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:
- 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.
- Print the Nth number having two set bits.
Efficient Approach:
- Find the leftmost set bit by finding the partition to which N belongs (Partition ‘i’ has ‘i’ numbers in it).
- 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.
k = k | (1<<(i))
- 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
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