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Count total set bits in all numbers from 1 to n

  • Difficulty Level : Hard
  • Last Updated : 12 Oct, 2021

Given a positive integer n, count the total number of set bits in binary representation of all numbers from 1 to n. 

Examples: 

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Input: n = 3
Output:  4

Input: n = 6
Output: 9

Input: n = 7
Output: 12

Input: n = 8
Output: 13

Source: Amazon Interview Question 



Method 1 (Simple) 
A simple solution is to run a loop from 1 to n and sum the count of set bits in all numbers from 1 to n.  

C++




// A simple program to count set bits
// in all numbers from 1 to n.
#include <iostream>
using namespace std;
 
// A utility function to count set bits
// in a number x
unsigned int countSetBitsUtil(unsigned int x);
 
// Returns count of set bits present in all
// numbers from 1 to n
unsigned int countSetBits(unsigned int n)
{
    int bitCount = 0; // initialize the result
 
    for (int i = 1; i <= n; i++)
        bitCount += countSetBitsUtil(i);
 
    return bitCount;
}
 
// A utility function to count set bits
// in a number x
unsigned int countSetBitsUtil(unsigned int x)
{
    if (x <= 0)
        return 0;
    return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(x / 2);
}
 
// Driver program to test above functions
int main()
{
    int n = 4;
    cout <<"Total set bit count is " <<countSetBits(n);
    return 0;
}
 
// This code is contributed by shivanisinghss2110.

C




// A simple program to count set bits
// in all numbers from 1 to n.
#include <stdio.h>
 
// A utility function to count set bits
// in a number x
unsigned int countSetBitsUtil(unsigned int x);
 
// Returns count of set bits present in all
// numbers from 1 to n
unsigned int countSetBits(unsigned int n)
{
    int bitCount = 0; // initialize the result
 
    for (int i = 1; i <= n; i++)
        bitCount += countSetBitsUtil(i);
 
    return bitCount;
}
 
// A utility function to count set bits
// in a number x
unsigned int countSetBitsUtil(unsigned int x)
{
    if (x <= 0)
        return 0;
    return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(x / 2);
}
 
// Driver program to test above functions
int main()
{
    int n = 4;
    printf("Total set bit count is %d", countSetBits(n));
    return 0;
}

Java




// A simple program to count set bits
// in all numbers from 1 to n.
 
class GFG{
 
    // Returns count of set bits present
    //  in all numbers from 1 to n
    static int countSetBits( int n)
    {
        // initialize the result
        int bitCount = 0;
     
        for (int i = 1; i <= n; i++)
            bitCount += countSetBitsUtil(i);
     
        return bitCount;
    }
     
    // A utility function to count set bits
    // in a number x
    static int countSetBitsUtil( int x)
    {
        if (x <= 0)
            return 0;
        return (x % 2 == 0 ? 0 : 1) +
               countSetBitsUtil(x / 2);
    }
     
    // Driver program
    public static void main(String[] args)
    {
        int n = 4;
        System.out.print("Total set bit count is ");
        System.out.print(countSetBits(n));
    }
}
 
// This code is contributed by
// Smitha Dinesh Semwal

Python3




# A simple program to count set bits
# in all numbers from 1 to n.
 
# Returns count of set bits present in all
# numbers from 1 to n
def countSetBits(n):
     
    # initialize the result
    bitCount = 0
 
    for i in range(1, n + 1):
        bitCount += countSetBitsUtil(i)
 
    return bitCount
 
 
# A utility function to count set bits
# in a number x
def countSetBitsUtil(x):
 
    if (x <= 0):
        return 0
    return (0 if int(x % 2) == 0 else 1) +  countSetBitsUtil(int(x / 2))
 
 
# Driver program
if __name__=='__main__':
    n = 4
    print("Total set bit count is", countSetBits(n))
      
# This code is contributed by
# Smitha Dinesh Semwal   

C#




// A simple C# program to count set bits
// in all numbers from 1 to n.
using System;
 
class GFG
{
    // Returns count of set bits present
    // in all numbers from 1 to n
    static int countSetBits( int n)
    {
        // initialize the result
        int bitCount = 0;
     
        for (int i = 1; i <= n; i++)
            bitCount += countSetBitsUtil(i);
     
        return bitCount;
    }
     
    // A utility function to count set bits
    // in a number x
    static int countSetBitsUtil( int x)
    {
        if (x <= 0)
            return 0;
        return (x % 2 == 0 ? 0 : 1) +
            countSetBitsUtil(x / 2);
    }
     
    // Driver program
    public static void Main()
    {
        int n = 4;
        Console.Write("Total set bit count is ");
        Console.Write(countSetBits(n));
    }
}
 
// This code is contributed by Sam007

PHP




<?php
// A simple program to count set bits
// in all numbers from 1 to n.
 
// Returns count of set bits present
// in all numbers from 1 to n
function countSetBits($n)
{
    $bitCount = 0; // initialize the result
 
    for ($i = 1; $i <= $n; $i++)
        $bitCount += countSetBitsUtil($i);
 
    return $bitCount;
}
 
// A utility function to count
// set bits in a number x
function countSetBitsUtil($x)
{
    if ($x <= 0)
        return 0;
    return ($x % 2 == 0 ? 0 : 1) +
            countSetBitsUtil($x / 2);
}
 
// Driver Code
$n = 4;
echo "Total set bit count is " .
               countSetBits($n);
 
// This code is contributed by ChitraNayal
?>

Javascript




<script>
// A simple program to count set bits
// in all numbers from 1 to n.
 
     
    // Returns count of set bits present
    //  in all numbers from 1 to n
    function countSetBits(n)
    {
        // initialize the result
        let bitCount = 0;
        for (let i = 1; i <= n; i++)
        {
            bitCount += countSetBitsUtil(i);
        }
        return bitCount;
    }
     
    // A utility function to count set bits
    // in a number x
    function countSetBitsUtil(x)
    {
        if (x <= 0)
        {
            return 0;
        }
        return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(Math.floor(x/2));
    }
     
    // Driver program
    let n = 4;
    document.write("Total set bit count is ");
    document.write(countSetBits(n));
     
    // This code is contributed by rag2127
</script>
Output
Total set bit count is 5

Time Complexity: O(nLogn) 

Method 2 (Simple and efficient than Method 1) 
If we observe bits from rightmost side at distance i than bits get inverted after 2^i position in vertical sequence. 
for example n = 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 (2^0 = 1) 
Observe the 3rd rightmost bit (i = 2) the bits get flipped after (2^2 = 4) 
So, We can count bits in vertical fashion such that at i’th right most position bits will be get flipped after 2^i iteration;

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function which counts set bits from 0 to n
int countSetBits(int n)
{
    int i = 0;
 
    // ans store sum of set bits from 0 to n 
    int ans = 0;
 
    // while n greater than equal 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) {
                k = !k; // When change = 1 flip the bit
                change = 1 << i; // again set change to 2^i
            }
            else {
                change--;
            }
        }
 
        // increment the position
        i++;
    }
 
    return ans;
}
 
// Main Function
int main()
{
    int n = 17;
    cout << countSetBits(n) << endl;
    return 0;
}

Java




public class GFG {
     
    // Function which counts set bits from 0 to n
    static int countSetBits(int n)
    {
        int i = 0;
 
        // ans store sum of set bits from 0 to n
        int ans = 0;
 
        // while n greater than equal to 2^i
        while ((1 << i) <= n) {
 
            // This k will get flipped after
            // 2^i iterations
            boolean k = false;
 
            // 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++) {
 
                if (k == true)
                    ans += 1;
                else
                    ans += 0;
 
                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;
    }
 
    // Driver program
    public static void main(String[] args)
    {
        int n = 17;
         
        System.out.println(countSetBits(n));
    }
}
 
// This code is contributed by Sam007.

Python3




# Function which counts set bits from 0 to n
def countSetBits(n) :
    i = 0
 
    # ans store sum of set bits from 0 to n 
    ans = 0
 
    # while n greater than equal to 2^i
    while ((1 << i) <= n) :
 
        # This k will get flipped after 
        # 2^i iterations
        k = 0
 
        # 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(0, n+1) :
            ans += k
 
            if change == 1 :
                 
                #  When change = 1 flip the bit
                k = not k
 
                # again set change to 2^i
                change = 1 << i
 
            else :
                change -= 1
 
        # increment the position
        i += 1
 
    return ans
 
 
 
# Driver code
if __name__ == "__main__" :
 
    n = 17
    print(countSetBits(n))
  
# This code is contributed by ANKITRAI1

C#




using System;
 
class GFG
{
    static int countSetBits(int n)
    {
        int i = 0;
 
        // ans store sum of set bits from 0 to n
        int ans = 0;
 
        // while n greater than equal to 2^i
        while ((1 << i) <= n) {
 
            // This k will get flipped after
            // 2^i iterations
            bool k = false;
 
            // 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++) {
 
                if (k == true)
                    ans += 1;
                else
                    ans += 0;
 
                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;
    }
 
    // Driver program
    static void Main()
    {
        int n = 17;
        Console.Write(countSetBits(n));
    }
}
 
// This code is contributed by Sam007

PHP




<?php
// Function which counts
// set bits from 0 to n
function countSetBits($n)
{
    $i = 0;
 
    // ans store sum of set
    // bits from 0 to n
    $ans = 0;
 
    // while n greater than
    // equal to 2^i
    while ((1 << $i) <= $n)
    {
 
        // This k will get flipped
        // after 2^i iterations
        $k = 0;
 
        // change is iterator
        // from 2^i to 1
        $change = 1 << $i;
 
        // This will loop from 0 to n
        // for every bit position
        for ($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;
}
 
// Driver code
$n = 17;
echo(countSetBits($n));
 
// This code is contributed by Smitha
?>

Javascript




<script>
 
// Javascript program for the above approach
 
// Function which counts set bits from 0 to n
function countSetBits(n)
{
    let i = 0;
 
    // ans store sum of set bits from 0 to n 
    let ans = 0;
 
    // while n greater than equal to 2^i
    while ((1 << i) <= n) {
 
        // This k will get flipped after
        // 2^i iterations
        let k = 0;
 
        // change is iterator from 2^i to 1
        let change = 1 << i;
 
        // This will loop from 0 to n for
        // every bit position
        for (let 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;
}
     
 
// Driver Code
 
    let n = 17;
    document.write(countSetBits(n));
 
</script>
Output
35

Time Complexity: O(k*n) 
where k = number of bits to represent number n 
k <= 64

Method 3 (Tricky) 
If the input number is of the form 2^b -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 (2^b)-1, if you complement and flip the list you end up with the same list (half the bits are on, half off). 
If the number does not have all set bits, then some position 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, and 
2) The 2^(m-1) numbers below that point, which is the closed form above.
An easy way to look at it is to consider the number 6: 

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

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.  



C++




#include <bits/stdc++.h>
 
// A O(Logn) complexity program to count
// set bits in all numbers from 1 to n
using namespace std;
 
/* Returns position of leftmost set bit.
The rightmost position is considered
as 0 */
unsigned int getLeftmostBit(int n)
{
    int m = 0;
    while (n > 1)
    {
        n = n >> 1;
        m++;
    }
    return m;
}
 
/* Given the position of previous leftmost
set bit in n (or an upper bound on
leftmost position) returns the new
position of 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;
}
 
// The main recursive function used by countSetBits()
unsigned int _countSetBits(unsigned int n, int m);
 
// Returns count of set bits present in
// all numbers from 1 to n
unsigned int countSetBits(unsigned int n)
{
    // Get the position of leftmost set
    // bit in n. This will be used as an
    // upper bound for next set bit function
    int m = getLeftmostBit(n);
 
    // Use the position
    return _countSetBits(n, m);
}
 
unsigned int _countSetBits(unsigned int n, int m)
{
    // Base Case: if n is 0, then set bit
    // count is 0
    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, i.e., if n
    // is like 1, 3, 7, 15, 31, .. etc,
    // then we are done.
    // Since positions are considered starting
    // from 0, 1 is added to m
    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) + countSetBits(n) + m * (1 << (m - 1));
}
 
// Driver code
int main()
{
    int n = 17;
    cout<<"Total set bit count is "<< countSetBits(n);
    return 0;
}
 
// This code is contributed by rathbhupendra

C




// A O(Logn) complexity program to count
// set bits in all numbers from 1 to n
#include <stdio.h>
 
/* Returns position of leftmost set bit.
   The rightmost position is considered
   as 0 */
unsigned int getLeftmostBit(int n)
{
    int m = 0;
    while (n > 1) {
        n = n >> 1;
        m++;
    }
    return m;
}
 
/* Given the position of previous leftmost
   set bit in n (or an upper bound on
   leftmost position) returns the new
   position of 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;
}
 
// The main recursive function used by countSetBits()
unsigned int _countSetBits(unsigned int n, int m);
 
// Returns count of set bits present in
// all numbers from 1 to n
unsigned int countSetBits(unsigned int n)
{
    // Get the position of leftmost set
    // bit in n. This will be used as an
    // upper bound for next set bit function
    int m = getLeftmostBit(n);
 
    // Use the position
    return _countSetBits(n, m);
}
 
unsigned int _countSetBits(unsigned int n, int m)
{
    // Base Case: if n is 0, then set bit
    // count is 0
    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, i.e., if n
    // is like 1, 3, 7, 15, 31, .. etc,
    // then we are done.
    // Since positions are considered starting
    // from 0, 1 is added to m
    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) + countSetBits(n) + m * (1 << (m - 1));
}
 
// Driver program to test above functions
int main()
{
    int n = 17;
    printf("Total set bit count is %d", countSetBits(n));
    return 0;
}

Java




// Java A O(Logn) complexity program to count
// set bits in all numbers from 1 to n
import java.io.*;
 
class GFG
{
     
/* Returns position of leftmost set bit.
The rightmost position is considered
as 0 */
static int getLeftmostBit(int n)
{
    int m = 0;
    while (n > 1)
    {
        n = n >> 1;
        m++;
    }
    return m;
}
 
/* Given the position of previous leftmost
set bit in n (or an upper bound on
leftmost position) returns the new
position of 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;
}
 
// The main recursive function used by countSetBits()
// Returns count of set bits present in
// all numbers from 1 to n
 
static int countSetBits(int n)
{
    // Get the position of leftmost set
    // bit in n. This will be used as an
    // upper bound for next set bit function
    int m = getLeftmostBit(n);
 
    // Use the position
    return countSetBits(n, m);
}
 
static int countSetBits( int n, int m)
{
    // Base Case: if n is 0, then set bit
    // count is 0
    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, i.e., if n
    // is like 1, 3, 7, 15, 31, .. etc,
    // then we are done.
    // Since positions are considered starting
    // from 0, 1 is added to m
    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) + countSetBits(n) + m * (1 << (m - 1));
}
 
// Driver code
public static void main (String[] args)
{
 
    int n = 17;
    System.out.println ("Total set bit count is " + countSetBits(n));
}
}
 
// This code is contributed by ajit..

Python3




# A O(Logn) complexity program to count
# set bits in all numbers from 1 to n
 
"""
/* Returns position of leftmost set bit.
The rightmost position is considered
as 0 */
"""
def getLeftmostBit(n):
 
    m = 0
    while (n > 1) :
 
        n = n >> 1
        m += 1
 
    return m
 
"""
/* Given the position of previous leftmost
set bit in n (or an upper bound on
leftmost position) returns the new
position of leftmost set bit in n */
"""
def getNextLeftmostBit(n, m) :
 
    temp = 1 << m
    while (n < temp) :
        temp = temp >> 1
        m -= 1
 
    return m
 
# The main recursive function used by countSetBits()
# def _countSetBits(n, m)
 
# Returns count of set bits present in
# all numbers from 1 to n
def countSetBits(n) :
 
    # Get the position of leftmost set
    # bit in n. This will be used as an
    # upper bound for next set bit function
    m = getLeftmostBit(n)
 
    # Use the position
    return _countSetBits(n, m)
 
def _countSetBits(n, m) :
 
    # Base Case: if n is 0, then set bit
    # count is 0
    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, i.e., if n
    # is like 1, 3, 7, 15, 31, .. etc,
    # then we are done.
    # Since positions are considered starting
    # from 0, 1 is added to m
    if (n == (1 << (m + 1)) - 1) :
        return ((m + 1) * (1 << m))
 
    # update n for next recursive call
    n = n - (1 << m)
    return (n + 1) + countSetBits(n) + m * (1 << (m - 1))
 
# Driver code
n = 17
print("Total set bit count is", countSetBits(n))
 
# This code is contributed by rathbhupendra

C#




// C# A O(Logn) complexity program to count
// set bits in all numbers from 1 to n
using System;
 
class GFG
{
     
/* Returns position of leftmost set bit.
The rightmost position is considered
as 0 */
static int getLeftmostBit(int n)
{
    int m = 0;
    while (n > 1)
    {
        n = n >> 1;
        m++;
    }
    return m;
}
 
/* Given the position of previous leftmost
set bit in n (or an upper bound on
leftmost position) returns the new
position of 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;
}
 
// The main recursive function used by countSetBits()
// Returns count of set bits present in
// all numbers from 1 to n
static int countSetBits(int n)
{
    // Get the position of leftmost set
    // bit in n. This will be used as an
    // upper bound for next set bit function
    int m = getLeftmostBit(n);
 
    // Use the position
    return countSetBits(n, m);
}
 
static int countSetBits( int n, int m)
{
    // Base Case: if n is 0,
    // then set bit count is 0
    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, i.e., if n
    // is like 1, 3, 7, 15, 31, .. etc,
    // then we are done.
    // Since positions are considered starting
    // from 0, 1 is added to m
    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) + countSetBits(n) +
                  m * (1 << (m - 1));
}
 
// Driver code
static public void Main ()
{
    int n = 17;
    Console.Write("Total set bit count is " +
                            countSetBits(n));
}
}
 
// This code is contributed by Tushil.

Javascript




<script>
 
// JavaScript A O(Logn) complexity program to count
// set bits in all numbers from 1 to n
 
/* Returns position of leftmost set bit.
The rightmost position is considered
as 0 */
function getLeftmostBit(n)
{
    let m = 0;
    while (n > 1)
    {
        n = n >> 1;
        m++;
    }
    return m;
}
 
/* Given the position of previous leftmost
set bit in n (or an upper bound on
leftmost position) returns the new
position of leftmost set bit in n */
function getNextLeftmostBit(n,m)
{
    let temp = 1 << m;
    while (n < temp)
    {
        temp = temp >> 1;
        m--;
    }
    return m;
}
 
// The main recursive function used by countSetBits()
// Returns count of set bits present in
// all numbers from 1 to n
function countSetBits(n)
{
    // Get the position of leftmost set
    // bit in n. This will be used as an
    // upper bound for next set bit function
    let m = getLeftmostBit(n);
  
    // Use the position
    return _countSetBits(n, m);
}
 
function _countSetBits(n,m)
{
    // Base Case: if n is 0, then set bit
    // count is 0
    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, i.e., if n
    // is like 1, 3, 7, 15, 31, .. etc,
    // then we are done.
    // Since positions are considered starting
    // from 0, 1 is added to m
    if (n == (1 << (m + 1)) - 1)
        return (m + 1) * (1 << m);
  
    // update n for next recursive call
    n = n - (1 << m);
    return (n + 1) + countSetBits(n) + m * (1 << (m - 1));
}
 
// Driver code
let n = 17;
document.write("Total set bit count is " + countSetBits(n));
     
// This code is contributed by ab2127
 
</script>
Output
Total set bit count is 35

Time Complexity: O(Logn). From the first look at the implementation, time complexity looks more. But if we take a closer look, statements inside while loop of getNextLeftmostBit() are executed for all 0 bits in n. And the number of times recursion is executed is less than or equal to set bits in n. In other words, if the control goes inside while loop of getNextLeftmostBit(), then it skips those many bits in recursion. 
Thanks to agatsu and IC for suggesting this solution.
Here is another solution suggested by Piyush Kapoor.

A simple solution , using the fact that for the ith least significant bit, answer will be  

(N/2^i)*2^(i-1)+ X

where 

X = N%(2^i)-(2^(i-1)-1)

iff  

N%(2^i)>=(2^(i-1)-1)

C++




int getSetBitsFromOneToN(int N){
    int two = 2,ans = 0;
    int n = N;
    while(n){
        ans += (N/two)*(two>>1);
        if((N&(two-1)) > (two>>1)-1) ans += (N&(two-1)) - (two>>1)+1;
        two <<= 1;
        n >>= 1;
    }
    return ans;
}

Java




static int getSetBitsFromOneToN(int N){
    int two = 2,ans = 0;
    int n = N;
    while(n != 0)
    {
        ans += (N / two)*(two >> 1);
        if((N&(two - 1)) > (two >> 1) - 1)
            ans += (N&(two - 1)) - (two >> 1) + 1;
        two <<= 1;
        n >>= 1;
    }
    return ans;
}
 
// This code is contributed by divyeshrabadiya07.

Python3




def getSetBitsFromOneToN(N):
    two = 2
    ans = 0
    n = N
    while(n != 0)
    {
        ans += int(N / two) * (two >> 1)
        if((N & (two - 1)) > (two >> 1) - 1):
            ans += (N&(two - 1)) - (two >> 1) + 1
        two <<= 1;
        n >>= 1;
    }
    return ans
 
# This code is contributed by avanitrachhadiya2155

C#




static int getSetBitsFromOneToN(int N){
    int two = 2,ans = 0;
    int n = N;
    while(n != 0)
    {
        ans += (N / two)*(two >> 1);
        if((N&(two - 1)) > (two >> 1) - 1)
            ans += (N&(two - 1)) - (two >> 1) + 1;
        two <<= 1;
        n >>= 1;
    }
    return ans;
}
 
// This code is contributed by divyesh072019.

Javascript




<script>
 
function getSetBitsFromOneToN(N)
{
    var two = 2
    var ans = 0
    var n = N
     
    while(n != 0)
    {
        ans += Math.floor(N / two) * (two >> 1)
         
        if ((N & (two - 1)) > (two >> 1) - 1)
            ans += (N&(two - 1)) - (two >> 1) + 1
             
        two <<= 1;
        n >>= 1;
    }
    return ans
}
 
// This code is contributed by AnkThon
 
</script>

Method 4 (Recursive) 

Approach:

For each number ‘n’, there will we a number a, a<=n and a is perfect power of two, like 1,2,4,8…..

Let n = 11, now we can see that



Numbers till n, are:
0  -> 0000000
1  -> 0000001
2  -> 0000010
3  -> 0000011
4  -> 0000100
5  -> 0000101
6  -> 0000110
7  -> 0000111
8  -> 0001000
9  -> 0001001
10 -> 0001010
11 -> 0001011

Now we can see that, from 0 to pow(2,1)-1 = 1, we can pair elements top-most with bottom-most, 
and count of set bit in a pair is 1

Similarly for pow(2,2)-1 = 4, pairs are:
00 and 11
01 and 10
here count of set bit in a pair is 2, so in both pairs is 4

Similarly we can see for 7, 15, ans soon.....

so we can generalise that, 
count(x) = (x*pow(2,(x-1))), 
here x is position of set bit of the largest power of 2 till n
for n = 8, x = 3
for n = 4, x = 2
for n = 5, x = 2

so now for n = 11,
we have added set bits count from 0 to 7 using count(x) = (x*pow(2,(x-1)))

for rest numbers 8 to 11, all will have a set bit at 3rd index, so we can add 
count of rest numbers to our ans, 
which can be calculated using 11 - 8 + 1 = (n-pow(2,x) + 1)

Now if notice that, after removing front bits from rest numbers, we get again number from 0 to some m
so we can recursively call our same function for next set of numbers, 
by calling countSetBits(n - pow(2,x))
8  -> 1000  -> 000 -> 0
9  -> 1001  -> 001 -> 1
10 -> 1010  -> 010 -> 2
11 -> 1011  -> 011 -> 3

Code: 

C++




#include <bits/stdc++.h>
using namespace std;
 
int findLargestPower(int n)
{
    int x = 0;
    while ((1 << x) <= n)
        x++;
    return x - 1;
}
 
int countSetBits(int n)
{
    if (n <= 1)
        return n;
    int x = findLargestPower(n);
    return (x * pow(2, (x - 1))) + (n - pow(2, x) + 1) + countSetBits(n - pow(2, x));
}
 
int main()
{
    int N = 17;
    cout << countSetBits(N) << endl;
    return 0;
}

Java




import java.io.*;
class GFG
{
static int findLargestPower(int n)
{
    int x = 0;
    while ((1 << x) <= n)
        x++;
    return x - 1;
}
 
static int countSetBits(int n)
{
    if (n <= 1)
        return n;
    int x = findLargestPower(n);
    return (x * (int)Math.pow(2, (x - 1))) + (n - (int)Math.pow(2, x) + 1)
    + countSetBits(n - (int)Math.pow(2, x));
}
 
public static void main(String[] args)
    {
    int N = 17;
    System.out.print(countSetBits(N));
   }
}
 
// This code is contributed by shivanisinghss2110

Python3




def findLargestPower(n):
 
    x = 0
    while ((1 << x) <= n):
        x += 1
    return x - 1
 
 
def countSetBits(n):
 
    if (n <= 1):
        return n
    x = findLargestPower(n)
    return (x * pow(2, (x - 1))) + (n - pow(2, x) + 1) + countSetBits(n - pow(2, x))
 
N = 17
print(countSetBits(N))
     
# This code is contributed by shivanisinghss2110

C#




using System;
class GFG
{
static int findLargestPower(int n)
{
    int x = 0;
    while ((1 << x) <= n)
        x++;
    return x - 1;
}
 
static int countSetBits(int n)
{
    if (n <= 1)
        return n;
    int x = findLargestPower(n);
    return (x * (int)Math.Pow(2, (x - 1))) + (n - (int)Math.Pow(2, x) + 1)
    + countSetBits(n - (int)Math.Pow(2, x));
}
 
public static void Main(string[] args)
    {
    int N = 17;
    Console.Write(countSetBits(N));
   }
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
function findLargestPower(n)
{
    var x = 0;
    while ((1 << x) <= n)
        x++;
    return x - 1;
}
 
function countSetBits(n)
{
    if (n <= 1)
        return n;
    var x = findLargestPower(n);
    return (x * Math.pow(2, (x - 1))) + (n - Math.pow(2, x) + 1) + countSetBits(n - Math.pow(2, x));
}
 
var N = 17;
document.write(countSetBits(N));
 
// this code is contributed by shivanisinghss2110
</script>
Output
35

Time Complexity: O(LogN)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

 Method 5(Iterative)

Example:

0  -> 0000000     8  -> 001000      16  -> 010000    24 -> 011000

1  -> 0000001     9  -> 001001      17  -> 010001    25 -> 011001

2  -> 0000010    10  -> 001010     18 -> 010010     26 -> 011010

3  -> 0000011    11  -> 001011     19  -> 010010    27 -> 011011



4  -> 0000100    12  -> 001100     20  -> 010100    28 -> 011100

5  -> 0000101    13  -> 001101     21 -> 010101     29 -> 011101

6  -> 0000110    14  -> 001110     22 -> 010110     30 -> 011110

7  -> 0000111    15  -> 001111     23 -> 010111     31 -> 011111

Input: N = 4

Output: 5

Input: N = 17

Output: 35

Approach : Pattern recognition 

Let ‘N’ be any arbitrary number and consider indexing from right to left(rightmost being 1); then  nearestPow = pow(2,i).

Now, when you write all numbers from 1 to N, you will observe the pattern mentioned below:

For every index i, there are exactly nearestPow/2 continuous elements that are unset followed by nearestPow/2 elements that are set.

Throughout the solution, i am going to use this concept.  

You can clearly observe above concept in the above table.

The general formula that i came up with:

   a. addRemaining = mod – (nearestPos/2) + 1 iff mod >= nearestPow/2;

   b. totalSetBitCount = totalRep*(nearestPow/2) + addRemaining

   where totalRep -> total number of times the pattern repeats at index i

             addRemaining -> total number of set bits left to be added after the pattern is exhausted

Eg:  let N = 17

       leftMostSetIndex = 5 (Left most set bit index, considering 1 based indexing)

       i = 1 => nearestPos = pow(2,1) = 2;   totalRep = (17+1)/2 = 9 (add 1 only for i=1)

                      mod = 17%2 = 1

                      addRemaining = 0 (only for base case)

                      totalSetBitCount = totalRep*(nearestPos/2) + addRemaining = 9*(2/2) + 0 = 9*1 + 0 = 9

      i = 2 => nearestPos = pow(2, 2)=4;       totalRep = 17/4 = 4

                    mod = 17%4 = 1

                    mod(1) < (4/2) => 1 < 2 => addRemaining = 0

                    totalSetBitCount  = 9 + 4*(2) + 0 = 9 + 8 = 17

      i = 3 => nearestPow = pow(2,3) = 8;    totalRep = 17/8 = 2



                    mod = 17%8 = 1

                    mod < 4 => addRemaining = 0

                    totalSetBitCount = 17 + 2*(4) + 0 = 17 + 8 + 0 = 25

    i = 4 => nearestPow = pow(2, 4) = 16; totalRep = 17/16 = 1

                  mod = 17%16 = 1

                   mod < 8 => addRemaining = 0

                 totalSetBitCount  = 25 + 1*(8) + 0 = 25 + 8 + 0 = 33

We cannot simply operate on the next power(32) as 32>17. Also, as the first half bits will be 0s only, we need to find the distance of the given number(17) from the last power to directly get the number of 1s to be added

   i = 5 => nearestPow = (2, 5) = 32 (base case 2)

                 lastPow = pow(2, 4) = 16

                 mod = 17%16 = 1

                 totalSetBit = 33 + (mod+1) = 33 + 1 + 1 = 35

Therefore, total num of set bits from 1 to 17 is 35

Try iterating with N = 30, for better understanding of the solution.

Solution

C++




#include <iostream>
#include <bits/stdc++.h>
 
using namespace std;
 
int GetLeftMostSetBit(int n){
    int pos = 0;
     
    while(n>0){
        pos++;
        n>>=1;
    }
     
    return pos;
}
 
int TotalSetBitsFrom1ToN(int n){
    int leftMostSetBitInd = GetLeftMostSetBit(n);   
     
    int totalRep, mod;
    int nearestPow;
    int totalSetBitCount = 0;
    int addRemaining=0;
     
    int curr=0; // denotes the number of set bits at index i
     
    //cout<<"leftMostSetBitInd: "<<leftMostSetBitInd<<endl;
     
    for(int i=1; i<=leftMostSetBitInd; ++i){
        nearestPow = pow(2, i);
        if(nearestPow>n){
            int lastPow = pow(2, i-1);
            mod = n%lastPow;
            totalSetBitCount += mod+1;
        }
        else{           
            if(i==1 && n%2==1){
                totalRep = (n+1)/nearestPow;
                mod = nearestPow%2;
                addRemaining =  0;
            }
            else{
                totalRep = n/nearestPow;
                mod = n%nearestPow;
                 
                if(mod >= (nearestPow/2)){
                    addRemaining = mod - (nearestPow/2) + 1;
                }else{
                    addRemaining = 0;
                }
            }
             
            curr = totalRep*(nearestPow/2) + addRemaining;
            totalSetBitCount += curr;
        }
        // debug output at each iteration
        //cout<<i<<" "<<nearestPow<<" "<<totalRep<<" "<<mod<<" "<<totalSetBitCount<<" "<<curr<<endl;
    }
     
    return totalSetBitCount;
}
 
int main(){
    std::cout<<TotalSetBitsFrom1ToN(4)<<endl;
    std::cout<<TotalSetBitsFrom1ToN(17)<<endl;
    std::cout<<TotalSetBitsFrom1ToN(30)<<endl;
    return 0;
}

Python3




def get_left_most_set_bit(n):
    left_most_set_bit_indx = 0
      
    while n > 0:
        left_most_set_bit_indx += 1
        n >>= 1
 
    return left_most_set_bit_indx
 
def total_set_bits_from_1_to_n(n):
    left_most_set_bit_indx = get_left_most_set_bit(n)
    total_rep = 0
    mod = 0
    nearest_pow = 0
    total_set_bit_count = 0
    add_remaining = 0
    curr=0 # denotes the number of set bits at index i
 
    for i in range(1, left_most_set_bit_indx + 1):
        nearest_pow = pow(2, i)
        if nearest_pow > n:
            last_pow = pow(2, i-1)
            mod = n % last_pow
            total_set_bit_count += mod + 1
         
        else:
            if i == 1 and n % 2 == 1:
                total_rep = (n + 1) / nearest_pow
                mod = nearest_pow % 2
                add_remaining = 0
            else:
                total_rep = int(n / nearest_pow)
                mod = n % nearest_pow
                add_remaining = int(mod - (nearest_pow / 2) + 1) if mod >= nearest_pow / 2 else 0
 
            curr = int(total_rep * (nearest_pow / 2) + add_remaining)
            total_set_bit_count += curr
     
    return total_set_bit_count
 
# Driver code
if __name__ == "__main__":
    print(total_set_bits_from_1_to_n(4))
    print(total_set_bits_from_1_to_n(17))
    print(total_set_bits_from_1_to_n(30))
 
    # This code is contributed by tssovi.

Output :

5

35

75

Time Complexity: O(log(n))

This code is contributed by rjkrsngh




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