Count set bits in an integer

Write an efficient program to count the number of 1s in the binary representation of an integer.
Examples : 

Input : n = 6
Output : 2
Binary representation of 6 is 110 and has 2 set bits

Input : n = 13
Output : 3
Binary representation of 13 is 1101 and has 3 set bits

[Naive Approach] - One by One Counting

This approach counts the number of set bits (1s) in the binary representation of a given integer. It works by repeatedly checking the least significant bit of the number using the bitwise AND operation (n & 1). If the least significant bit is 1, the count is incremented. Then, the number is right-shifted by 1 (n >>= 1), effectively removing the least significant bit and moving on to the next one. This process continues until the number becomes 0, and the function returns the total count of set bits.

// C++ program to Count set // bits in an integer #include <bits/stdc++.h> using namespace std; /* Function to get no of set bits in binary representation of positive integer n */ unsigned int countSetBits(unsigned int n) {  unsigned int count = 0;  while (n) {  count += n & 1;  n >>= 1;  }  return count; } /* Program to test function countSetBits */ int main() {  int i = 9;  cout << countSetBits(i);  return 0; } 

// C program to Count set // bits in an integer #include <stdio.h> /* Function to get no of set bits in binary  representation of positive integer n */ unsigned int countSetBits(unsigned int n) {  unsigned int count = 0;  while (n) {  count += n & 1;  n >>= 1;  }  return count; } /* Program to test function countSetBits */ int main() {  int i = 9;  printf("%d", countSetBits(i));  return 0; } 

// Java program to Count set // bits in an integer import java.io.*; class countSetBits {  /* Function to get no of set   bits in binary representation   of positive integer n */  static int countSetBits(int n)  {  int count = 0;  while (n > 0) {  count += n & 1;  n >>= 1;  }  return count;  }  // driver program  public static void main(String args[])  {  int i = 9;  System.out.println(countSetBits(i));  } } 

# Python3 program to Count set # bits in an integer  # Function to get no of set bits in binary # representation of positive integer n */ def countSetBits(n): count = 0 while (n): count += n & 1 n >>= 1 return count # Program to test function countSetBits */ i = 9 print(countSetBits(i)) 

// C# program to Count set // bits in an integer using System; class GFG {  // Function to get no of set  // bits in binary representation  // of positive integer n  static int countSetBits(int n)  {  int count = 0;  while (n > 0) {  count += n & 1;  n >>= 1;  }  return count;  }  // Driver Code  public static void Main()  {  int i = 9;  Console.Write(countSetBits(i));  } } 

   // Javascript program to Count set  // bits in an integer  /* Function to get no of set bits in binary  representation of positive integer n */  function countSetBits(n)  {  var count = 0;  while (n)  {  count += n & 1;  n >>= 1;  }  return count;  }  /* Program to test function countSetBits */  var i = 9;  console.log(countSetBits(i));   

<?php // PHP program to Count set // bits in an integer  // Function to get no of set  // bits in binary representation  // of positive integer n  function countSetBits($n) { $count = 0; while ($n) { $count += $n & 1; $n >>= 1; } return $count; } // Driver Code $i = 9; echo countSetBits($i); ?> 

2

Time Complexity: O(log n) 
Auxiliary Space: O(1)

Recursive Implementation The function checks if the least significant bit is set (i.e., if it's 1). It does this by using a bitwise operation that isolates the least significant bit. If the bit is 1, it adds 1 to the count; otherwise, it adds 0. After checking the bit, the function recursively calls itself with the number shifted right by one position, essentially removing the checked bit. This process continues until all bits are checked, and the final count of set bits is returned.

// cpp implementation of recursive approach to find the // number of set bits in binary representation of positive // integer n #include <bits/stdc++.h> using namespace std; // recursive function to count set bits int countSetBits(int n) {  // base case  if (n == 0)  return 0;  else  // if last bit set add 1 else add 0  return (n & 1) + countSetBits(n >> 1); } // driver code int main() {  int n = 9;  // function calling  cout << countSetBits(n);  return 0; } 

// cpp implementation of recursive approach to find the // number of set bits in binary representation of positive // integer n #include <stdio.h> // recursive function to count set bits int countSetBits(int n) {  // base case  if (n == 0)  return 0;  else  // if last bit set add 1 else add 0  return (n & 1) + countSetBits(n >> 1); } // driver code int main() {  int n = 9;  // function calling  printf("%d", countSetBits(n));  return 0; } 

// Java implementation of recursive // approach to find the number // of set bits in binary representation // of positive integer n import java.io.*; class GFG {  // recursive function to count set bits  public static int countSetBits(int n)  {  // base case  if (n == 0)  return 0;  else  // if last bit set add 1 else add 0  return (n & 1) + countSetBits(n >> 1);  }  // Driver code  public static void main(String[] args)  {  // get value from user  int n = 9;  // function calling  System.out.println(countSetBits(n));  } } 

# Python3 implementation of recursive # approach to find the number of set # bits in binary representation of  # positive integer n def countSetBits( n): # base case if (n == 0): return 0 else: # if last bit set add 1 else # add 0 return (n & 1) + countSetBits(n >> 1) # Get value from user n = 9 # Function calling print( countSetBits(n)) 

// C# implementation of recursive // approach to find the number of // set bits in binary representation // of positive integer n using System; class GFG {  // recursive function  // to count set bits  public static int countSetBits(int n)  {  // base case  if (n == 0)  return 0;  else  // if last bit set  // add 1 else add 0  return (n & 1) + countSetBits(n >> 1);  }  // Driver code  static public void Main()  {  // get value  // from user  int n = 9;  // function calling  Console.WriteLine(countSetBits(n));  } } 

// Javascript implementation of recursive // approach to find the number // of set bits in binary representation // of positive integer n // recursive function to count set bits function countSetBits(n) {  // base case  if (n == 0)  return 0;  else  // if last bit set add 1 else add 0  return (n & 1) + countSetBits(n >> 1); } // driver code  // get value from user  let n = 9;  // function calling  console.log (countSetBits(n)); 

<?php // PHP implementation of recursive // approach to find the number of  // set bits in binary representation // of positive integer n // recursive function  // to count set bits function countSetBits($n) { // base case if ($n == 0) return 0; else // if last bit set  // add 1 else add 0 return ($n & 1) + countSetBits($n >> 1); } // Driver code // get value from user $n = 9; // function calling echo countSetBits($n); ?> 

2

Time Complexity: O(log n)
Auxiliary Space: O(log n)for recursive stack space

[Expected Approach 1] - Brian Kernighan's Algorithm

This approach uses the Brian Kernighan's algorithm to count the number of set bits (1s) in the binary representation of a given integer. The key idea is that the expression n &= (n - 1) clears the least significant set bit in the number n. By repeatedly applying this operation, we reduce the number and count how many set bits are cleared, which directly gives us the count of set bits. This method is more efficient than checking each bit individually, as it only processes the set bits in the number, making it faster in cases where the number has fewer set bits.  

Subtracting 1 from a decimal number flips all the bits after the rightmost set bit(which is 1) including the rightmost set bit. 
for example : 
10 in binary is 00001010 
9 in binary is 00001001 
8 in binary is 00001000 
7 in binary is 00000111 
So if we subtract a number by 1 and do it bitwise & with itself (n & (n-1)), we unset the rightmost set bit. If we do n & (n-1) in a loop and count the number of times the loop executes, we get the set bit count. 

The beauty of this solution is the number of times it loops is equal to the number of set bits in a given integer. 

1 Initialize count: = 0
2 If integer n is not zero
(a) Do bitwise & with (n-1) and assign the value back to n
n: = n&(n-1)
(b) Increment count by 1
(c) go to step 2
3 Else return count

Example for Brian Kernighan's Algorithm:  

n = 9 (1001)
count = 0

Since 9 > 0, subtract by 1 and do bitwise & with (9-1)
n = 9&8 (1001 & 1000)
n = 8
count = 1

Since 8 > 0, subtract by 1 and do bitwise & with (8-1)
n = 8&7 (1000 & 0111)
n = 0
count = 2

Since n = 0, return count which is 2 now.

// C++ program to Count set // bits in an integer #include <iostream> using namespace std; class gfg {  /* Function to get no of set bits in binary representation of passed binary no. */ public:  unsigned int countSetBits(int n)  {  unsigned int count = 0;  while (n) {  n &= (n - 1);  count++;  }  return count;  } }; /* Program to test function countSetBits */ int main() {  gfg g;  int i = 9;  cout << g.countSetBits(i);  return 0; } 

// C program to Count set // bits in an integer #include <stdio.h> /* Function to get no of set bits in binary  representation of passed binary no. */ unsigned int countSetBits(int n) {  unsigned int count = 0;  while (n) {  n &= (n - 1);  count++;  }  return count; } /* Program to test function countSetBits */ int main() {  int i = 9;  printf("%d", countSetBits(i));  getchar();  return 0; } 

// Java program to Count set // bits in an integer import java.io.*; class countSetBits {  /* Function to get no of set   bits in binary representation   of passed binary no. */  static int countSetBits(int n)  {  int count = 0;  while (n > 0) {  n &= (n - 1);  count++;  }  return count;  }  // driver program  public static void main(String args[])  {  int i = 9;  System.out.println(countSetBits(i));  } } // This code is contributed by Anshika Goyal. 

# Function to get no of set bits in binary # representation of passed binary no. */ def countSetBits(n): count = 0 while (n): n &= (n-1) count+= 1 return count # Program to test function countSetBits  i = 9 print(countSetBits(i)) 

// C# program to Count set // bits in an integer using System; class GFG {  /* Function to get no of set   bits in binary representation   of passed binary no. */  static int countSetBits(int n)  {  int count = 0;  while (n > 0) {  n &= (n - 1);  count++;  }  return count;  }  // Driver Code  static public void Main()  {  int i = 9;  Console.WriteLine(countSetBits(i));  } } // This code is contributed by ajit 

// JavaScript program to Count set // bits in an integerclass  /* Function to get no of set  bits in binary representation  of passed binary no. */ function countSetBits(n) {  var count = 0;  while (n > 0)  {  n &= (n - 1);  count++;  }  return count; } // driver program var i = 9; console.log (countSetBits(i)); 

<?php /* Function to get no of  set bits in binary representation of passed  binary no. */ function countSetBits($n) { $count = 0; while ($n) { $n &= ($n - 1) ; $count++; } return $count; } // Driver Code $i = 9; echo countSetBits($i); ?> 

2

Time Complexity: O(log n)
Auxiliary Space: O(1)

[Expected Approach 2] - Using Lookup table

This approach uses a lookup table to count the number of set bits (1s) in a number efficiently.

The program first initializes a table, BitsSetTable256[], where each entry at index i contains the number of set bits in the 8-bit binary representation of i. This is done by iterating through all numbers from 0 to 255 and calculating the set bits for each.

In the function countSetBits, the number is divided into 4 bytes (8 bits each), and the lookup table is used to quickly get the number of set bits for each byte. The result is the sum of the set bits in all 4 bytes, giving the total number of set bits in the integer.

This method is fast because it avoids looping through individual bits by using precomputed values stored in the lookup table. It is particularly useful when there are many set bits to count.

// C++ implementation of the approach  #include <bits/stdc++.h> using namespace std; int BitsSetTable256[256]; // Function to initialise the lookup table  void initialize()  {   // To initially generate the   // table algorithmically   BitsSetTable256[0] = 0;   for (int i = 0; i < 256; i++)  {   BitsSetTable256[i] = (i & 1) +   BitsSetTable256[i / 2];   }  }  // Function to return the count  // of set bits in n  int countSetBits(int n)  {   return (BitsSetTable256[n & 0xff] +   BitsSetTable256[(n >> 8) & 0xff] +   BitsSetTable256[(n >> 16) & 0xff] +   BitsSetTable256[n >> 24]);  }  // Driver code  int main()  {   // Initialise the lookup table   initialize();   int n = 9;   cout << countSetBits(n); }  

// Java implementation of the approach import java.util.*; class GFG {  // Lookup table  static int[] BitsSetTable256 = new int[256];  // Function to initialise the lookup table  public static void initialize()  {  // To initially generate the  // table algorithmically  BitsSetTable256[0] = 0;  for (int i = 0; i < 256; i++) {  BitsSetTable256[i] = (i & 1) + BitsSetTable256[i / 2];  }  }  // Function to return the count  // of set bits in n  public static int countSetBits(int n)  {  return (BitsSetTable256[n & 0xff]  + BitsSetTable256[(n >> 8) & 0xff]  + BitsSetTable256[(n >> 16) & 0xff]  + BitsSetTable256[n >> 24]);  }  // Driver code  public static void main(String[] args)  {  // Initialise the lookup table  initialize();  int n = 9;  System.out.print(countSetBits(n));  } } 

# Python implementation of the approach  BitsSetTable256 = [0] * 256 # Function to initialise the lookup table  def initialize(): # To initially generate the  # table algorithmically  BitsSetTable256[0] = 0 for i in range(256): BitsSetTable256[i] = (i & 1) + BitsSetTable256[i // 2] # Function to return the count  # of set bits in n  def countSetBits(n): return (BitsSetTable256[n & 0xff] + BitsSetTable256[(n >> 8) & 0xff] + BitsSetTable256[(n >> 16) & 0xff] + BitsSetTable256[n >> 24]) # Driver code  # Initialise the lookup table  initialize() n = 9 print(countSetBits(n)) 

// C# implementation of the approach using System; using System.Collections.Generic; class GFG  {  // Lookup table  static int[] BitsSetTable256 = new int[256];  // Function to initialise the lookup table  public static void initialize()  {  // To initially generate the  // table algorithmically  BitsSetTable256[0] = 0;  for (int i = 0; i < 256; i++)  {  BitsSetTable256[i] = (i & 1) + BitsSetTable256[i / 2];  }  }  // Function to return the count  // of set bits in n  public static int countSetBits(int n)  {  return (BitsSetTable256[n & 0xff]  + BitsSetTable256[(n >> 8) & 0xff]  + BitsSetTable256[(n >> 16) & 0xff]  + BitsSetTable256[n >> 24]);  }  // Driver code  public static void Main(String[] args)  {  // Initialise the lookup table  initialize();  int n = 9;  Console.Write(countSetBits(n));  } } 

// javascript implementation of the approach  var BitsSetTable256 = Array.from({length: 256}, (_, i) => 0); // Function to initialise the lookup table function initialize() {  // To initially generate the  // table algorithmically  BitsSetTable256[0] = 0;  for (var i = 0; i < 256; i++) {  BitsSetTable256[i] = (i & 1) +   BitsSetTable256[parseInt(i / 2)];  } } // Function to return the count // of set bits in n function countSetBits(n) {  return (BitsSetTable256[n & 0xff]  + BitsSetTable256[(n >> 8) & 0xff]  + BitsSetTable256[(n >> 16) & 0xff]  + BitsSetTable256[n >> 24]); } // Driver code  // Initialise the lookup table initialize(); var n = 9; console.log (countSetBits(n)); 

2

Time Complexity: O(1)
Auxiliary Space: O(1)

[Expected Approach 3] - Mapping Nibbles

It simply maintains a map(or array) of numbers to bits for a nibble. A Nibble contains 4 bits. So we need an array of up to 15. 
int num_to_bits[16] = {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4}; 
Now we just need to get nibbles of a given long/int/word etc recursively.

// C++ program to count set bits by pre-storing // count set bits in nibbles. #include <bits/stdc++.h> using namespace std; int num_to_bits[16] = { 0, 1, 1, 2, 1, 2, 2, 3,  1, 2, 2, 3, 2, 3, 3, 4 }; /* Recursively get nibble of a given number  and map them in the array */ unsigned int countSetBitsRec(unsigned int num) {  int nibble = 0;  if (0 == num)  return num_to_bits[0];  // Find last nibble  nibble = num & 0xf;  // Use pre-stored values to find count  // in last nibble plus recursively add  // remaining nibbles.  return num_to_bits[nibble] + countSetBitsRec(num >> 4); } // Driver code int main() {  int num = 31;  cout << countSetBitsRec(num);  return 0; } 

// C program to count set bits by pre-storing // count set bits in nibbles. #include <stdio.h> int num_to_bits[16] = { 0, 1, 1, 2, 1, 2, 2, 3,  1, 2, 2, 3, 2, 3, 3, 4 }; /* Recursively get nibble of a given number   and map them in the array */ unsigned int countSetBitsRec(unsigned int num) {  int nibble = 0;  if (0 == num)  return num_to_bits[0];  // Find last nibble  nibble = num & 0xf;  // Use pre-stored values to find count  // in last nibble plus recursively add  // remaining nibbles.  return num_to_bits[nibble] + countSetBitsRec(num >> 4); } // Driver code int main() {  int num = 31;  printf("%d\n", countSetBitsRec(num)); } 

// Java program to count set bits by pre-storing // count set bits in nibbles. import java.util.*; class GFG {  static int[] num_to_bits = new int[] { 0, 1, 1, 2, 1, 2, 2,  3, 1, 2, 2, 3, 2, 3, 3, 4 };  /* Recursively get nibble of a given number  and map them in the array */  static int countSetBitsRec(int num)  {  int nibble = 0;  if (0 == num)  return num_to_bits[0];  // Find last nibble  nibble = num & 0xf;  // Use pre-stored values to find count  // in last nibble plus recursively add  // remaining nibbles.  return num_to_bits[nibble] + countSetBitsRec(num >> 4);  }  // Driver code  public static void main(String[] args)  {  int num = 31;  System.out.println(countSetBitsRec(num));  } } // this code is contributed by mits 

# Python3 program to count set bits by pre-storing  # count set bits in nibbles.  num_to_bits =[0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4]; # Recursively get nibble of a given number  # and map them in the array def countSetBitsRec(num): nibble = 0; if(0 == num): return num_to_bits[0]; # Find last nibble nibble = num & 0xf; # Use pre-stored values to find count # in last nibble plus recursively add # remaining nibbles. return num_to_bits[nibble] + countSetBitsRec(num >> 4); # Driver code  num = 31; print(countSetBitsRec(num)); # this code is contributed by mits 

// C# program to count set bits by pre-storing // count set bits in nibbles. class GFG {  static int[] num_to_bits = new int[16] { 0, 1, 1, 2, 1, 2, 2,  3, 1, 2, 2, 3, 2, 3, 3, 4 };  /* Recursively get nibble of a given number  and map them in the array */  static int countSetBitsRec(int num)  {  int nibble = 0;  if (0 == num)  return num_to_bits[0];  // Find last nibble  nibble = num & 0xf;  // Use pre-stored values to find count  // in last nibble plus recursively add  // remaining nibbles.  return num_to_bits[nibble] + countSetBitsRec(num >> 4);  }  // Driver code  static void Main()  {  int num = 31;  System.Console.WriteLine(countSetBitsRec(num));  } } // this code is contributed by mits 

// Javascript program to count set bits by pre-storing // count set bits in nibbles.  var num_to_bits =[ 0, 1, 1, 2, 1, 2, 2,  3, 1, 2, 2, 3, 2, 3, 3, 4 ]; /* Recursively get nibble of a given number  and map them in the array */ function countSetBitsRec(num) {  var nibble = 0;  if (0 == num)  return num_to_bits[0];  // Find last nibble  nibble = num & 0xf;  // Use pre-stored values to find count  // in last nibble plus recursively add  // remaining nibbles.  return num_to_bits[nibble] + countSetBitsRec(num >> 4); } // Driver code var num = 31; console.log (countSetBitsRec(num)); 

<?php // PHP program to count set bits by  // pre-storing count set bits in nibbles.  $num_to_bits = array(0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4); /* Recursively get nibble of a given  number and map them in the array */ function countSetBitsRec( $num) { global $num_to_bits; $nibble = 0; if (0 == $num) return $num_to_bits[0]; // Find last nibble  $nibble = $num & 0xf; // Use pre-stored values to find count  // in last nibble plus recursively add  // remaining nibbles.  return $num_to_bits[$nibble] + countSetBitsRec($num >> 4); } // Driver code  $num = 31; echo (countSetBitsRec($num)); // This code is contributed by mits ?> 

5

Time Complexity: O(log n), because we have log(16, n) levels of recursion.
Storage Complexity: O(1) Whether the given number is short, int, long, or long long we require an array of 16 sizes only, which is constant.

Approach: Using power of 2 (efficient method to find for large value also)

Iterate from k to 0 , where k is the largest power of 2 such that pow(2, k) <= num . And check if the Bitwise AND of num and pow(2, i) is greater than zero or not. If it is greater than zero , Then i-th bit is set ,then increase the count by 1.

// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to find largest power of 2 such that // pow(2,k) <= N  int findk(int n) { int k; int i=0; int val=pow(2,i);    while(val<=n)  {   k=i; i++;  val=pow(2,i);  }  return k; } // Function to count set bits in a number int countSetBits(int N) {  int count = 0;  int k=findk(N);  int val , x;    // Iterating from largest power to 2 such that   // pow(2,k) to 0  for (int i = k; i >= 0; i--)  {  val=pow(2,i);  x=val & N; //x will store Bitwise AND of N & val    if(x > 0)  { count++;  }  }  return count;//return count of set bits } // Drive Code int main() {  int N = 15;    // Function call  cout << countSetBits(N) << endl;  return 0; } 

/*package whatever //do not write package name here */ import java.io.*; // import java.lang.Math; public class GFG {  // Function to find largest power of 2 such that  // pow(2,k) <= N  static int findk(int n)  {  int k = 0;  int i = 0;  int val = (int)Math.pow(2, i);  while (val <= n) {  k = i;  i++;  val = (int)Math.pow(2, i);  }  return k;  }  // Function to count set bits in a number  static int countSetBits(int N)  {  int count = 0;  int k = findk(N);  int val, x;  // Iterating from largest power to 2 such that  // pow(2,k) to 0  for (int i = k; i >= 0; i--) {  val = (int)Math.pow(2, i);  x = val  & N; // x will store Bitwise AND of N & val  if (x > 0) {  count++;  }  }  return count; // return count of set bits  }  // Driver code  public static void main(String[] args)  {  int N = 15;  // Function call  System.out.println(countSetBits(N));  } } 

# Python implementation of the above approach import math # Function to find largest power of 2 such that # pow(2,k) <= N def findk(n): i = 0 val = math.pow(2, i) while val <= n: k = i i += 1 val = math.pow(2, i) return k # Function to count set bits in a number def countSetBits(N): count = 0 k = findk(N) for i in range(k, -1, -1): val = int(math.pow(2, i)) x = val & N # x will store Bitwise AND of N & val if x > 0: count += 1 return count # Drive Code if __name__ == '__main__': N = 15 # Function call print(countSetBits(N)) 

using System; class Program {  // Function to find largest power of 2 such that  // pow(2,k) <= N  static int FindK(int n)  {  int k = 0;  int i = 0;  int val = (int)Math.Pow(2, i);  while (val <= n)  {  k = i;  i++;  val = (int)Math.Pow(2, i);  }  return k;  }  // Function to count set bits in a numnber  static int CountSetBits(int N)  {  int count = 0;  int k = FindK(N);  int val, x;  // Iterating from largest power to 2 such that  // pow(2,k) to 0  for (int i = k; i >= 0; i--)  {  val = (int)Math.Pow(2, i);  x = val & N; //x will store Bitwise AND of N & val  if (x > 0)  {  count++;  }  }  return count; //return count of set bits  }  // Drive Code  static void Main()  {  int N = 15;  // Function call  Console.WriteLine(CountSetBits(N));  } } //this code is contributed by ajay 

// JavaScript implementation of the above approach // function to find largest power of 2 such that // pow(2,k) <= N function findk(n){  let k;  let i = 0;  let val = Math.pow(2,i);  while(val <= n){  k=i;  i++;  val = Math.pow(2,i);  }  return k; } // function to count set bits in a number function countSetBits(N){  let count = 0;  let k = findk(N);  let val;  let x;    // iterating from largest power to 2 such that  // pow(2,k) to 0  for(let i = k; i>=0; i--){  val = Math.pow(2,i);  x = val & N; // x will store bitwise and of N and val    if(x > 0) count++;  }  return count; // return count of set bits } // driver program let N = 15; // function call console.log(countSetBits(N)); 

4 

Time Complexity: O(logn) 
Auxiliary Space: O(1)

Using Library or Quick Syntax

In C++ GCC, we can directly count set bits using __builtin_popcount(). So we can avoid a separate function for counting set bits. Similarly, we have direct syntax in other programming languages.

// C++ program to demonstrate __builtin_popcount() #include <iostream> using namespace std; int main() {  cout << __builtin_popcount(4) << endl;  cout << __builtin_popcount(15);  return 0; } 

// java program to demonstrate // __builtin_popcount() import java.io.*; class GFG {  // Driver code  public static void main(String[] args)  {  System.out.println(Integer.bitCount(4));  System.out.println(Integer.bitCount(15));  } } 

print(bin(4).count('1')); print(bin(15).count('1')); # This code is Contributed by mits 

using System; class Program {  static void Main(string[] args)  {  Console.WriteLine(Convert.ToString(4, 2).Replace("0", "").Length);  Console.WriteLine(Convert.ToString(15, 2).Replace("0", "").Length);  } } 

// Javascript program to demonstrate // __builtin_popcount() console.log ((4).toString(2).split('').  filter(x => x == '1').length + "<br>"); console.log ((15).toString(2).split('').  filter(x => x == '1').length); 

<?php // PHP program to demonstrate // __builtin_popcount() // Driver code $t = log10(4); $x = log(15, 2); $tt = ceil($t); $xx = ceil($x); echo ($tt), "\n"; echo ($xx), "\n"; ?> 

1 4