Programa para encontrar el N-ésimo número natural con exactamente dos bits configurados

Dado un número entero N , la tarea es encontrar el N-ésimo número natural con exactamente dos bits establecidos. 
Ejemplos: 
 

Entrada: N = 4 
Salida: 9 
Explicación: 
Números con exactamente dos bits establecidos: 3, 5, 6, 9, 10, 12, …  El
cuarto número es 9
Entrada: N = 15 
Salida: 48 
 

Enfoque ingenuo: 
 

1. Ejecute un bucle a través de todos los números naturales y, para cada número, verifique si tiene dos bits establecidos o no contando los bits establecidos en un número .
2. Imprime el número N con dos bits establecidos.

Enfoque eficiente: 
 

1. Encuentre el bit establecido más a la izquierda encontrando la partición a la que pertenece N (la partición ‘i’ tiene números ‘i’).
2. Para encontrar el otro bit establecido, primero tendremos que encontrar la distancia de N desde el último número de la partición anterior. En función de su diferencia, establecemos el bit correspondiente.
    

1. Nota: Para configurar i-ésimo bit (i = 0, 1, 2…) en un número K : 
    

k = k | (1<<(i))

1. A continuación se muestra la implementación del enfoque anterior:
    

// C++ Code to  find the Nth number
// with exactly two bits set
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the Nth number
// with exactly two bits set
void findNthNum(long long int N)
{
 
    long long int bit_L = 1, last_num = 0;
 
    // Keep incrementing until
    // we reach the partition of 'N'
    // stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N) {
        last_num = last_num + bit_L;
        bit_L++;
    }
 
    // set the rightmost bit
    // based on bit_R
    int bit_R = N - last_num - 1;
 
    cout << (1 << bit_L) + (1 << bit_R)
         << endl;
}
 
// Driver code
int main()
{
    long long int N = 13;
 
    findNthNum(N);
 
    return 0;
}

// Java Code to  find the Nth number
// with exactly two bits set
class GFG{
  
// Function to find the Nth number
// with exactly two bits set
static void findNthNum(int N)
{
  
    int bit_L = 1, last_num = 0;
  
    // Keep incrementing until
    // we reach the partition of 'N'
    // stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N) {
        last_num = last_num + bit_L;
        bit_L++;
    }
  
    // set the rightmost bit
    // based on bit_R
    int bit_R = N - last_num - 1;
  
    System.out.print((1 << bit_L) + (1 << bit_R)
         +"\n");
}
  
// Driver code
public static void main(String[] args)
{
    int N = 13;
  
    findNthNum(N);
}
}
 
// This code is contributed by Princi Singh

# Python Code to  find the Nth number
# with exactly two bits set
 
 
# Function to find the Nth number
# with exactly two bits set
def findNthNum(N):
 
    bit_L = 1;
    last_num = 0;
 
    # Keep incrementing until
    # we reach the partition of 'N'
    # stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N):
        last_num = last_num + bit_L;
        bit_L+=1;
     
 
    # set the rightmost bit
    # based on bit_R
    bit_R = N - last_num - 1;
 
    print((1 << bit_L) + (1 << bit_R));
 
 
# Driver code
if __name__ == '__main__':
    N = 13;
 
    findNthNum(N);
 
 
# This code contributed by PrinciRaj1992

// C# Code to  find the Nth number
// with exactly two bits set
using System;
 
class GFG{
   
// Function to find the Nth number
// with exactly two bits set
static void findNthNum(int N)
{
   
    int bit_L = 1, last_num = 0;
   
    // Keep incrementing until
    // we reach the partition of 'N'
    // stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N) {
        last_num = last_num + bit_L;
        bit_L++;
    }
   
    // set the rightmost bit
    // based on bit_R
    int bit_R = N - last_num - 1;
   
    Console.Write((1 << bit_L) + (1 << bit_R)
         +"\n");
}
   
// Driver code
public static void Main(String[] args)
{
    int N = 13;
   
    findNthNum(N);
}
}
 
// This code is contributed by Princi Singh

<script>
 
// JavaScript Code to find the Nth number
// with exactly two bits set
 
// Function to find the Nth number
// with exactly two bits set
function findNthNum(N)
{
 
    let bit_L = 1, last_num = 0;
 
    // Keep incrementing until
    // we reach the partition of 'N'
    // stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N) {
        last_num = last_num + bit_L;
        bit_L++;
    }
 
    // set the rightmost bit
    // based on bit_R
    let bit_R = N - last_num - 1;
 
    document.write((1 << bit_L) + (1 << bit_R)
        + "<br>");
}
 
// Driver code
let N = 13;
findNthNum(N);
 
// This code is contributed by Mayank Tyagi
</script>

36

1. Complejidad de tiempo: O (Partición de número)