# Count total unset bits in all the numbers from 1 to N

Given a positive integer **N**, the task is to count the total number of unset bits in the binary representation of all the numbers from **1** to **N**. **Note** that leading zeroes will not be counted as unset bits.**Examples:**

Input:N = 5Output:4

Integer Binary Representation Count of unset bits 1 1 0 2 10 1 3 11 0 4 100 2 5 101 1 0 + 1 + 0 + 2 + 1 = 4

Input:N = 14Output:17

**Approach:**

- Iterate the loop from
**1**to**N**. - While number is greater than
**0**divide it by**2**and check the remainder. - If remainder is equal to
**0**then increase the value of**count**by**1**.

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 unset` `// bits in the binary representation of` `// all the numbers from 1 to n` `int` `countUnsetBits(` `int` `n)` `{` ` ` `// To store the count of unset bits` ` ` `int` `cnt = 0;` ` ` `// For every integer from the range [1, n]` ` ` `for` `(` `int` `i = 1; i <= n; i++) {` ` ` `// A copy of the current integer` ` ` `int` `temp = i;` ` ` `// Count of unset bits in` ` ` `// the current integer` ` ` `while` `(temp) {` ` ` `// If current bit is unset` ` ` `if` `(temp % 2 == 0)` ` ` `cnt++;` ` ` `temp = temp / 2;` ` ` `}` ` ` `}` ` ` `return` `cnt;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 5;` ` ` `cout << countUnsetBits(n);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `class` `GFG` `{` ` ` `// Function to return the count of unset` ` ` `// bits in the binary representation of` ` ` `// all the numbers from 1 to n` ` ` `static` `int` `countUnsetBits(` `int` `n)` ` ` `{` ` ` `// To store the count of unset bits` ` ` `int` `cnt = ` `0` `;` ` ` `// For every integer from the range [1, n]` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i++) ` ` ` `{` ` ` `// A copy of the current integer` ` ` `int` `temp = i;` ` ` `// Count of unset bits in` ` ` `// the current integer` ` ` `while` `(temp > ` `0` `)` ` ` `{` ` ` `// If current bit is unset` ` ` `if` `(temp % ` `2` `== ` `0` `)` ` ` `{` ` ` `cnt = cnt + ` `1` `;` ` ` `}` ` ` `temp = temp / ` `2` `;` ` ` `}` ` ` `}` ` ` `return` `cnt;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{` ` ` `int` `n = ` `5` `;` ` ` `System.out.println(countUnsetBits(n));` ` ` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python3 implementation of the approach ` `# Function to return the count of unset ` `# bits in the binary representation of ` `# all the numbers from 1 to n ` `def` `countUnsetBits(n) : ` ` ` `# To store the count of unset bits ` ` ` `cnt ` `=` `0` `; ` ` ` `# For every integer from the range [1, n] ` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `) :` ` ` ` ` `# A copy of the current integer ` ` ` `temp ` `=` `i; ` ` ` `# Count of unset bits in ` ` ` `# the current integer ` ` ` `while` `(temp) :` ` ` `# If current bit is unset ` ` ` `if` `(temp ` `%` `2` `=` `=` `0` `) :` ` ` `cnt ` `+` `=` `1` `; ` ` ` `temp ` `=` `temp ` `/` `/` `2` `; ` ` ` `return` `cnt; ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` `n ` `=` `5` `; ` ` ` `print` `(countUnsetBits(n)); ` ` ` `# This code is contributed by AnkitRai01` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{ ` ` ` `// Function to return the count of unset` ` ` `// bits in the binary representation of` ` ` `// all the numbers from 1 to n` ` ` `static` `int` `countUnsetBits(` `int` `n)` ` ` `{` ` ` ` ` `// To store the count of unset bits` ` ` `int` `cnt = 0;` ` ` ` ` `// For every integer from the range [1, n]` ` ` `for` `(` `int` `i = 1; i <= n; i++) ` ` ` `{` ` ` ` ` `// A copy of the current integer` ` ` `int` `temp = i;` ` ` ` ` `// Count of unset bits in` ` ` `// the current integer` ` ` `while` `(temp > 0) ` ` ` `{` ` ` ` ` `// If current bit is unset` ` ` `if` `(temp % 2 == 0)` ` ` `cnt = cnt + 1;` ` ` ` ` `temp = temp / 2;` ` ` `}` ` ` `}` ` ` `return` `cnt;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main() ` ` ` `{` ` ` `int` `n = 5;` ` ` `Console.Write(countUnsetBits(n)); ` ` ` `} ` `}` `// This code is contributed by Sanjit_Prasad` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to return the count of unset` `// bits in the binary representation of` `// all the numbers from 1 to n` `function` `countUnsetBits(n)` `{` ` ` `// To store the count of unset bits` ` ` `var` `cnt = 0;` ` ` `// For every integer from the range [1, n]` ` ` `for` `(` `var` `i = 1; i <= n; i++) {` ` ` `// A copy of the current integer` ` ` `var` `temp = i;` ` ` ` ` `// Count of unset bits in` ` ` `// the current integer` ` ` `while` `(temp) {` ` ` `// If current bit is unset` ` ` `if` `(temp % 2 == 0)` ` ` `cnt++;` ` ` `temp = parseInt(temp / 2);` ` ` `}` ` ` `}` ` ` `return` `cnt;` `}` `// Driver code` `var` `n = 5;` `document.write( countUnsetBits(n));` `</script>` |

**Output:**

4

Time Complexity: O(n * log n)

Auxiliary Space: O(1)