, R].
Examples:
Input : l = 6, r = 10 Output : 4 Explanation : 6 -> 110 (2 set bits, 2 is prime) 7 -> 111 (3 set bits, 3 is prime) 9 -> 1001 (2 set bits, 2 is prime) 10 -> 1010 (2 set bits, 2 is prime) Hence count is 4 Input : l = 10, r = 15 Output : 5 10 -> 1010 (2 set bits, 2 is prime) 11 -> 1011 (3 set bits, 3 is prime) 12 -> 1100 (2 set bits, 2 is prime) 13 -> 1101 (3 set bits, 3 is prime) 14 -> 1110 (3 set bits, 3 is prime) Hence count is 5
Explanation: In this program we find a total number, that’s having prime number of set bit. so we use a CPP predefined function __builtin_popcount() these functions provide a total set bit in number. as well as be check the total bit’s is prime or not if prime we increase the counter these process repeat till given range.
C++
// C++ program to count total prime// number of set bits in given range#include <bits/stdc++.h>using namespace std;bool isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n%2 == 0 || n%3 == 0) return false; for (int i=5; i*i<=n; i=i+6) if (n%i == 0 || n%(i+2) == 0) return false; return true;}// count number, that contains prime number of set bitint primeBitsInRange(int l, int r){ // tot_bit store number of bit in number int tot_bit, count = 0; // iterate loop from l to r for (int i = l; i <= r; i++) { // use predefined function for finding // set bit it is return number of set bit tot_bit = __builtin_popcount(i); // check tot_bit prime or, not if (isPrime(tot_bit)) count++; } return count;}// Driven Programint main(){ int l = 6, r = 10; cout << primeBitsInRange(l, r); return 0;} |
Java
// Java program to count total prime// number of set bits in given rangeimport java.lang.*;class GFG{static boolean isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n%2 == 0 || n%3 == 0) return false; for (int i=5; i*i<=n; i=i+6) if (n%i == 0 || n%(i+2) == 0) return false; return true;}// count number, that contains prime number of set bitstatic int primeBitsInRange(int l, int r){ // tot_bit store number of bit in number int tot_bit, count = 0; // iterate loop from l to r for (int i = l; i <= r; i++) { // use predefined function for finding // set bit it is return number of set bit tot_bit = Integer.bitCount(i); // check tot_bit prime or, not if (isPrime(tot_bit)) count++; } return count;}// Driven Programpublic static void main(String[] args){ int l = 6, r = 10; System.out.println(primeBitsInRange(l, r)); }}// This code is Contributed by mits |
Python3
# Python3 program to count total prime# number of set bits in given rangedef isPrime(n): # Corner cases if (n <= 1): return False; if (n <= 3): return True; # This is checked so that we can skip # middle five numbers in below loop if (n % 2 == 0 or n % 3 == 0): return False; i = 5; while (i * i <= n): if(n % i == 0 or n % (i + 2) == 0): return False; i = i + 6; return True;# count number, that contains# prime number of set bitdef primeBitsInRange(l, r): # tot_bit store number of # bit in number count = 0; # iterate loop from l to r for i in range(l, r + 1): # use predefined function for finding # set bit it is return number of set bit tot_bit = bin(i).count('1'); # check tot_bit prime or, not if (isPrime(tot_bit)): count += 1; return count;# Driver Codel = 6;r = 10;print(primeBitsInRange(l, r));# This code is contributed by mits |
C#
// C# program to count total prime// number of set bits in given rangeclass GFG{ // To count the bitsstatic int BitCount(int n){ int count = 0; while (n != 0) { count++; n &= (n - 1); } return count;} static bool isPrime(int n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n%2 == 0 || n%3 == 0) return false; for (int i=5; i*i<=n; i=i+6) if (n%i == 0 || n%(i+2) == 0) return false; return true;}// count number, that contains prime number of set bitstatic int primeBitsInRange(int l, int r){ // tot_bit store number of bit in number int tot_bit, count = 0; // iterate loop from l to r for (int i = l; i <= r; i++) { // use predefined function for finding // set bit it is return number of set bit tot_bit = BitCount(i); // check tot_bit prime or, not if (isPrime(tot_bit)) count++; } return count;}// Driven Programpublic static void Main(){ int l = 6, r = 10; System.Console.WriteLine(primeBitsInRange(l, r)); }}// This code is Contributed by mits |
PHP
<?php// PHP program to count total prime// number of set bits in given rangefunction BitCount($n){ $count = 0; while ($n != 0) { $count++; $n &= ($n - 1); } return $count;}function isPrime($n){ // Corner cases if ($n <= 1) return false; if ($n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if ($n % 2 == 0 || $n % 3 == 0) return false; for ($i = 5; $i * $i <= $n; $i = $i + 6) if ($n % $i == 0 || $n % ($i + 2) == 0) return false; return true;}// count number, that contains// prime number of set bitfunction primeBitsInRange($l, $r){ // tot_bit store number of // bit in number $count = 0; // iterate loop from l to r for ($i = $l; $i <= $r; $i++) { // use predefined function for finding // set bit it is return number of set bit $tot_bit = BitCount($i); // check tot_bit prime or, not if (isPrime($tot_bit)) $count++; } return $count;}// Driver Code$l = 6;$r = 10;echo primeBitsInRange($l, $r);// This code is contributed by mits?> |
Javascript
<script>// Javascript program to count total prime// number of set bits in given range// To count the bitsfunction BitCount(n){ count = 0; while (n != 0) { count++; n &= (n - 1); } return count;}function isPrime(n){ // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for(i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true;}// Count number, that contains// prime number of set bitfunction primeBitsInRange(l, r){ // tot_bit store number of bit in number var tot_bit, count = 0; // Iterate loop from l to r for(i = l; i <= r; i++) { // Use predefined function for finding // set bit it is return number of set bit tot_bit = BitCount(i); // Check tot_bit prime or, not if (isPrime(tot_bit)) count++; } return count;}// Driver codevar l = 6, r = 10;document.write(primeBitsInRange(l, r));// This code is contributed by Rajput-Ji</script> |
Output:
4
Time Complexity : Let’s n = (r-l)
so overall time complexity is N*sqrt(N)
We can optimize above solution using Sieve of Eratosthenes.
Prime Number of Set Bits in Binary Representation | Set 2
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