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.
C++
#include<bits/stdc++.h>
using namespace std;
bool isPowerOfTwo( int n)
{
if (n==0)
return false ;
return ( ceil (log2(n)) == floor (log2(n)));
}
int main()
{
isPowerOfTwo(31)? cout<< "Yes" <<endl: cout<< "No" <<endl;
isPowerOfTwo(64)? cout<< "Yes" <<endl: cout<< "No" <<endl;
return 0;
}
|
C
#include<stdio.h>
#include<stdbool.h>
#include<math.h>
bool isPowerOfTwo( int n)
{
if (n==0)
return false ;
return ( ceil (log2(n)) == floor (log2(n)));
}
int main()
{
isPowerOfTwo(31)? printf ( "Yes
" ): printf ( "No
" );
isPowerOfTwo(64)? printf ( "Yes
" ): printf ( "No
" );
return 0;
}
|
Java
class GFG
{
static boolean isPowerOfTwo( int n)
{
if (n== 0 )
return false ;
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" );
}
}
|
/div>
Python3
import math
def Log2(x):
if x = = 0 :
return false;
return (math.log10(x) /
math.log10( 2 ));
def isPowerOfTwo(n):
return (math.ceil(Log2(n)) = =
math.floor(Log2(n)));
if (isPowerOfTwo( 31 )):
print ( "Yes" );
else :
print ( "No" );
if (isPowerOfTwo( 64 )):
print ( "Yes" );
else :
print ( "No" );
|
C#
using System;
class GFG
{
static bool isPowerOfTwo( int n)
{
if (n==0)
return false ;
return ( int )(Math.Ceiling((Math.Log(n) /
Math.Log(2)))) ==
( int )(Math.Floor(((Math.Log(n) /
Math.Log(2)))));
}
public static void Main()
{
if (isPowerOfTwo(31))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
if (isPowerOfTwo(64))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
PHP
<?php
function Log2( $x )
{
return (log10( $x ) /
log10(2));
}
function isPowerOfTwo( $n )
{
return ( ceil (Log2( $n )) ==
floor (Log2( $n )));
}
if (isPowerOfTwo(31))
echo "Yes
" ;
else
echo "No
" ;
if (isPowerOfTwo(64))
echo "Yes
" ;
else
echo "No
" ;
?>
|
Output:
No
Yes
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.
C
#include<stdio.h>
#include<stdbool.h>
bool isPowerOfTwo( int n)
{
if (n == 0)
return 0;
while (n != 1)
{
if (n%2 != 0)
return 0;
n = n/2;
}
return 1;
}
int main()
{
isPowerOfTwo(31)? printf ( "Yes
" ): printf ( "No
" );
isPowerOfTwo(64)? printf ( "Yes
" ): printf ( "No
" );
return 0;
}
|
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" );
}
}
|
Python3
def isPowerOfTwo(n):
if (n = = 0 ):
return False
while (n ! = 1 ):
if (n % 2 ! = 0 ):
return False
n = n / / 2
return True
if (isPowerOfTwo( 31 )):
print ( 'Yes' )
else :
print ( 'No' )
if (isPowerOfTwo( 64 )):
print ( 'Yes' )
else :
print ( 'No' )
|
C#
using System;
class GFG
{
static bool 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()
{
Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" );
Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" );
}
}
|
PHP
<?php
function isPowerOfTwo( $n )
{
if ( $n == 0)
return 0;
while ( $n != 1)
{
if ( $n % 2 != 0)
return 0;
$n = $n / 2;
}
return 1;
}
if (isPowerOfTwo(31))
echo "Yes
" ;
else
echo "No
" ;
if (isPowerOfTwo(64))
echo "Yes
" ;
else
echo "No
" ;
?>
|
Output :
No
Yes
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://tutorialspoint.dev/slugresolver/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).
Below is the implementation of this method.
C
#include<stdio.h>
#define bool int
bool isPowerOfTwo ( int x)
{
return x && (!(x&(x-1)));
}
int main()
{
isPowerOfTwo(31)? printf ( "Yes
" ): printf ( "No
" );
isPowerOfTwo(64)? printf ( "Yes
" ): printf ( "No
" );
return 0;
}
|
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" );
}
}
|
Python
def isPowerOfTwo (x):
return (x and ( not (x & (x - 1 ))) )
if (isPowerOfTwo( 31 )):
print ( 'Yes' )
else :
print ( 'No' )
if (isPowerOfTwo( 64 )):
print ( 'Yes' )
else :
print ( 'No' )
|
C#
using System;
class GFG
{
static bool isPowerOfTwo ( int x)
{
return x != 0 && ((x & (x - 1)) == 0);
}
public static void Main()
{
Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" );
Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" );
}
}
|
PHP
<?php
function isPowerOfTwo ( $x )
{
return $x && (!( $x & ( $x - 1)));
}
if (isPowerOfTwo(31))
echo "Yes
" ;
else
echo "No
" ;
if (isPowerOfTwo(64))
echo "Yes
" ;
else
echo "No
" ;
?>
|
Output :
No
Yes
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
0
0
You Might Also Like
Subscribe to Our Newsletter
leave a comment
0 Comments