Conteo de números que tienen solo 1 bit establecido en el rango [0, n]

Dado un entero n , la tarea es contar los números que tienen solo 1 bit establecido en el rango [0, n] .
Ejemplos: 

Entrada: n = 7 
Salida: 3 
Explicación: 000, 001, 010, 011, 100, 101, 110 y 111 son la representación binaria de todos los números hasta el 7. Y solo hay 3 números (001, 010 y 100) que tienen solo 1 juego de bits.

Entrada: n = 3 
Salida: 2 

Enfoque: si se requieren k bits para representar n , entonces hay k números posibles ya que 1 puede colocarse en k posiciones diferentes cada vez.
A continuación se muestra la implementación del enfoque anterior.

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the required count
int count(int n)
{
 
    // To store the count of numbers
    int cnt = 0;
    int p = 1;
    while (p <= n) {
        cnt++;
 
        // Every power of 2 contains
        // only 1 set bit
        p *= 2;
    }
    return cnt;
}
 
// Driver code
int main()
{
    int n = 7;
    cout << count(n);
 
    return 0;
}

// Java implementation of the approach
class GFG {
 
    // Function to return the required count
    static int count(int n)
    {
 
        // To store the count of numbers
        int cnt = 0;
        int p = 1;
        while (p <= n) {
            cnt++;
 
            // Every power of 2 contains
            // only 1 set bit
            p *= 2;
        }
        return cnt;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 7;
        System.out.print(count(n));
    }
}

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the required count
    static int count(int n)
    {
 
        // To store the count of numbers
        int cnt = 0;
        int p = 1;
        while (p <= n) {
            cnt++;
 
            // Every power of 2 contains
            // only 1 set bit
            p *= 2;
        }
        return cnt;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 7;
        Console.Write(count(n));
    }
}

# Python3 implementation of the approach
 
# Function to return the required count
def count(n):
     
    # To store the count of numbers
    cnt = 0
    p = 1
    while (p <= n):
        cnt = cnt + 1
         
        # Every power of 2 contains
        # only 1 set bit
        p *= 2
    return cnt
 
# Driver code
n = 7
print(count(n));

<?php
// PHP implementation of the approach
 
// Function to return the required count
function count_t($n)
{
 
    // To store the count of numbers
    $cnt = 0;
    $p = 1;
    while ($p <= $n)
    {
        $cnt++;
 
        // Every power of 2 contains
        // only 1 set bit
        $p *= 2;
    }
    return $cnt;
}
 
// Driver code
$n = 7;
echo count_t($n);
 
// This Code is contributed by ajit.
?>

<script>
// Javascript implementation of the approach
 
// Function to return the required count
function count(n)
{
 
    // To store the count of numbers
    var cnt = 0;
    var p = 1;
    while (p <= n) {
        cnt++;
 
        // Every power of 2 contains
        // only 1 set bit
        p *= 2;
    }
    return cnt;
}
 
// Driver code
var n = 7;
document.write(count(n));
 
// This code is contributed by noob2000.
</script>

3

Complejidad temporal: O(log n)
Espacio auxiliar: O(1)