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 <bits/stdc++.h>
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

<?php
// PHP implementation of the approach

// Function to count the lucky number
function countLuckyNum($n)
{

  $count = 0;
  $i = 1;

  while(1)
  {
      if ($n >= ((1 << $i) - 1))
        $count += 1;
      else
        break;
      $i += 1;
  }
  return $count;
}

// Driver code
$n = 7;
echo countLuckyNum($n) ;
?>

Javascript

<script>

    // Function to return the count of lucky number
    function countLuckyNum(n) {

        var count = 0, i = 1;
        while (true) {
            if (n >= ((1 << i) - 1))
                count++;
            else
                break;
            i++;
        }
        return count;
    }

    // Driver code
    var n = 7;
    document.write(countLuckyNum(n));

// This code is contributed by aashish1995
</script>

output:3

Time Complexity: O(logn)

Auxiliary Space: O(1)