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 = 3Output:4

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

Input:N = 6Output:9

**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 (**101**1), 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 <bits/stdc++.h>` `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] = 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] = 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` ` ` `static` `public` `void` `Main ()` ` ` `{` ` ` `int` `n = 6;` ` ` ` ` `Console.WriteLine(countSetBits(n));` ` ` `}` `}` `// This code is contributed by AnkitRai01` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to return the count of` `// set bits in all the integers` `// from the range [1, n]` `function` `countSetBits(n)` `{` ` ` ` ` `// To store the required count` ` ` `// of the set bits` ` ` `var` `cnt = 0;` ` ` `// To store the count of set` ` ` `// bits in every integer` ` ` `var` `setBits = Array.from(` ` ` `{length: n + 1}, (_, i) => 0);` ` ` `// 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 = 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` `(i = 0; i <= n; i++)` ` ` `{` ` ` `cnt = cnt + setBits[i];` ` ` `}` ` ` `return` `cnt;` `}` `// Driver code` `var` `n = 6;` `document.write(countSetBits(n));` `// This code is contributed by 29AjayKumar` `</script>` |

**Output:**

9

**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 <bits/stdc++.h>` `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` |

**Output**

9

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

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