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

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: 4
setBits(1) + setBits(2) + setBits(3) = 1 + 1 + 2 = 4

Input: N = 6
Output: 9

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 and 1 are 0 and 1 respectively.
• Now for every element i from the range [2, N], if i is even then it will have the same number of set bits as i / 2 because to get the number i we just shift the number i / 2 by one. While shifting, the number of set bits does not change.
• Similarly, if i is odd then it will have 1 additional set bit at 0th position than i – 1 which was even.

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; ` ` `  `    ``// 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 ` `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``; ` ` `  `    ``// 1 has a single set bit ` `    ``setBits[``1``] = ``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 ` `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` ` `  `    ``# 1 has a single set bit ` `    ``setBits[``1``] ``=` `1` ` `  `    ``# For the rest of the elements ` `    ``for` `i ``in` `range``(``2``, n ``+` `1``):  ` ` `  `        ``# 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` `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  `

Output:

```9
```

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