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

  • Difficulty Level : Medium
  • Last Updated : 01 Apr, 2021

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

Input: N = 4 
Output:
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>
  1.  
Output: 
36

 

  1. Time Complexity : O(Partition of Number)
     

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