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# Count numbers have all 1s together in binary representation

Given an integer n, the task is to count the total lucky numbers smaller than or equal to n. A number is said to be lucky if it has all contagious number of 1’s in binary representation from the beginning. For example 1, 3, 7, 15 are lucky numbers, and 2, 5 and 9 are not lucky numbers.
Examples:

```Input :n = 7
Output :3
1, 3 and 7 are lucky numbers

Input :n = 17
Output :4```

Approach:one approach is that first we find out the binary representation of each number and than check for contagious number of 1’s for each number, but this approach is time consuming and can give tle if the constraints are two large, Efficient approach can be find out by observing the numbers, we can say that every ith lucky number can be found by the formula 2i-1, and by iterating a loop upto number less than equal to n we can find out the total lucky numbers.
Below is the implementation of above approach

## CPP

 `#include ``using` `namespace` `std;` `int` `countLuckyNum(``int` `n)``{``    ``int` `count = 0, i = 1;` `    ``while` `(1) {``        ``if` `(n >= ((1 << i) - 1))``            ``count++;``        ``else``            ``break``;``        ``i++;``    ``}` `    ``return` `count;``}` `// Driver code``int` `main()``{``    ``int` `n = 7;``    ``cout << countLuckyNum(n);``    ``return` `0;``}`

## Java

 `import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;` `public` `class` `GFG {` `    ``// Function to return the count of lucky number``    ``static` `int` `countLuckyNum(``int` `n)``    ``{` `        ``int` `count = ``0``, i = ``1``;``        ``while` `(``true``) {``            ``if` `(n >= ((``1` `<< i) - ``1``))``                ``count++;``            ``else``                ``break``;``            ``i++;``        ``}``        ``return` `count;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``7``;``        ``System.out.println(countLuckyNum(n));``    ``}``}`

## Python

 `# python3 code of above problem` `# function to count the lucky number` `def` `countLuckyNum(n):``    ` `    ``count, i ``=` `0``, ``1``    ` `    ``while` `True``:``        ``if` `n>``=` `2``*``*``i``-``1``:``            ``count``+``=` `1``        ``else``:``            ``break``        ``i``+``=` `1``;``    ``return` `count    ` `# driver code``n ``=` `7` `print``(countLuckyNum(n))`

## C#

 `// C# implementation of the approach``using` `System;` `public` `class` `GFG {` `    ``// Function to return the count of lucky number``    ``static` `int` `countLuckyNum(``int` `n)``    ``{` `        ``int` `count = 0, i = 1;``        ``while` `(``true``) {``            ``if` `(n >= ((1 << i) - 1))``                ``count++;``            ``else``                ``break``;``            ``i++;``        ``}``        ``return` `count;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 7;``        ``Console.WriteLine(countLuckyNum(n));``    ``}``}`

## PHP

 `= ((1 << ``\$i``) - 1))``        ``\$count` `+= 1;``      ``else``        ``break``;``      ``\$i` `+= 1;``  ``}``  ``return` `\$count``;                    ``}  ``    ` `// Driver code  ``\$n` `= 7;``echo` `countLuckyNum(``\$n``) ;  ``?>`

## Javascript

 ``

`output:3`

Time Complexity: O(logn)

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up