Count total set bits in all numbers from range L to R

Given two positive integer L and R, the task is to count the total number of set bits in the binary representation of all the numbers from L to R

Examples:

Input: L = 3, R = 5 
Output:
Explanation: (3)10 = (11)2, (4)10 = (100)2, (5)10 = (101)2 
So, Total set bit in range 3 to 5 is 5
Input: L = 10, R = 15 
Output: 17

Method 1 – Naive Approach: The idea is to run a loop from L to R and sum the count of set bits in all numbers from L to R.
Below is the implementation of the above approach:
 

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count set bits in x
unsigned int
countSetBitsUtil(unsigned int x)
{
    // Base Case
    if (x <= 0)
        return 0;
 
    // Recursive Call
    return ((x % 2 == 0 ? 0 : 1)
            + countSetBitsUtil(x / 2));
}
 
// Function that returns count of set bits
// present in all numbers from 1 to N
unsigned int countSetBits(unsigned int L,
                          unsigned int R)
{
    // Initialize the result
    int bitCount = 0;
 
    for (int i = L; i <= R; i++) {
        bitCount += countSetBitsUtil(i);
    }
 
    // Return the setbit count
    return bitCount;
}
 
// Driver Code
int main()
{
    // Given L and R
    int L = 3, R = 5;
 
    // Function Call
    printf("Total set bit count is %d",
           countSetBits(L, R));
    return 0;
}

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// Java program for the above approach
class GFG{
 
// Function to count set bits in x
static int countSetBitsUtil(int x)
{
    // Base Case
    if (x <= 0)
        return 0;
 
    // Recursive Call
    return ((x % 2 == 0 ? 0 : 1) +
             countSetBitsUtil(x / 2));
}
 
// Function that returns count of set bits
// present in all numbers from 1 to N
static int countSetBits(int L, int R)
{
    // Initialize the result
    int bitCount = 0;
 
    for (int i = L; i <= R; i++)
    {
        bitCount += countSetBitsUtil(i);
    }
 
    // Return the setbit count
    return bitCount;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given L and R
    int L = 3, R = 5;
 
    // Function Call
    System.out.printf("Total set bit count is %d",
                                 countSetBits(L, R));
}
}
 
// This code is contributed by Rajput-Ji

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Python3

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# Python3 program for the above approach
 
# Function to count set bits in x
def countSetBitsUtil(x):
   
    # Base Case
    if (x < 1):
        return 0;
 
    # Recursive Call
    if (x % 2 == 0):
        return 0;
    else:
        return 1 + (countSetBitsUtil(x / 2));
 
# Function that returns count of set bits
# present in all numbers from 1 to N
def countSetBits(L, R):
   
    # Initialize the result
    bitCount = 0;
 
    for i in range(L, R + 1):
        bitCount += countSetBitsUtil(i);
 
    # Return the setbit count
    return bitCount;
 
# Driver Code
if __name__ == '__main__':
   
    # Given L and R
    L = 3;
    R = 5;
 
    # Function Call
    print("Total set bit count is ",
                countSetBits(L, R));
 
# This code is contributed by Princi Singh

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// C# program for the above approach
using System;
class GFG{
 
// Function to count set bits in x
static int countSetBitsUtil(int x)
{
    // Base Case
    if (x <= 0)
        return 0;
 
    // Recursive Call
    return ((x % 2 == 0 ? 0 : 1) +
             countSetBitsUtil(x / 2));
}
 
// Function that returns count of set bits
// present in all numbers from 1 to N
static int countSetBits(int L, int R)
{
    // Initialize the result
    int bitCount = 0;
 
    for (int i = L; i <= R; i++)
    {
        bitCount += countSetBitsUtil(i);
    }
 
    // Return the setbit count
    return bitCount;
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given L and R
    int L = 3, R = 5;
 
    // Function Call
    Console.Write("Total set bit count is {0}",
                           countSetBits(L, R));
}
}
 
// This code is contributed by Rajput-Ji

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Output: 



Total set bit count is 5





 

Time Complexity: O(N*Log N) 
Auxiliary Space: O(1)
Method 2 – Better Approach: The idea is to observe bits from rightmost side at distance i than bits get inverted after 2i position in vertical sequence. 
For Example: 
 

L = 3, R = 5
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101





Observe the right most bit (i = 0) the bits get flipped after (20 = 1)
Observe the 3rd rightmost bit (i = 2) the bits get flipped after (22 = 4). 
Therefore, We can count bits in vertical fashion such that at i’th right most position bits will be get flipped after 2i iteration.
Below is the implementation of the above approach:
 

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that counts the set bits
// from 0 to N
int countSetBit(int n)
{
    int i = 0;
 
    // To store sum of set bits from 0 - N
    int ans = 0;
 
    // Untill n >= to 2^i
    while ((1 << i) <= n) {
 
        // This k will get flipped after
        // 2^i iterations
        bool k = 0;
 
        // Change is iterator from 2^i to 1
        int change = 1 << i;
 
        // This will loop from 0 to n for
        // every bit position
        for (int j = 0; j <= n; j++) {
 
            ans += k;
 
            if (change == 1) {
 
                // When change = 1 flip the bit
                k = !k;
 
                // Again set change to 2^i
                change = 1 << i;
            }
            else {
                change--;
            }
        }
 
        // Increment the position
        i++;
    }
 
    return ans;
}
 
// Function that counts the set bit
// in the range (L, R)
int countSetBits(int L, int R)
{
 
    // Return the count
    return abs(countSetBit(R)
               - countSetBit(L - 1));
}
 
// Driver Code
int main()
{
    // Given L and R
    int L = 3, R = 5;
 
    // Function Call
    cout << "Total set bit count is "
         << countSetBits(L, R) << endl;
    return 0;
}

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// Java program for the above approach
import java.util.*;
class GFG{
 
// Function that counts the set bits
// from 0 to N
static int countSetBit(int n)
{
    int i = 0;
 
    // To store sum of set bits from 0 - N
    int ans = 0;
 
    // Untill n >= to 2^i
    while ((1 << i) <= n)
    {
 
        // This k will get flipped after
        // 2^i iterations
        boolean k = true;
 
        // Change is iterator from 2^i to 1
        int change = 1 << i;
 
        // This will loop from 0 to n for
        // every bit position
        for (int j = 0; j <= n; j++)
        {
            ans += k==true?0:1;
 
            if (change == 1)
            {
 
                // When change = 1 flip the bit
                k = !k;
 
                // Again set change to 2^i
                change = 1 << i;
            }
            else
            {
                change--;
            }
        }
 
        // Increment the position
        i++;
    }
 
    return ans;
}
 
// Function that counts the set bit
// in the range (L, R)
static int countSetBits(int L, int R)
{
 
    // Return the count
    return Math.abs(countSetBit(R) -
                    countSetBit(L - 1));
}
 
// Driver Code
public static void main(String[] args)
{
    // Given L and R
    int L = 3, R = 5;
 
    // Function Call
    System.out.print("Total set bit count is " +
                      countSetBits(L, R) +"\n");
}
}
 
// This code is contributed by gauravrajput1

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# Python3 program for the above approach
 
# Function that counts the set bits
# from 0 to N
def countSetBit(n):
    i = 0;
 
    # To store sum of set bits from 0 - N
    ans = 0;
 
    # Untill n >= to 2^i
    while ((1 << i) <= n):
 
        # This k will get flipped after
        # 2^i iterations
        k = True;
 
        # Change is iterator from 2^i to 1
        change = 1 << i;
 
        # This will loop from 0 to n for
        # every bit position
        for j in range(n+1):
            ans += 0 if k == True else 1;
 
            if (change == 1):
 
                # When change = 1 flip the bit
                k = False if k == True else True;
 
                # Again set change to 2^i
                change = 1 << i;
            else:
                change -=1;
 
        # Increment the position
        i += 1;
 
    return ans;
 
# Function that counts the set bit
# in the range (L, R)
def countSetBits(L, R):
   
    # Return the count
    return abs(countSetBit(R) -
               countSetBit(L - 1));
 
# Driver Code
if __name__ == '__main__':
   
    # Given L and R
    L = 3;
    R = 5;
 
    # Function Call
    print("Total set bit count is "
          countSetBits(L, R));
 
# This code is contributed by Rajput-Ji

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// C# program for the above approach
using System;
class GFG{
 
// Function that counts the set bits
// from 0 to N
static int countSetBit(int n)
{
    int i = 0;
 
    // To store sum of set bits from 0 - N
    int ans = 0;
 
    // Untill n >= to 2^i
    while ((1 << i) <= n)
    {
 
        // This k will get flipped after
        // 2^i iterations
        bool k = true;
 
        // Change is iterator from 2^i to 1
        int change = 1 << i;
 
        // This will loop from 0 to n for
        // every bit position
        for (int j = 0; j <= n; j++)
        {
            ans += k==true?0:1;
 
            if (change == 1)
            {
 
                // When change = 1 flip the bit
                k = !k;
 
                // Again set change to 2^i
                change = 1 << i;
            }
            else
            {
                change--;
            }
        }
 
        // Increment the position
        i++;
    }
 
    return ans;
}
 
// Function that counts the set bit
// in the range (L, R)
static int countSetBits(int L, int R)
{
 
    // Return the count
    return Math.Abs(countSetBit(R) -
                    countSetBit(L - 1));
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given L and R
    int L = 3, R = 5;
 
    // Function Call
    Console.Write("Total set bit count is " +
                   countSetBits(L, R) +"\n");
}
}
 
// This code is contributed by Rajput-Ji

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Output: 

Total set bit count is 5





 

 

Time Complexity: O((L + R)*K), where K is the number of bits in L and R. 
Auxiliary Space: O(1)
 

 

Method 3 – Tricky If the input number is of the form 2b – 1 e.g., 1, 3, 7, 15, … etc, the number of set bits is b * 2(b-1). This is because for all the numbers 0 to 2b – 1, if you complement and flip the list you end up with the same list (half the bits are set and half bits are unset).



 

If the number does not have all set bits, then let m is the position of leftmost set bit. The number of set bits in that position is n – (1 << m) + 1. The remaining set bits are in two parts:

 

  1. The bits in the (m – 1) positions down to the point where the leftmost bit becomes 0
  2. The 2(m – 1) numbers below that point, which is the closed form above.

 

For Example: N = 6 
 

0|0 0
0|0 1
0|1 0
0|1 1
-|--
1|0 0
1|0 1
1|1 0





 

From the above we have: 

 

  • The leftmost set bit is in position 2 (positions are considered starting from 0).
  • If we mask that off what remains is 2 (the “1 0” in the right part of the last row), So the number of bits in the 2nd position (the lower left box) is 3 (that is, 2 + 1).
  • The set bits from 0-3 (the upper right box above) is 2*2(2 – 1) = 4.
  • The box in the lower right is the remaining bits we haven’t yet counted, and is the number of set bits for all the numbers up to 2 (the value of the last entry in the lower right box) which can be figured recursively.

 

Below is the implementation of the above approach:
 



 

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
unsigned int countSetBit(unsigned int n);
 
// Returns position of leftmost set bit
// The rightmost position is taken as 0
unsigned int getLeftmostBit(int n)
{
    int m = 0;
    while (n > 1) {
        n = n >> 1;
        m++;
    }
    return m;
}
 
// Function that gives the position of
// previous leftmost set bit in n
unsigned int getNextLeftmostBit(int n, int m)
{
    unsigned int temp = 1 << m;
    while (n < temp) {
        temp = temp >> 1;
        m--;
    }
    return m;
}
 
// Function to count the set bits between
// the two numbers N and M
unsigned int _countSetBit(unsigned int n,
                          int m)
{
    // Base Case
    if (n == 0)
        return 0;
 
    // Get position of next leftmost set bit
    m = getNextLeftmostBit(n, m);
 
    // If n is of the form 2^x-1
    if (n == ((unsigned int)1 << (m + 1)) - 1)
        return (unsigned int)(m + 1) * (1 << m);
 
    // Update n for next recursive call
    n = n - (1 << m);
    return ((n + 1)
            + countSetBit(n)
            + m * (1 << (m - 1)));
}
 
// Function that returns count of set
// bits present in all numbers from 1 to n
unsigned int countSetBit(unsigned int n)
{
    // Get the position of leftmost set
    // bit in n
    int m = getLeftmostBit(n);
 
    // Use the position
    return _countSetBit(n, m);
}
 
// Function that counts the set bits
// between L and R
int countSetBits(int L, int R)
{
    return abs(countSetBit(R)
               - countSetBit(L - 1));
}
 
// Driver Code
int main()
{
    // Given L and R
    int L = 3, R = 5;
 
    // Function Call
    cout << "Total set bit count is "
         << countSetBits(L, R);
    return 0;
}

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// Java program for the above approach
import java.util.*;
class GFG{
 
// Returns position of leftmost set bit
// The rightmost position is taken as 0
static  int getLeftmostBit(int n)
{
    int m = 0;
    while (n > 1)
    {
        n = n >> 1;
        m++;
    }
    return m;
}
 
// Function that gives the position of
// previous leftmost set bit in n
static  int getNextLeftmostBit(int n, int m)
{
     int temp = 1 << m;
    while (n < temp)
    {
        temp = temp >> 1;
        m--;
    }
    return m;
}
 
// Function that returns count of set
// bits present in all numbers from 1 to n
static int countSetBit( int n)
{
   // Get the position of leftmost set
   // bit in n
   int m = getLeftmostBit(n);
 
   // Use the position
   return _countSetBit(n, m);
}
 
// Function to count the set bits between
// the two numbers N and M
static int _countSetBit(int n, int m)
{
    // Base Case
    if (n == 0)
        return 0;
     
    // Get position of next leftmost set bit
    m = getNextLeftmostBit(n, m);
 
    // If n is of the form 2^x-1
    if (n == (( int)1 << (m + 1)) - 1)
        return ( int)(m + 1) * (1 << m);
 
    // Update n for next recursive call
    n = n - (1 << m);
    return ((n + 1) +
            countSetBit(n) +
            m * (1 << (m - 1)));
}
 
// Function that counts the set bits
// between L and R
static int countSetBits(int L, int R)
{
    return Math.abs(countSetBit(R) -
                    countSetBit(L - 1));
}
 
// Driver Code
public static void main(String[] args)
{
    // Given L and R
    int L = 3, R = 5;
 
    // Function Call
    System.out.print("Total set bit count is " +
                             countSetBits(L, R));
}
}
 
// This code is contributed by sapnasingh4991

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# Python program for the above approach
 
 
# Returns position of leftmost set bit
# The rightmost position is taken as 0
def getLeftmostBit(n):
    m = 0;
    while (n > 1):
        n = n >> 1;
        m += 1;
 
    return m;
 
 
# Function that gives the position of
# previous leftmost set bit in n
def getNextLeftmostBit(n, m):
    temp = 1 << m;
    while (n < temp):
        temp = temp >> 1;
        m -=1;
 
    return m;
 
 
# Function that returns count of set
# bits present in all numbers from 1 to n
def countSetBit(n):
    # Get the position of leftmost set
    # bit in n
    m = getLeftmostBit(n);
 
    # Use the position
    return _countSetBit(n, m);
 
 
# Function to count the set bits between
# the two numbers N and M
def _countSetBit(n, m):
    # Base Case
    if (n == 0):
        return 0;
 
    # Get position of next leftmost set bit
    m = getNextLeftmostBit(n, m);
 
    # If n is of the form 2^x-1
    if (n == int(1 << (m + 1)) - 1):
        return int(m + 1) * (1 << m);
 
 
    # Update n for next recursive call
    n = n - (1 << m);
    return ((n + 1) + countSetBit(n) + m * (1 << (m - 1)));
 
 
# Function that counts the set bits
# between L and R
def countSetBits(L, R):
    return abs(countSetBit(R) - countSetBit(L - 1));
 
 
# Driver Code
if __name__ == '__main__':
    # Given L and R
    L = 3;
    R = 5;
 
    # Function Call
    print("Total set bit count is " , countSetBits(L, R));
 
# This code contributed by shikhasingrajput

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// C# program for the above approach
using System;
 
class GFG{
 
// Returns position of leftmost set bit
// The rightmost position is taken as 0
static int getLeftmostBit(int n)
{
    int m = 0;
    while (n > 1)
    {
        n = n >> 1;
        m++;
    }
    return m;
}
 
// Function that gives the position of
// previous leftmost set bit in n
static int getNextLeftmostBit(int n, int m)
{
    int temp = 1 << m;
    while (n < temp)
    {
        temp = temp >> 1;
        m--;
    }
    return m;
}
 
// Function that returns count of set
// bits present in all numbers from 1 to n
static int countSetBit(int n)
{
     
    // Get the position of leftmost set
    // bit in n
    int m = getLeftmostBit(n);
     
    // Use the position
    return _countSetBit(n, m);
}
 
// Function to count the set bits between
// the two numbers N and M
static int _countSetBit(int n, int m)
{
     
    // Base Case
    if (n == 0)
        return 0;
     
    // Get position of next leftmost set bit
    m = getNextLeftmostBit(n, m);
 
    // If n is of the form 2^x-1
    if (n == (( int)1 << (m + 1)) - 1)
        return ( int)(m + 1) * (1 << m);
 
    // Update n for next recursive call
    n = n - (1 << m);
    return ((n + 1) +
            countSetBit(n) +
            m * (1 << (m - 1)));
}
 
// Function that counts the set bits
// between L and R
static int countSetBits(int L, int R)
{
    return Math.Abs(countSetBit(R) -
                    countSetBit(L - 1));
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given L and R
    int L = 3, R = 5;
 
    // Function call
    Console.Write("Total set bit count is " +
                          countSetBits(L, R));
}
}
 
// This code is contributed by Amit Katiyar

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Output: 

Total set bit count is 5





 

Time Complexity: O(log N) 
Auxiliary Space: O(1)
Method 4 – using setbit: In setbit method count one by one set bit of each number in range L to R using last bit, check to last bit and if it is set then increase the count and finally sum up over it.
Below is the implementation of the above approach:
 

C++

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// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to count set bit in range
int countSetBits(int L, int R)
{
    // Count variable
    int count = 0;
 
    for (int i = L; i <= R; i++) {
 
        // Find the set bit in Nth number
        int n = i;
        while (n > 0) {
 
            // If last bit is set
            count += (n & 1);
 
            // Left sift by one bit
            n = n >> 1;
        }
    }
 
    // Return count
    return count;
}
 
// Driver Code
int main()
{
    // Given Range L and R
    int L = 3, R = 5;
 
    // Function Call
    cout << "Total set Bit count is "
         << countSetBits(L, R);
 
    return 0;
}

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// Java program for the above approach
class GFG{
   
// Function to count set bit in range
static int countSetBits(int L, int R)
{
    // Count variable
    int count = 0;
  
    for (int i = L; i <= R; i++)
    {
  
        // Find the set bit in Nth number
        int n = i;
        while (n > 0)
        {
  
            // If last bit is set
            count += (n & 1);
  
            // Left sift by one bit
            n = n >> 1;
        }
    }
  
    // Return count
    return count;
}
  
// Driver Code
public static void main(String[] args)
{
    // Given Range L and R
    int L = 3, R = 5;
  
    // Function Call
    System.out.print("Total set Bit count is " +
                             countSetBits(L, R));
}
}
 
// This code is contributed by Ritik Bansal

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Python3

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# Python3 program for the above approach
 
# Function to count set bit in range
def countSetBits(L, R):
     
    # Count variable
    count = 0;
 
    for i in range(L, R + 1):
 
        # Find the set bit in Nth number
        n = i;
        while (n > 0):
 
            # If last bit is set
            count += (n & 1);
 
            # Left sift by one bit
            n = n >> 1;
 
    # Return count
    return count;
 
# Driver Code
if __name__ == '__main__':
     
    # Given range L and R
    L = 3; R = 5;
 
    # Function call
    print("Total set Bit count is ",
           countSetBits(L, R));
     
# This code is contributed by Amit Katiyar

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C#

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// C# program for the above approach
using System;
 
class GFG{
 
// Function to count set bit in range
static int countSetBits(int L, int R)
{
     
    // Count Variable
    int count = 0;
 
    for(int i = L; i <= R; i++)
    {
         
        // Find the set bit in Nth number
        int n = i;
         
        while (n > 0)
        {
             
            // If last bit is set
            count += (n & 1);
 
            // Left sift by one bit
            n = n >> 1;
        }
    }
 
    // Return count
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given Range L and R
    int L = 3, R = 5;
 
    // Function Call
    Console.Write("Total set Bit count is " +
                   countSetBits(L, R));
}
}
 
// This code is contributed by Amit Katiyar

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Output: 

Total set Bit count is 5





 

Time Complexity: O(N) 
Auxiliary Space: O(1)
Method 5 – Using STL __builtin_popcount() function: STL provides an inbuilt function for counting set a bit in an integer, so here call that function for every number in range L to R and count set bits.

C++

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// C++ program for the above approach
#include <iostream>
using namespace std;
 
// Function to count set bit in [L, R]
int countSetBits(int L, int R)
{
    // Variable for count set
    // bit in range
    int count = 0;
 
    // Count set bit for all
    // number in range
    for (int i = L; i <= R; i++) {
 
        // Use inbuilt function
        count += __builtin_popcount(i);
    }
 
    return count;
}
 
// Driver Code
int main()
{
    // Given range L and R
    int L = 3, R = 5;
 
    // Function Call
    cout << "Total set bit count is "
         << countSetBits(L, R);
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
 
// Function to count set bit in [L, R]
static int countSetBits(int L, int R)
{
    // Variable for count set
    // bit in range
    int count = 0;
 
    // Count set bit for all
    // number in range
    for (int i = L; i <= R; i++)
    {
 
        // Use inbuilt function
        count += Integer.bitCount(i);
    }
 
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
    // Given range L and R
    int L = 3, R = 5;
 
    // Function Call
    System.out.print("Total set bit count is " +
                            countSetBits(L, R));
}
}
 
// This code is contributed by Rajput-Ji

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Python3

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# Python3 program for the above approach
 
# Function to count set bit in [L, R]
def countSetBits(L, R):
   
    # Variable for count set
    # bit in range
    count = 0;
 
    # Count set bit for all
    # number in range
    for i in range(L, R + 1):
       
        # Use inbuilt function
        count += countSetBit(i);
 
    return count;
 
def  countSetBit(n):
    count = 0
    while (n):
        count += n & 1
        n >>= 1
    return count
   
# Driver Code
if __name__ == '__main__':
   
    # Given range L and R
    L = 3;
    R = 5;
 
    # Function Call
    print("Total set bit count is " ,
          countSetBits(L, R));
 
# This code is contributed by sapnasingh4991

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C#

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// C# program for the above approach
using System;
class GFG{
 
// Function to count set bit in [L, R]
static int countSetBits(int L, int R)
{
    // Variable for count set
    // bit in range
    int count = 0;
 
    // Count set bit for all
    // number in range
    for (int i = L; i <= R; i++)
    {
 
        // Use inbuilt function
        count += countSetBits(i);
    }
 
    return count;
}
static int countSetBits(long x)
{
    int setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}
// Driver Code
public static void Main(String[] args)
{
    // Given range L and R
    int L = 3, R = 5;
 
    // Function Call
    Console.Write("Total set bit count is " +
                         countSetBits(L, R));
}
}
 
// This code is contributed by gauravrajput1

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Output: 

Total set bit count is 5





 

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

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