# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` `} `

## 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 `

## 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 `

## 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  `

## PHP

 ` `

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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