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# Count total set bits in all numbers from 1 to N | Set 3

• Difficulty Level : Medium
• Last Updated : 27 Sep, 2022

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

Examples:

Input: N = 3
Output:
setBits(1) + setBits(2) + setBits(3) = 1 + 1 + 2 = 4

Input: N = 6
Output:

Approach: Solution to this problem has been published in the Set 1 and the Set 2 of this article. Here, a dynamic programming based approach is discussed.

• Base case: Number of set bits in 0 is 0.
• For any number n: n and n>>1 has same no of set bits except for the rightmost bit.

Example: n = 11 (1011),  n >> 1 = 5 (101)… same bits in 11 and 5 are marked bold. So assuming we already know set bit count of 5, we only need to take care for the rightmost bit of 11 which is 1. setBit(11) = setBit(5) + 1 = 2 + 1 = 3

Rightmost bit is 1 for odd and 0 for even number.

Recurrence Relation: setBit(n) = setBit(n>>1) + (n & 1) and setBit(0) = 0

We can use bottom-up dynamic programming approach to solve this.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the count of` `// set bits in all the integers` `// from the range [1, n]` `int` `countSetBits(``int` `n)` `{`   `    ``// To store the required count` `    ``// of the set bits` `    ``int` `cnt = 0;`   `    ``// To store the count of set` `    ``// bits in every integer` `    ``vector<``int``> setBits(n + 1);`   `    ``// 0 has no set bit` `    ``setBits = 0;`   `    ``for` `(``int` `i = 1; i <= n; i++) {`   `        ``setBits[i] = setBits[i >> 1] + (i & 1);` `    ``}`   `    ``// Sum all the set bits` `    ``for` `(``int` `i = 0; i <= n; i++) {` `        ``cnt = cnt + setBits[i];` `    ``}`   `    ``return` `cnt;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 6;`   `    ``cout << countSetBits(n);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to return the count of` `    ``// set bits in all the integers` `    ``// from the range [1, n]` `    ``static` `int` `countSetBits(``int` `n)` `    ``{`   `        ``// To store the required count` `        ``// of the set bits` `        ``int` `cnt = ``0``;`   `        ``// To store the count of set` `        ``// bits in every integer` `        ``int``[] setBits = ``new` `int``[n + ``1``];`   `        ``// 0 has no set bit` `        ``setBits[``0``] = ``0``;`   `        ``// For the rest of the elements` `        ``for` `(``int` `i = ``1``; i <= n; i++) {`   `            ``setBits[i] = setBits[i >> ``1``] + (i & ``1``);` `        ``}`   `        ``// Sum all the set bits` `        ``for` `(``int` `i = ``0``; i <= n; i++) {` `            ``cnt = cnt + setBits[i];` `        ``}` `        ``return` `cnt;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``6``;`   `        ``System.out.println(countSetBits(n));` `    ``}` `}`   `// This code is contributed by Princi Singh`

## Python3

 `# Python3 implementation of the approach`   `# Function to return the count of` `# set bits in all the integers` `# from the range [1, n]`     `def` `countSetBits(n):`   `    ``# To store the required count` `    ``# of the set bits` `    ``cnt ``=` `0`   `    ``# To store the count of set` `    ``# bits in every integer` `    ``setBits ``=` `[``0` `for` `x ``in` `range``(n ``+` `1``)]`   `    ``# 0 has no set bit` `    ``setBits[``0``] ``=` `0`   `    ``# For the rest of the elements` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):` `        ``setBits[i] ``=` `setBits[i ``/``/` `2``] ``+` `(i & ``1``)`   `    ``# Sum all the set bits` `    ``for` `i ``in` `range``(``0``, n ``+` `1``):` `        ``cnt ``=` `cnt ``+` `setBits[i]`   `    ``return` `cnt`     `# Driver code` `n ``=` `6` `print``(countSetBits(n))`   `# This code is contributed by Sanjit Prasad`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG {`   `    ``// Function to return the count of` `    ``// set bits in all the integers` `    ``// from the range [1, n]` `    ``static` `int` `countSetBits(``int` `n)` `    ``{`   `        ``// To store the required count` `        ``// of the set bits` `        ``int` `cnt = 0;`   `        ``// To store the count of set` `        ``// bits in every integer` `        ``int``[] setBits = ``new` `int``[n + 1];`   `        ``// 0 has no set bit` `        ``setBits = 0;`   `        ``// 1 has a single set bit` `        ``setBits = 1;`   `        ``// For the rest of the elements` `        ``for` `(``int` `i = 2; i <= n; i++) {`   `            ``// If current element i is even then` `            ``// it has set bits equal to the count` `            ``// of the set bits in i / 2` `            ``if` `(i % 2 == 0) {` `                ``setBits[i] = setBits[i / 2];` `            ``}`   `            ``// Else it has set bits equal to one` `            ``// more than the previous element` `            ``else` `{` `                ``setBits[i] = setBits[i - 1] + 1;` `            ``}` `        ``}`   `        ``// Sum all the set bits` `        ``for` `(``int` `i = 0; i <= n; i++) {` `            ``cnt = cnt + setBits[i];` `        ``}` `        ``return` `cnt;` `    ``}`   `    ``// Driver code` `    ``static` `public` `void` `Main()` `    ``{` `        ``int` `n = 6;`   `        ``Console.WriteLine(countSetBits(n));` `    ``}` `}`   `// This code is contributed by AnkitRai01`

## Javascript

 ``

Output:

`9`

Time Complexity: O(N), where N is the given number.
Auxiliary Space: O(N), for creating an additional array of size N + 1.

Another simple and easy-to-understand solution:

A simple easy to implement and understand solution would be not using bits operations.  The solution is to directly count set bits using __builtin_popcount(). The solution is explained in code using comments.

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the count of` `// set bits in all the integers` `// from the range [1, n]` `int` `countSetBits(``int` `n)` `{`   `    ``// To store the required count` `    ``// of the set bits` `    ``int` `cnt = 0;`   `    ``// Calculate set bits in each number using` `    ``// __builtin_popcount() and  Sum all the set bits` `    ``for` `(``int` `i = 1; i <= n; i++) {` `        ``cnt = cnt + __builtin_popcount(i);` `    ``}`   `    ``return` `cnt;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 6;`   `    ``cout << countSetBits(n);`   `    ``return` `0;` `}`   `// This article is contributed by Abhishek`

## Java

 `// Java implementation of the approach` `import` `java.util.*;`   `class` `GFG {`   `    ``// Function to return the count of` `    ``// set bits in all the integers` `    ``// from the range [1, n]` `    ``static` `int` `countSetBits(``int` `n)` `    ``{`   `        ``// To store the required count` `        ``// of the set bits` `        ``int` `cnt = ``0``;`   `        ``// Calculate set bits in each number using` `        ``// Integer.bitCount() and  Sum all the set bits` `        ``for` `(``int` `i = ``1``; i <= n; i++) {` `            ``cnt = cnt + Integer.bitCount(i);` `        ``}`   `        ``return` `cnt;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``6``;` `        ``System.out.print(countSetBits(n));` `    ``}` `}`   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python implementation of the approach`   `# Function to return the count of` `# set bits in all the integers` `# from the range [1, n]`     `def` `countSetBits(n):`   `    ``# To store the required count` `    ``# of the set bits` `    ``cnt ``=` `0`   `    ``# Calculate set bits in each number using` `    ``# Integer.bitCount() and Sum all the set bits` `    ``for` `i ``in` `range``(``1``, n``+``1``):` `        ``cnt ``=` `cnt ``+` `(``bin``(i)[``2``:]).count(``'1'``)`   `    ``return` `cnt`     `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``n ``=` `6` `    ``print``(countSetBits(n))`   `# This code is contributed by Rajput-Ji`

## C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Linq;`   `public` `class` `GFG {`   `    ``// Function to return the count of` `    ``// set bits in all the integers` `    ``// from the range [1, n]` `    ``static` `int` `countSetBits(``int` `n)` `    ``{`   `        ``// To store the required count` `        ``// of the set bits` `        ``int` `cnt = 0;`   `        ``// Calculate set bits in each number using` `        ``// int.bitCount() and  Sum all the set bits` `        ``for` `(``int` `i = 1; i <= n; i++) {` `            ``cnt = cnt` `                  ``+ (Convert.ToString(i, 2).Count(` `                      ``c = > c == ``'1'``));` `        ``}`   `        ``return` `cnt;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``int` `n = 6;` `        ``Console.Write(countSetBits(n));` `    ``}` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

`9`

Time Complexity: O(NlogN), where N is the given number.
Auxiliary Space: O(1)