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Program to find the Nth natural number with exactly two bits set | Set 2
• Last Updated : 21 Apr, 2021

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

Examples:

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 ``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

 ``
Output:
`48`

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

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