Java Program to find whether a no is power of two
Last Updated :
15 Nov, 2022
Given a positive integer, write a function to find if it is a power of two or not.
Examples :
Input : n = 4
Output : Yes
22 = 4
Input : n = 7
Output : No
Input : n = 32
Output : Yes
25 = 32
1. A simple method for this is to simply take the log of the number on base 2 and if you get an integer then number is power of 2.
Java
class GFG {
static boolean isPowerOfTwo( int n)
{
return ( int )(Math.ceil((Math.log(n) / Math.log( 2 ))))
== ( int )(Math.floor(((Math.log(n) / Math.log( 2 )))));
}
public static void main(String[] args)
{
if (isPowerOfTwo( 31 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
if (isPowerOfTwo( 64 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Time Complexity: O(log2n)
Auxiliary Space: O(1)
2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2.
Java
import java.io.*;
class GFG {
static boolean isPowerOfTwo( int n)
{
if (n == 0 )
return false ;
while (n != 1 ) {
if (n % 2 != 0 )
return false ;
n = n / 2 ;
}
return true ;
}
public static void main(String args[])
{
if (isPowerOfTwo( 31 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
if (isPowerOfTwo( 64 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Time Complexity: O(log2n)
Auxiliary Space: O(1)
3. All power of two numbers have only one bit set. So count the no. of set bits and if you get 1 then number is a power of 2. Please see Count set bits in an integer for counting set bits.
4. If we subtract a power of 2 numbers by 1 then all unset bits after the only set bit become set; and the set bit become unset.
For example for 4 ( 100) and 16(10000), we get following after subtracting 1
3 –> 011
15 –> 01111
So, if a number n is a power of 2 then bitwise & of n and n-1 will be zero. We can say n is a power of 2 or not based on value of n&(n-1). The expression n&(n-1) will not work when n is 0. To handle this case also, our expression will become n& (!n&(n-1)) (thanks to https://www.geeksforgeeks.org/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).
Below is the implementation of this method.
Java
class Test {
static boolean isPowerOfTwo( int x)
{
return x != 0 && ((x & (x - 1 )) == 0 );
}
public static void main(String[] args)
{
System.out.println(isPowerOfTwo( 31 ) ? "Yes" : "No" );
System.out.println(isPowerOfTwo( 64 ) ? "Yes" : "No" );
}
}
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Please refer complete article on Program to find whether a no is power of two for more details!
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