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++
// CPP 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 bit 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 = __builtin_popcount(i); // check tot_bit prime or, not if (isPrime(tot_bit)) count++; } return count; } // Driven Program int main() { int l = 6, r = 10; cout << primeBitsInRange(l, r); return 0; } |
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Java
// Java program to count total prime // number of set bits in given range import 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 bit static 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 Program public static void main(String[] args) { int l = 6, r = 10; System.out.println(primeBitsInRange(l, r)); } } // This code is Contributed by mits |
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Python3
# Python3 program to count total prime # number of set bits in given range def 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 bit def 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 Code l = 6; r = 10; print(primeBitsInRange(l, r)); # This code is contributed by mits |
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C#
// C# program to count total prime // number of set bits in given range class GFG{ // To count the bits static 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 bit static 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 Program public static void Main() { int l = 6, r = 10; System.Console.WriteLine(primeBitsInRange(l, r)); } } // This code is Contributed by mits |
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PHP
<?php // PHP program to count total prime // number of set bits in given range function 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 bit function 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 ?> |
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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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