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

1. Simple Method Loop through all bits in an integer, check if a bit is set and if it is, then increment the set bit count. See the program below. 

C++

// 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;
}

// This code is contributed
// by Akanksha Rai

C

// 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

// 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));
    }
}

// This code is contributed by Anshika Goyal.

Python3

# 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))

# This code is contributed by
# Smitha Dinesh Semwal

C#

// 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));
    }
}

// This code is contributed by Sam007

PHP

<?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);

// This code is contributed by ajit
?>

Javascript

<script>
   // 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;
   document.write(countSetBits(i));

     // This code is contributed by noob2000.
 </script>

Output

2

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

Recursive Approach:

C++

// 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;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

C

// 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;
}

// This code is contributed by Sania Kumari Gupta (kriSania804)

Java

// 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));
    }
}

// This code is contributes by sunnysingh

Python3

# 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))

# This code is contributed by sunnysingh

C#

// 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));
    }
}

// This code is contributed by aj_36

PHP

<?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);

// This code is contributed by m_kit.
?>

Javascript

<script>

// 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
    document.write(countSetBits(n));


// This code is contributed by Mayank Tyagi
</script>

Output

2

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

2. Brian Kernighan’s Algorithm:
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.

Implementation of Brian Kernighan’s Algorithm:

C++

// 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

// 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

// 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.

Python3

# 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))

# This code is contributed by
# Smitha Dinesh Semwal

C#

// 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

PHP

<?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);

// This code is contributed
// by akt_mit
?>

Javascript

<script>

// 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;
document.write(countSetBits(i));

// This code is contributed by 29AjayKumar

</script>

Output

2

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

Recursive Approach:

C++

// CPP implementation for recursive
// approach to find the number of set
// bits using Brian Kernighan’s Algorithm
#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
        return 1 + countSetBits(n & (n - 1));
}

// driver code
int main()
{
    // get value from user
    int n = 9;

    // function calling
    cout << countSetBits(n);

    return 0;
}

// This code is contributed by Raj.

Java

// Java implementation for recursive
// approach to find the number of set
// bits using Brian Kernighan Algorithm
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
            return 1 + countSetBits(n & (n - 1));
    }

    // Driver function
    public static void main(String[] args)
    {

        // get value from user
        int n = 9;

        // function calling
        System.out.println(countSetBits(n));
    }
}

// This code is contributed by sunnysingh

Python3

# Python3 implementation for
# recursive approach to find
# the number of set bits using
# Brian Kernighan’s Algorithm

# recursive function to count
# set bits
def countSetBits(n):

    # base case
    if (n == 0):
        return 0
    else:
        return 1 + countSetBits(n & (n - 1))


# Get value from user
n = 9

# function calling
print(countSetBits(n))

# This code is contributed by sunnysingh

C#

// C# implementation for recursive
// approach to find the number of set
// bits using Brian Kernighan Algorithm
using System;

class GFG {

    // recursive function
    // to count set bits
    public static int countSetBits(int n)
    {

        // base case
        if (n == 0)
            return 0;
        else
            return 1 + countSetBits(n & (n - 1));
    }

    // Driver Code
    static public void Main()
    {

        // get value from user
        int n = 9;

        // function calling
        Console.WriteLine(countSetBits(n));
    }
}

// This code is contributed by aj_36

PHP

<?php
// PHP implementation for
// recursive approach to
// find the number of set
// bits using Brian
// Kernighan’s Algorithm

// recursive function to
// count set bits
function countSetBits($n)
{
    // base case
    if ($n == 0)
        return 0;
    else
        return 1 +
          countSetBits($n &
                      ($n - 1));
}

// Driver Code

// get value from user
$n = 9;

// function calling
echo countSetBits($n);

// This code is contributed by ajit.
?>

Javascript

<script>

// Javascript implementation for recursive
// approach to find the number of set
// bits using Brian Kernighan’s Algorithm

// recursive function to count set bits
function countSetBits(n)
{
    // base case
    if (n == 0)
        return 0;
    else
        return 1 + countSetBits(n & (n - 1));
}

// driver code
// get value from user

var n = 9;
// function calling

document.write(countSetBits(n));

</script>

Output

2

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

3. Using Lookup table: We can count bits in O(1) time using the lookup table.
Below is the implementation of the above approach:

C++

// 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);
}

// This code is contributed by Sanjit_Kumar

Java

// 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

# 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))

# This code is contributed by SHUBHAMSINGH10

C#

// 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));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// 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;
document.write(countSetBits(n));

// This code is contributed by 29AjayKumar

</script>

Output

2

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

We can find one use of counting set bits at Count number of bits to be flipped to convert A to B
Note: In GCC, we can directly count set bits using __builtin_popcount(). So we can avoid a separate function for counting set bits. 

C++

// 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

// 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));
    }
}

// This code is contributed by Raj

Python3

print(bin(4).count('1'));
print(bin(15).count('1'));

# This code is Contributed by mits

C#

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);
    }
}

PHP

<?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";

// This code is contributed
// by jit_t
?>

Javascript

<script>

// Javascript program to demonstrate
// __builtin_popcount()

document.write((4).toString(2).split('').
  filter(x => x == '1').length + "<br>");
document.write((15).toString(2).split('').
  filter(x => x == '1').length);

</script>

Output

1
4

Time complexity: O(1)
Auxiliary space: O(1)

4. Mapping numbers with the bit. 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++

// 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;
}

// This code is contributed by rathbhupendra

C

// 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

// 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

# 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#

// 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

PHP

<?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
?>

Javascript

<script>

// 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;
document.write(countSetBitsRec(num));

// This code is contributed by 29AjayKumar

</script>

Output

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.

5. Checking each bit in a number:

Each bit in the number is checked for whether it is set or not. The number is bitwise AND with powers of 2, so if the result is not equal to zero, we come to know that the particular bit in the position is set.

C

#include <stdio.h>

// Check each bit in a number is set or not
// and return the total count of the set bits.
int countSetBits(int N)
{
    int count = 0;

    // (1 << i) = pow(2, i)
    for (int i = 0; i < sizeof(int) * 8; i++) {
        if (N & (1 << i))
            count++;
    }
    return count;
}

// Driver Code
int main()
{
    int N = 15;

    printf("%d", countSetBits(N));
    return 0;
}

C++

#include <iostream>
using namespace std;

// Check each bit in a number is set or not
// and return the total count of the set bits.
int countSetBits(int N)
{
    int count = 0;
    // (1 << i) = pow(2, i)
    for (int i = 0; i < sizeof(int) * 8; i++) {
        if (N & (1 << i))
            count++;
    }
    return count;
}

int main()
{

    int N = 15;

    cout << countSetBits(N) << endl;
    return 0;
}

Java

import java.util.*;
public class GFG
{

  // Check each bit in a number is set or not
  // and return the total count of the set bits.
  static int countSetBits(int N)
  {
    int count = 0;
    // (1 << i) = pow(2, i)
    for (int i = 0; i < 4 * 8; i++)
    {
      if ((N & (1 << i)) != 0)
        count++;
    }
    return count;
  }

  // Driver code
  public static void main(String[] args)
  {
    int N = 15;
    System.out.println(countSetBits(N));
  }
}

// This code is contributed by divyeshrabadiya07.

Python3

# Check each bit in a number is set or not
# and return the total count of the set bits


def countSetBits(N):
    count = 0

    # (1 << i) = pow(2, i)
    for i in range(4*8):
        if(N & (1 << i)):
            count += 1

    return count

    # Driver code
N = 15
print(countSetBits(N))

# This code is contributed by avanitrachhadiya2155

C#

using System;
class GFG
{

  // Check each bit in a number is set or not
  // and return the total count of the set bits.
  static int countSetBits(int N)
  {
    int count = 0;

    // (1 << i) = pow(2, i)
    for (int i = 0; i < 4 * 8; i++)
    {
      if ((N & (1 << i)) != 0)
        count++;
    }
    return count;
  }

  // Driver code
  static void Main()
  {
    int N = 15;
    Console.WriteLine(countSetBits(N));
  }
}

// This code is contributed by divyesh072019.

Javascript

<script>

  // Check each bit in a number is set or not
  // and return the total count of the set bits.
  function countSetBits(N)
  {
      var count = 0;
    // (1 << i) = pow(2, i)
    for (i = 0; i < 4 * 8; i++)
    {
        if ((N & (1 << i)) != 0)
        count++;
    }
    return count;
  }

  // Driver code
  var N = 15;
  document.write(countSetBits(N));

// This code is contributed by Amit Katiyar
</script>

Output

4

Time complexity: O(log n)
Auxiliary space: O(1)

6. 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++

// 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;
}

// This code is contributed by nikhilsainiofficial546

Java

/*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));
    }
}

Python3

# 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))

C#

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

// 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));

// THIS CODE IS CONTRIUBTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)

Output

4

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

Count set bits in an integer Using Lookup Table

Asked in: Adobe, Brocade, Cisco, Juniper Networks, Qualcomm