Given a positive integer n, write a function to find if it is a power of 2 or not
Examples:
Input : n = 4
Output : Yes
Explanation: 22 = 4Input : n = 32
Output : Yes
Explanation: 25 = 32
Finding whether a given number is a power of 2 using Log operator:
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
Below is the implementation of the above approach:
// C++ Program to find whether a // no is a power of two #include <bits/stdc++.h> using namespace std;
// Function to check if x is power of 2 bool isPowerOfTwo( int n)
{ if (n == 0)
return false ;
return ( ceil (log2(n)) == floor (log2(n)));
} // Driver code int main()
{ // Function call
isPowerOfTwo(31) ? cout << "Yes" << endl
: cout << "No" << endl;
isPowerOfTwo(64) ? cout << "Yes" << endl
: cout << "No" << endl;
return 0;
} // This code is contributed by Surendra_Gangwar |
// C Program to find whether a // no is power of two #include <math.h> #include <stdbool.h> #include <stdio.h> /* Function to check if x is power of 2*/ bool isPowerOfTwo( int n)
{ if (n == 0)
return false ;
return ( ceil (log2(n)) == floor (log2(n)));
} // Driver code int main()
{ // Function call
isPowerOfTwo(31) ? printf ( "Yes\n" ) : printf ( "No\n" );
isPowerOfTwo(64) ? printf ( "Yes\n" ) : printf ( "No\n" );
return 0;
} // This code is contributed by bibhudhendra |
// Java Program to find whether a // no is power of two import java.lang.Math;
class GFG {
/* Function to check if x is power of 2*/
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 )))));
}
// Driver Code
public static void main(String[] args)
{
// Function call
if (isPowerOfTwo( 31 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
if (isPowerOfTwo( 64 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
} // This code is contributed by mits |
# Python3 Program to find # whether a no is # power of two import math
# Function to check # Log base 2 def Log2(x):
if x = = 0 :
return false
return (math.log10(x) /
math.log10( 2 ))
# Function to check # if x is power of 2 def isPowerOfTwo(n):
return (math.ceil(Log2(n)) = =
math.floor(Log2(n)))
# Driver Code if __name__ = = "__main__" :
# Function call
if (isPowerOfTwo( 31 )):
print ( "Yes" )
else :
print ( "No" )
if (isPowerOfTwo( 64 )):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed # by mits |
// C# Program to find whether // a no is power of two using System;
class GFG {
/* Function to check if
x is power of 2*/
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)))));
}
// Driver Code
public static void Main()
{
// Function call
if (isPowerOfTwo(31))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
if (isPowerOfTwo(64))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
} // This code is contributed // by Akanksha Rai(Abby_akku) |
<script> // javascript Program to find whether a // no is power of two /* Function to check if x is power of 2 */
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))))));
}
// Driver Code
if (isPowerOfTwo(31))
document.write( "Yes<br/>" );
else
document.write( "No<br/>" );
if (isPowerOfTwo(64))
document.write( "Yes<br/>" );
else
document.write( "No<br/>" );
// This code is contributed by shikhasingrajput. </script> |
<?php // PHP Program to find // whether a no is // power of two // Function to check // Log base 2 function Log2( $x )
{ return (log10( $x ) /
log10(2));
} // Function to check // if x is power of 2 function isPowerOfTwo( $n )
{ return ( ceil (Log2( $n )) ==
floor (Log2( $n )));
} // Driver Code // Function call if (isPowerOfTwo(31))
echo "Yes\n" ;
else echo "No\n" ;
if (isPowerOfTwo(64))
echo "Yes\n" ;
else echo "No\n" ;
// This code is contributed // by Sam007 ?> |
No Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
Finding whether a given number is a power of 2 using the modulo & division operator:
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.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
/* Function to check if x is power of 2*/ bool isPowerOfTwo( int n)
{ if (n == 0)
return 0;
while (n != 1) {
if (n % 2 != 0)
return 0;
n = n / 2;
}
return 1;
} // Driver code int main()
{ // Function call
isPowerOfTwo(31) ? cout << "Yes\n" : cout << "No\n" ;
isPowerOfTwo(64) ? cout << "Yes\n" : cout << "No\n" ;
return 0;
} // This code is contributed by rathbhupendra |
// C program for the above approach #include <stdbool.h> #include <stdio.h> /* Function to check if x is power of 2*/ bool isPowerOfTwo( int n)
{ if (n == 0)
return 0;
while (n != 1) {
if (n % 2 != 0)
return 0;
n = n / 2;
}
return 1;
} // Driver code int main()
{ // Function call
isPowerOfTwo(31) ? printf ( "Yes\n" ) : printf ( "No\n" );
isPowerOfTwo(64) ? printf ( "Yes\n" ) : printf ( "No\n" );
return 0;
} |
// Java program to find whether // a no is power of two import java.io.*;
class GFG {
// Function to check if
// x is power of 2
static boolean isPowerOfTwo( int n)
{
if (n == 0 )
return false ;
while (n != 1 ) {
if (n % 2 != 0 )
return false ;
n = n / 2 ;
}
return true ;
}
// Driver code
public static void main(String args[])
{
// Function call
if (isPowerOfTwo( 31 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
if (isPowerOfTwo( 64 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
} // This code is contributed by Nikita tiwari. |
# Python program to check if given # number is power of 2 or not # Function to check if x is power of 2 def isPowerOfTwo(n):
if (n = = 0 ):
return False
while (n ! = 1 ):
if (n % 2 ! = 0 ):
return False
n = n / / 2
return True
# Driver code if __name__ = = "__main__" :
# Function call
if (isPowerOfTwo( 31 )):
print ( 'Yes' )
else :
print ( 'No' )
if (isPowerOfTwo( 64 )):
print ( 'Yes' )
else :
print ( 'No' )
# This code is contributed by Danish Raza |
// C# program to find whether // a no is power of two using System;
class GFG {
// Function to check if
// x is power of 2
static bool isPowerOfTwo( int n)
{
if (n == 0)
return false ;
while (n != 1) {
if (n % 2 != 0)
return false ;
n = n / 2;
}
return true ;
}
// Driver code
public static void Main()
{
// Function call
Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" );
Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" );
}
} // This code is contributed by Sam007 |
<script> /* Function to check if x is power of 2*/
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> |
<?php // php program for the above approach // Function to check if // x is power of 2 function isPowerOfTwo( $n )
{ if ( $n == 0)
return 0;
while ( $n != 1)
{ if ( $n % 2 != 0)
return 0;
$n = $n / 2;
} return 1;
} // Driver Code // Function call if (isPowerOfTwo(31))
echo "Yes\n" ;
else echo "No\n" ;
if (isPowerOfTwo(64))
echo "Yes\n" ;
else echo "No\n" ;
// This code is contributed // by Sam007 ?> |
No Yes
Time Complexity: O(log N)
Auxiliary Space: O(1)
Finding whether a given number is a power of 2 by checking the count of set bits:
To solve the problem follow the below idea:
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.
Below is the implementation of the above approach:
// C++ program of the above approach #include <bits/stdc++.h> using namespace std;
/* Function to check if x is power of 2*/ bool isPowerOfTwo( int n)
{ /* First x in the below expression is for the case when
* x is 0 */
int cnt = 0;
while (n > 0) {
if ((n & 1) == 1) {
cnt++;
}
n = n >> 1; // keep dividing n by 2 using right
// shift operator
}
if (cnt == 1) { // if cnt = 1 only then it is power of 2
return true ;
}
return false ;
} // Driver code int main()
{ // Function call
isPowerOfTwo(31) ? cout << "Yes\n" : cout << "No\n" ;
isPowerOfTwo(64) ? cout << "Yes\n" : cout << "No\n" ;
return 0;
} // This code is contributed by devendra salunke |
// Java program of the above approach import java.io.*;
class GFG {
// Function to check if x is power of 2
static boolean isPowerofTwo( int n)
{
int cnt = 0 ;
while (n > 0 ) {
if ((n & 1 ) == 1 ) {
cnt++; // if n&1 == 1 keep incrementing cnt
// variable
}
n = n >> 1 ; // keep dividing n by 2 using right
// shift operator
}
if (cnt == 1 ) {
// if cnt = 1 only then it is power of 2
return true ;
}
return false ;
}
// Driver code
public static void main(String[] args)
{
// Function call
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" );
}
} // This code is contributed by devendra salunke. |
# Python3 program to check if given # number is power of 2 or not # Function to check if x is power of 2 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
# Driver code if __name__ = = "__main__" :
# Function call
if (isPowerOfTwo( 31 )):
print ( 'Yes' )
else :
print ( 'No' )
if (isPowerOfTwo( 64 )):
print ( 'Yes' )
else :
print ( 'No' )
# This code is contributed by devendra salunke |
// C# program to check for power for 2 using System;
class GFG {
// Method to check if x is power of 2
static bool isPowerOfTwo( int n)
{
int cnt = 0; // initialize count to 0
while (n > 0) {
// run loop till n > 0
if ((n & 1) == 1) {
// if n&1 == 1 keep incrementing cnt
// variable
cnt++;
}
n = n >> 1; // keep dividing n by 2 using right
// shift operator
}
if (cnt
== 1) // if cnt = 1 only then it is power of 2
return true ;
return false ;
}
// Driver code
public static void Main()
{
// Function call
Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" );
Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" );
}
} // This code is contributed by devendra salunke |
<script> // JavaScript code for the above approach
// Function to check if x is power of 2
function isPowerofTwo(n)
{
let cnt = 0;
while (n > 0) {
if ((n & 1) == 1) {
cnt++; // if n&1 == 1 keep incrementing cnt
// variable
}
n = n >> 1; // keep dividing n by 2 using right
// shift operator
}
if (cnt == 1) {
// if cnt = 1 only then it is power of 2
return true ;
}
return false ;
}
// Driver code
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/>" );
// This code is contributed by sanjoy_62.
</script>
|
No Yes
Time complexity: O(log N)
Auxiliary Space: O(1)
Finding whether a given number is a power of 2 using the AND(&) operator:
To solve the problem follow the below idea:
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 –> 01111So, 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).
Note: With any method which involves subtraction by 1 (just as below, ‘n-1’), some online platforms may give an error if n is given the minimum value of integer or -2147483648. Subtraction by 1 gives integer overflow error in C++.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
/* Function to check if x is power of 2*/ bool isPowerOfTwo( int x)
{ if (x<0) return false ;
/* First x in the below expression is for the case when
* x is 0 */
return x && (!(x & (x - 1)));
} // Driver code int main()
{ // Function call
isPowerOfTwo(31) ? cout << "Yes\n" : cout << "No\n" ;
isPowerOfTwo(64) ? cout << "Yes\n" : cout << "No\n" ;
return 0;
} // This code is contributed by rathbhupendra |
// C program for the above approach #include <stdio.h> #define bool int /* Function to check if x is power of 2*/ bool isPowerOfTwo( int x)
{ /* First x in the below expression is for the case when
* x is 0 */
return x && (!(x & (x - 1)));
} // Driver code int main()
{ // Function call
isPowerOfTwo(31) ? printf ( "Yes\n" ) : printf ( "No\n" );
isPowerOfTwo(64) ? printf ( "Yes\n" ) : printf ( "No\n" );
return 0;
} |
// Java program for the above approach class Test {
/* Method to check if x is power of 2*/
static boolean isPowerOfTwo( int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x != 0 && ((x & (x - 1 )) == 0 );
}
// Driver code
public static void main(String[] args)
{
// Function call
System.out.println(isPowerOfTwo( 31 ) ? "Yes" : "No" );
System.out.println(isPowerOfTwo( 64 ) ? "Yes" : "No" );
}
} // This program is contributed by Gaurav Miglani |
# Python3 program for the above approach # Function to check if x is power of 2 def isPowerOfTwo(x):
# First x in the below expression
# is for the case when x is 0
return (x and ( not (x & (x - 1 ))))
# Driver code if __name__ = = "__main__" :
# Function call
if (isPowerOfTwo( 31 )):
print ( 'Yes' )
else :
print ( 'No' )
if (isPowerOfTwo( 64 )):
print ( 'Yes' )
else :
print ( 'No' )
# This code is contributed by Danish Raza |
// C# program for the above approach using System;
class GFG {
// Method to check if x is power of 2
static bool isPowerOfTwo( int x)
{
// First x in the below expression
// is for the case when x is 0
return x != 0 && ((x & (x - 1)) == 0);
}
// Driver code
public static void Main()
{
// Function call
Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" );
Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" );
}
} // This code is contributed by Sam007 |
<script> // JavaScript program to efficiently // check for power for 2 /* Method to check if x is power of 2*/ function isPowerOfTwo (x)
{
/* First x in the below expression is
for the case when x is 0 */
return x!=0 && ((x&(x-1)) == 0);
}
// Driver method document.write(isPowerOfTwo(31) ? "Yes" : "No" );
document.write( "<br>" +(isPowerOfTwo(64) ? "Yes" : "No" ));
// This code is contributed by 29AjayKumar </script> |
<?php // php program for the above approach // Function to check if // x is power of 2 function isPowerOfTwo ( $x )
{ // First x in the below expression // is for the case when x is 0 return $x && (!( $x & ( $x - 1)));
} // Driver Code // Function call if (isPowerOfTwo(31))
echo "Yes\n" ;
else echo "No\n" ;
if (isPowerOfTwo(64))
echo "Yes\n" ;
else echo "No\n" ;
// This code is contributed by Sam007 ?> |
No Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
Finding whether a given number is a power of 2 using the AND(&) and NOT(~) operator:
To solve the problem follow the below idea:
Another way is to use the logic to find the rightmost bit set of a given number and then check if (n & (~(n-1))) is equal to n or not
Note: With any method which involves subtraction by 1 (just as below, ‘n-1’), some online platforms may give an error if n is given the minimum value of integer or -2147483648. Subtraction by 1 gives integer overflow error in C++.
Below is the implementation of the above approach:
// C++ program of the above approach #include <bits/stdc++.h> using namespace std;
/* Function to check if x is power of 2*/ bool isPowerofTwo( long long n)
{ if (n <= 0)
return 0;
if ((n & (~(n - 1))) == n)
return 1;
return 0;
} // Driver code int main()
{ // Function call
isPowerofTwo(30) ? cout << "Yes\n" : cout << "No\n" ;
isPowerofTwo(128) ? cout << "Yes\n" : cout << "No\n" ;
return 0;
} // This code is contributed by Sachin |
// Java program of the above approach import java.io.*;
class GFG {
// Function to check if x is power of 2
static boolean isPowerofTwo( int n)
{
if (n == 0 )
return false ;
if ((n & (~(n - 1 ))) == n)
return true ;
return false ;
}
// Driver code
public static void main(String[] args)
{
// Function call
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" );
}
} // This code is contributed by rajsanghavi9. |
# Python program of the above approach # Function to check if x is power of 2*/ def isPowerofTwo(n):
if (n = = 0 ):
return 0
if ((n & (~(n - 1 ))) = = n):
return 1
return 0
# Driver code if __name__ = = "__main__" :
# Function call
if (isPowerofTwo( 30 )):
print ( 'Yes' )
else :
print ( 'No' )
if (isPowerofTwo( 128 )):
print ( 'Yes' )
else :
print ( 'No' )
# This code is contributed by shivanisinghss2110 |
// C# program of the above approach using System;
public class GFG {
// Function to check if x is power of 2
static bool isPowerofTwo( int n)
{
if (n == 0)
return false ;
if ((n & (~(n - 1))) == n)
return true ;
return false ;
}
// Driver code
public static void Main(String[] args)
{
// Function call
if (isPowerofTwo(30) == true )
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
if (isPowerofTwo(128) == true )
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
} // This code contributed by gauravrajput1 |
<script> // javascript program of the above approach // Function to check if x is power of 2
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/>" );
// This code is contributed by umadevi9616 </script> |
No Yes
Time complexity: O(1)
Auxiliary Space: O(1)