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Count total set bits in all numbers from 1 to n
• Difficulty Level : Hard
• Last Updated : 15 Jun, 2021

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

Examples:

```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 ` `// 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

 ``

## Javascript

 ``
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 ``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

 ``

## Javascript

 ``
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 ` `// 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 ` `/* 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

 ``
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

 ``

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 ``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;``}`
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

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