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.

   

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

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

   Below is the implementation of the above approach:

   C++

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   // 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; }

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   Java

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   // 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

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   Python3

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   # 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

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   C#

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   // 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

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   Output:

   36
   

   Time Complexity : O(Partition of Number)

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