Program to find the Nth natural number with exactly two bits set

Given an integer N, the task is to find the Nth natural number with exactly two bits set. 
Examples:

Input: N = 4 
Output: 9 
Explanation:
Numbers with exactly two bits set: 3, 5, 6, 9, 10, 12, … 
4th number in this is 9
Input: N = 15 
Output: 48 

Naive Approach:

1. Run a loop through all natural numbers, and for each number, check if it has two bits set or not by counting set bits in a number.
2. Print the Nth number having two set bits.

Efficient Approach: 

1. Find the leftmost set bit by finding the partition to which N belongs (Partition ‘i’ has ‘i’ numbers in it).
2. To find the other set bit, we’ll have to first find the distance of N from the last number of the previous partition. Based on their difference, we set the corresponding bit.
   

1. Note: To set ith bit (i = 0, 1, 2…) in a number K: 
   

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

1. Below is the implementation of the above approach:
   

C++

// 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

// 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

Python3

# 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#

// 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

Javascript

<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>

Output:

36

Time Complexity : O(Partition of Number)

Auxiliary space: O(1) as it is using constant variables