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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