# Count of numbers in range [L, R] with LSB as 0 in their Binary representation

Given two integers L and R. The task is to find the count of all numbers in the range [L, R] whose Least Significant Bit in binary representation is 0.

Examples:

Input: L = 10, R = 20
Output: 6

Input: L = 7, R = 11
Output: 2

Naive approach: The simplest approach is to solve this problem is to check for every number in the range [L, R],  if Least Significant Bit in binary representation is 0.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach`   `#include ` `using` `namespace` `std;`   `// Function to return the count` `// of required numbers` `int` `countNumbers(``int` `l, ``int` `r)` `{` `    ``int` `count = 0;`   `    ``for` `(``int` `i = l; i <= r; i++) {` `        ``// If rightmost bit is 0` `        ``if` `((i & 1) == 0) {` `            ``count++;` `        ``}` `    ``}` `    ``// Return the required count` `    ``return` `count;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `l = 10, r = 20;`   `    ``// Call function countNumbers` `    ``cout << countNumbers(l, r);` `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.io.*;`   `class` `GFG{` `    `  `// Function to return the count` `// of required numbers` `static` `int` `countNumbers(``int` `l, ``int` `r)` `{` `    ``int` `count = ``0``;` `    ``for``(``int` `i = l; i <= r; i++) ` `    ``{` `        `  `        ``// If rightmost bit is 0` `        ``if` `((i & ``1``) == ``0``)` `            ``count += ``1``;` `    ``}` `    `  `    ``// Return the required count` `    ``return` `count;` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `l = ``10``, r = ``20``;` `    `  `    ``// Call function countNumbers` `    ``System.out.println(countNumbers(l, r));` `}` `}`   `// This code is contributed by MuskanKalra1`

## Python3

 `# Python3 implementation of the approach`   `# Function to return the count` `# of required numbers` `def` `countNumbers(l, r):` `    `  `    ``count ``=` `0` `    `  `    ``for` `i ``in` `range``(l, r ``+` `1``):` `        `  `        ``# If rightmost bit is 0` `        ``if` `((i & ``1``) ``=``=` `0``):` `            ``count ``+``=` `1` `            `  `    ``# Return the required count` `    ``return` `count`   `# Driver code` `l ``=` `10` `r ``=` `20`   `# Call function countNumbers` `print``(countNumbers(l, r))`   `# This code is contributed by amreshkumar3`

## C#

 `// C# implementation of the approach` `using` `System;`   `class` `GFG {`   `    ``// Function to return the count` `    ``// of required numbers` `    ``static` `int` `countNumbers(``int` `l, ``int` `r)` `    ``{` `        ``int` `count = 0;` `        ``for` `(``int` `i = l; i <= r; i++) {`   `            ``// If rightmost bit is 0` `            ``if` `((i & 1) == 0)` `                ``count += 1;` `        ``}`   `        ``// Return the required count` `        ``return` `count;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `l = 10, r = 20;`   `        ``// Call function countNumbers` `        ``Console.WriteLine(countNumbers(l, r));` `    ``}` `}`   `// This code is contributed by subham348.`

## Javascript

 ``

Output

`6`

Time Complexity: O(r – l)
Auxiliary Space: O(1)

Efficient approach: This problem can be solved by using properties of bits. Only even numbers have rightmost bit as 0. The count can be found using this formula ((R / 2) – (L – 1) / 2) in O(1) time.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the count` `// of required numbers` `int` `countNumbers(``int` `l, ``int` `r)` `{` `    ``// Count of numbers in range` `    ``// which are divisible by 2` `    ``return` `((r / 2) - (l - 1) / 2);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `l = 10, r = 20;` `    ``cout << countNumbers(l, r);` `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.io.*;`   `class` `GFG{` `    `  `// Function to return the count` `// of required numbers` `static` `int` `countNumbers(``int` `l, ``int` `r)` `{` `    `  `    ``// Count of numbers in range` `    ``//  which are divisible by 2` `    ``return` `((r / ``2``) - (l - ``1``) / ``2``);` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `l = ``10``;` `    ``int` `r = ``20``;` `    `  `    ``System.out.println(countNumbers(l, r));` `}` `}`   `// This code is contributed by MuskanKalra1`

## Python3

 `# Python3 implementation of the approach`   `# Function to return the count` `# of required numbers` `def` `countNumbers(l, r):`   `    ``# Count of numbers in range` `    ``#  which are divisible by 2` `    ``return` `((r ``/``/` `2``) ``-` `(l ``-` `1``) ``/``/` `2``)`   `# Driver code` `l ``=` `10` `r ``=` `20`   `print``(countNumbers(l, r))`   `# This code is contributed by amreshkumar3`

## C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `// Function to return the count` `// of required numbers` `static` `int` `countNumbers(``int` `l, ``int` `r)` `{` `  `  `    ``// Count of numbers in range` `    ``// which are divisible by 2` `    ``return` `((r / 2) - (l - 1) / 2);` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int` `l = 10, r = 20;` `    ``Console.Write(countNumbers(l, r));` `}` `}`   `// This code is contributed by SURENDRA_GANGWAR.`

## Javascript

 ``

Output

`6`

Time Complexity: O(1)
Auxiliary Space: O(1)

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