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Program to find the Nth natural number with exactly two bits set | Set 2

  • Last Updated : 23 Nov, 2021

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

Examples:

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Input: N = 4
Output: 9
Explanation: Numbers with exactly two set bits are 3, 5, 6, 9, 10, 12, …. 
The 4th term in this series is 9.



Input: N = 15
Output: 48
Explanation: Numbers with exactly two set bits are 3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 33, 34, 36, 40, 48….
The 15th term in this series is 48.

Naive Approach: Refer to the previous post of this article for the simplest solution possible for this problem. 

Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea to use Binary Search to find the Nth number. Follow the steps below to solve the problem:

  • Any number in the given series is in the form (2a + 2b) where a > b.
  • The Nth term in a series can be identified by identifying values a and b.
  • Find the value of a such that (a * (a + 1)) / 2 ≥ N and keeping a constraint that a must be minimum
  • Therefore, find the value of a for constraints in the above steps using Binary Search.
  • After the above steps, find the value of b using a and N and print the result as (2a + 2b).

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the Nth number
// with exactly two bits set
void findNthNum(long long int N)
{
    // Initialize variables
    long long int a, b, left;
    long long int right, mid;
    long long int t, last_num = 0;
 
    // Initialize the range in which
    // the value of 'a' is present
    left = 1, right = N;
 
    // Perform Binary Search
    while (left <= right) {
 
        // Find the mid value
        mid = left + (right - left) / 2;
 
        t = (mid * (mid + 1)) / 2;
 
        // Update the range using the
        // mid value t
        if (t < N) {
            left = mid + 1;
        }
        else if (t == N) {
            a = mid;
            break;
        }
        else {
            a = mid;
            right = mid - 1;
        }
    }
 
    // Find b value using a and N
    t = a - 1;
    b = N - (t * (t + 1)) / 2 - 1;
 
    // Print the value 2^a + 2^b
    cout << (1 << a) + (1 << b);
}
 
// Driver Code
int main()
{
    long long int N = 15;
 
    // Function Call
    findNthNum(N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the Nth number
// with exactly two bits set
static void findNthNum(int N)
{
    // Initialize variables
    int a = 0, b, left;
    int right, mid;
    int t, last_num = 0;
 
    // Initialize the range in which
    // the value of 'a' is present
    left = 1;
    right = N;
 
    // Perform Binary Search
    while (left <= right) {
 
        // Find the mid value
        mid = left + (right - left) / 2;
 
        t = (mid * (mid + 1)) / 2;
 
        // Update the range using the
        // mid value t
        if (t < N) {
            left = mid + 1;
        }
        else if (t == N) {
            a = mid;
            break;
        }
        else {
            a = mid;
            right = mid - 1;
        }
    }
 
    // Find b value using a and N
    t = a - 1;
    b = N - (t * (t + 1)) / 2 - 1;
 
    // Print the value 2^a + 2^b
    System.out.print((1 << a) + (1 << b));
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 15;
 
    // Function Call
    findNthNum(N);
}
}
 
// This code contributed by shikhasingrajput

Python3




# Python3 program for the above approach
 
# Function to find the Nth number
# with exactly two bits set
def findNthNum(N):
     
    # Initialize variables
    last_num = 0
 
    # Initialize the range in which
    # the value of 'a' is present
    left = 1
    right = N
 
    # Perform Binary Search
    while (left <= right):
         
        # Find the mid value
        mid = left + (right - left) // 2
 
        t = (mid * (mid + 1)) // 2
 
        # Update the range using the
        # mid value t
        if (t < N):
            left = mid + 1
 
        elif (t == N):
            a = mid
            break
 
        else:
            a = mid
            right = mid - 1
 
    # Find b value using a and N
    t = a - 1
    b = N - (t * (t + 1)) // 2 - 1
 
    # Print the value 2^a + 2^b
    print((1 << a) + (1 << b))
 
# Driver Code
if __name__ == "__main__":
 
    N = 15
 
    # Function Call
    findNthNum(N)
 
# This code is contributed by chitranayal

C#




// C# program for the above approach
using System;
   
class GFG{
   
// Function to find the Nth number
// with exactly two bits set
static void findNthNum(int N)
{
   
    // Initialize variables
    int a = 0, b, left;
    int right, mid;
    int t;
    //int last_num = 0;
  
    // Initialize the range in which
    // the value of 'a' is present
    left = 1; right = N;
  
    // Perform Binary Search
    while (left <= right)
    {
       
        // Find the mid value
        mid = left + (right - left) / 2;
  
        t = (mid * (mid + 1)) / 2;
  
        // Update the range using the
        // mid value t
        if (t < N)
        {
            left = mid + 1;
        }
        else if (t == N)
        {
            a = mid;
            break;
        }
        else
        {
            a = mid;
            right = mid - 1;
        }
    }
  
    // Find b value using a and N
    t = a - 1;
    b = N - (t * (t + 1)) / 2 - 1;
  
    // Print the value 2^a + 2^b
    Console.Write((1 << a) + (1 << b));
}
   
// Driver Code
public static void Main()
{
    int N = 15;
  
    // Function Call
    findNthNum(N);
}
}
 
// This code is contributed by sanjoy_62

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Function to find the Nth number
// with exactly two bits set
function findNthNum(N)
{
    // Initialize variables
    let a = 0, b, left;
    let right, mid;
    let t, last_num = 0;
  
    // Initialize the range in which
    // the value of 'a' is present
    left = 1;
    right = N;
  
    // Perform Binary Search
    while (left <= right) {
  
        // Find the mid value
        mid = left + (right - left) / 2;
  
        t = (mid * (mid + 1)) / 2;
  
        // Update the range using the
        // mid value t
        if (t < N) {
            left = mid + 1;
        }
        else if (t == N) {
            a = mid;
            break;
        }
        else {
            a = mid;
            right = mid - 1;
        }
    }
  
    // Find b value using a and N
    t = a - 1;
    b = N - (t * (t + 1)) / 2 - 1;
  
    // Print the value 2^a + 2^b
    document.write((1 << a) + (1 << b));
}
 
// Driver Code
 
    let N = 15;
  
    // Function Call
    findNthNum(N);
     
</script>
Output: 
48

 

Time Complexity: O(log N)
Auxiliary Space: O(1)




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