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 the number is the 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\n" ): printf ( "No\n" );
isPowerOfTwo(64)? printf ( "Yes\n" ): printf ( "No\n" );
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" );
}
}
|
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\n" ;
else
echo "No\n" ;
if (isPowerOfTwo(64))
echo "Yes\n" ;
else
echo "No\n" ;
?>
|
Javascript
<script>
function isPowerOfTwo(n)
{
if (n == 0)
return false ;
return parseInt( (Math.ceil((Math.log(n) / Math.log(2))))) == parseInt( (Math.floor(((Math.log(n) / Math.log(2))))));
}
if (isPowerOfTwo(31))
document.write( "Yes<br/>" );
else
document.write( "No<br/>" );
if (isPowerOfTwo(64))
document.write( "Yes<br/>" );
else
document.write( "No<br/>" );
</script>
|
Output:
No
Yes
Time Complexity: O(1)
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.
C++
#include <bits/stdc++.h>
using namespace std;
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)? cout<< "Yes\n" : cout<< "No\n" ;
isPowerOfTwo(64)? cout<< "Yes\n" : cout<< "No\n" ;
return 0;
}
|
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\n" ): printf ( "No\n" );
isPowerOfTwo(64)? printf ( "Yes\n" ): printf ( "No\n" );
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\n" ;
else
echo "No\n" ;
if (isPowerOfTwo(64))
echo "Yes\n" ;
else
echo "No\n" ;
?>
|
Javascript
<script>
function isPowerOfTwo(n)
{
if (n == 0)
return 0;
while (n != 1)
{
if (n%2 != 0)
return 0;
n = n/2;
}
return 1;
}
isPowerOfTwo(31)? document.write( "Yes" + "</br>" ): document.write( "No" + "</br>" );
isPowerOfTwo(64)? document.write( "Yes" ): document.write( "No" );
</script>
|
Output :
No
Yes
Time Complexity: O(log2n)
Auxiliary Space: O(1)
3. Another way is to use this simple recursive solution. It uses the same logic as the above iterative solution but uses recursion instead of iteration.
C++
#include <bits/stdc++.h>
using namespace std;
bool powerOf2( int n)
{
if (n == 1)
return true ;
else if (n % 2 != 0 || n == 0)
return false ;
return powerOf2(n / 2);
}
int main()
{
int n = 64;
int m = 12;
if (powerOf2(n) == 1)
cout << "True" << endl;
else
cout << "False" << endl;
if (powerOf2(m) == 1)
cout << "True" << endl;
else
cout << "False" << endl;
}
|
C
#include <stdbool.h>
#include <stdio.h>
bool powerOf2( int n)
{
if (n == 1)
return true ;
else if (n % 2 != 0 || n == 0)
return false ;
return powerOf2(n / 2);
}
int main()
{
int n = 64;
int m = 12;
if (powerOf2(n) == 1)
printf ( "True\n" );
else
printf ( "False\n" );
if (powerOf2(m) == 1)
printf ( "True\n" );
else
printf ( "False\n" );
}
|
Java
import java.util.*;
class GFG{
static boolean powerOf2( int n)
{
if (n == 1 )
return true ;
else if (n % 2 != 0 ||
n == 0 )
return false ;
return powerOf2(n / 2 );
}
public static void main(String[] args)
{
int n = 64 ;
int m = 12 ;
if (powerOf2(n) == true )
System.out.print( "True" + "\n" );
else System.out.print( "False" + "\n" );
if (powerOf2(m) == true )
System.out.print( "True" + "\n" );
else
System.out.print( "False" + "\n" );
}
}
|
Python3
def powerof2(n):
if n = = 1 :
return True
elif n % 2 ! = 0 or n = = 0 :
return False
return powerof2(n / 2 )
if __name__ = = "__main__" :
print (powerof2( 64 ))
print (powerof2( 12 ))
|
C#
using System;
class GFG{
static bool powerOf2( int n)
{
if (n == 1)
return true ;
else if (n % 2 != 0 || n == 0)
return false ;
return powerOf2(n / 2);
}
static void Main()
{
int n = 64;
int m = 12;
if (powerOf2(n))
{
Console.Write( "True" + "\n" );
}
else
{
Console.Write( "False" + "\n" );
}
if (powerOf2(m))
{
Console.Write( "True" );
}
else
{
Console.Write( "False" );
}
}
}
|
Javascript
<script>
function powerOf2(n)
{
if (n == 1)
return true ;
else if (n % 2 != 0 ||
n ==0)
return false ;
return powerOf2(n / 2);
}
var n = 64;
var m = 12;
if (powerOf2(n) == true )
document.write( "True" + "\n" );
else document.write( "False" + "\n" );
if (powerOf2(m) == true )
document.write( "True" + "\n" );
else
document.write( "False" + "\n" );
</script>
|
Time Complexity: O(log2n)
Auxiliary Space: O(log2n)
4. All power of two numbers has only a one-bit set. So count the no. of set bits and if you get 1 then the number is a power of 2. Please see Count set bits in an integer for counting set bits.
C++
#include <bits/stdc++.h>
using namespace std;
#define bool int
bool isPowerOfTwo( int n)
{
int cnt = 0;
while (n > 0) {
if ((n & 1) == 1) {
cnt++;
}
n = n >> 1;
}
if (cnt == 1) {
return true ;
}
return false ;
}
int main()
{
isPowerOfTwo(31) ? cout << "Yes\n" : cout << "No\n" ;
isPowerOfTwo(64) ? cout << "Yes\n" : cout << "No\n" ;
return 0;
}
|
Java
import java.io.*;
class GFG {
static boolean isPowerofTwo( int n)
{
int cnt = 0 ;
while (n > 0 ) {
if ((n & 1 ) == 1 ) {
cnt++;
}
n = n >> 1 ;
}
if (cnt == 1 ) {
return true ;
}
return false ;
}
public static void main(String[] args)
{
if (isPowerofTwo( 30 ) == true )
System.out.println( "Yes" );
else
System.out.println( "No" );
if (isPowerofTwo( 128 ) == true )
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
C#
using System;
class GFG {
static bool isPowerOfTwo( int n)
{
int cnt = 0;
while (n > 0) {
if ((n & 1) == 1) {
cnt++;
}
n = n >> 1;
}
if (cnt == 1)
return true ;
return false ;
}
public static void Main()
{
Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" );
Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" );
}
}
|
Python3
def isPowerOfTwo(n):
cnt = 0
while n > 0 :
if n & 1 = = 1 :
cnt = cnt + 1
n = n >> 1
if cnt = = 1 : return 1
return 0
if (isPowerOfTwo( 31 )):
print ( 'Yes' )
else :
print ( 'No' )
if (isPowerOfTwo( 64 )):
print ( 'Yes' )
else :
print ( 'No' )
|
Javascript
<script>
function isPowerofTwo(n)
{
let cnt = 0;
while (n > 0) {
if ((n & 1) == 1) {
cnt++;
}
n = n >> 1;
}
if (cnt == 1) {
return true ;
}
return false ;
}
if (isPowerofTwo(30) == true )
document.write( "Yes" + "<br/>" );
else
document.write( "No" + "<br/>" );
if (isPowerofTwo(128) == true )
document.write( "Yes" + "<br/>" );
else
document.write( "No" + "<br/>" );
</script>
|
Output :
No
Yes
Time complexity : O(N)
Space Complexity : O(1)
5. 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 becomes unset.
For example for 4 ( 100) and 16(10000), we get the 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 the 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.
Time complexity : O(1)
Space complexity : O(1)
C++
#include <bits/stdc++.h>
using namespace std;
#define bool int
bool isPowerOfTwo ( int x)
{
return x && (!(x&(x-1)));
}
int main()
{
isPowerOfTwo(31)? cout<< "Yes\n" : cout<< "No\n" ;
isPowerOfTwo(64)? cout<< "Yes\n" : cout<< "No\n" ;
return 0;
}
|
C
#include<stdio.h>
#define bool int
bool isPowerOfTwo ( int x)
{
return x && (!(x&(x-1)));
}
int main()
{
isPowerOfTwo(31)? printf ( "Yes\n" ): printf ( "No\n" );
isPowerOfTwo(64)? printf ( "Yes\n" ): printf ( "No\n" );
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" );
}
}
|
Python3
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\n" ;
else
echo "No\n" ;
if (isPowerOfTwo(64))
echo "Yes\n" ;
else
echo "No\n" ;
?>
|
Javascript
<script>
function isPowerOfTwo (x)
{
return x!=0 && ((x&(x-1)) == 0);
}
document.write(isPowerOfTwo(31) ? "Yes" : "No" );
document.write( "<br>" +(isPowerOfTwo(64) ? "Yes" : "No" ));
</script>
|
Output :
No
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
6. Another way is to use the logic to find the rightmost bit set of a given number.
C++
#include <iostream>
using namespace std;
bool isPowerofTwo( long long n)
{
if (n == 0)
return 0;
if ((n & (~(n - 1))) == n)
return 1;
return 0;
}
int main()
{
isPowerofTwo(30) ? cout << "Yes\n" : cout << "No\n" ;
isPowerofTwo(128) ? cout << "Yes\n" : cout << "No\n" ;
return 0;
}
|
Java
import java.io.*;
class GFG {
static boolean isPowerofTwo( int n)
{
if (n == 0 )
return false ;
if ((n & (~(n - 1 ))) == n)
return true ;
return false ;
}
public static void main(String[] args)
{
if (isPowerofTwo( 30 ) == true )
System.out.println( "Yes" );
else
System.out.println( "No" );
if (isPowerofTwo( 128 ) == true )
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
def isPowerofTwo(n):
if (n = = 0 ):
return 0
if ((n & (~(n - 1 ))) = = n):
return 1
return 0
if (isPowerofTwo( 30 )):
print ( 'Yes' )
else :
print ( 'No' )
if (isPowerofTwo( 128 )):
print ( 'Yes' )
else :
print ( 'No' )
|
C#
using System;
public class GFG {
static bool isPowerofTwo( int n)
{
if (n == 0)
return false ;
if ((n & (~(n - 1))) == n)
return true ;
return false ;
}
public static void Main(String[] args)
{
if (isPowerofTwo(30) == true )
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
if (isPowerofTwo(128) == true )
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
Javascript
<script>
function isPowerofTwo(n)
{
if (n == 0)
return false ;
if ((n & (~(n - 1))) == n)
return true ;
return false ;
}
if (isPowerofTwo(30) == true )
document.write( "Yes<br/>" );
else
document.write( "No<br/>" );
if (isPowerofTwo(128) == true )
document.write( "Yes<br/>" );
else
document.write( "No<br/>" );
</script>
|
Time complexity : O(1)
Space complexity : O(1)
7. Brian Kernighan’s algorithm ( Efficient Method )
Approach :
As we know that the number which will be the power of two have only one set bit , therefore when we do bitwise and with the number which is just less than the number which can be represented as the power of (2) then the result will be 0 .
Example : 4 can be represented as (2^2 ) ,
(4 & 3)=0 or in binary (100 & 011=0)
Here is the code of the given approach :
C++
#include <iostream>
using namespace std;
bool isPowerofTwo( long long n)
{
return (n != 0) && ((n & (n - 1)) == 0);
}
int main()
{
isPowerofTwo(30) ? cout << "Yes\n" : cout << "No\n" ;
isPowerofTwo(128) ? cout << "Yes\n" : cout << "No\n" ;
return 0;
}
|
Java
import java.io.*;
class GFG {
public static boolean isPowerofTwo( long n)
{
return (n != 0 ) && ((n & (n - 1 )) == 0 );
}
public static void main (String[] args) {
if (isPowerofTwo( 30 ))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
if (isPowerofTwo( 128 ))
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
}
}
|
Python3
def isPowerofTwo(n) :
return (n ! = 0 ) and ((n & (n - 1 )) = = 0 )
if __name__ = = "__main__" :
if isPowerofTwo( 30 ) :
print ( "Yes" )
else :
print ( "No" )
if isPowerofTwo( 128 ) :
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
public class GFG{
static public bool isPowerofTwo( ulong n)
{
return (n != 0) && ((n & (n - 1)) == 0);
}
static public void Main (){
if (isPowerofTwo(30))
{
System.Console.WriteLine( "Yes" );
}
else
{
System.Console.WriteLine( "No" );
}
if (isPowerofTwo(128))
{
System.Console.WriteLine( "Yes" );
}
else
{
System.Console.WriteLine( "No" );
}
}
}
|
Output :
No
Yes
Time Complexity : O(1)
Auxiliary Space: O(1)
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