Program to find whether a given number is power of 2
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
To solve the problem follow the below idea:
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++
// 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
// 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
// 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
# 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#
// 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) |
PHP
<?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 ?> |
Javascript
<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> |
No Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
Find whether a given number is a power of 2 using the division operator:
To solve the problem follow the below idea:
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.
Below is the implementation of the above approach:
C++
// 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
// 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
// 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. |
Python3
# 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#
// 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 |
PHP
<?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 ?> |
Javascript
<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> |
No Yes
Time Complexity: O(log N)
Auxiliary Space: O(1)
Below is the recursive implementation of the above approach:
C++
// C++ program for above approach #include <bits/stdc++.h> using namespace std; // Function which checks whether a number is a power of 2 bool powerOf2( int n) { // base cases // '1' is the only odd number which is a power of 2(2^0) if (n == 1) return true ; // all other odd numbers are not powers of 2 else if (n % 2 != 0 || n == 0) return false ; // recursive function call return powerOf2(n / 2); } // Driver Code int main() { int n = 64; // True int m = 12; // False // Function call if (powerOf2(n) == 1) cout << "True" << endl; else cout << "False" << endl; if (powerOf2(m) == 1) cout << "True" << endl; else cout << "False" << endl; } // This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program for above approach #include <stdbool.h> #include <stdio.h> // Function which checks whether a number is a power of 2 bool powerOf2( int n) { // base cases // '1' is the only odd number which is a power of 2(2^0) if (n == 1) return true ; // all other odd numbers are not powers of 2 else if (n % 2 != 0 || n == 0) return false ; // recursive function call return powerOf2(n / 2); } // Driver Code int main() { int n = 64; // True int m = 12; // False // Function call if (powerOf2(n) == 1) printf ( "True\n" ); else printf ( "False\n" ); if (powerOf2(m) == 1) printf ( "True\n" ); else printf ( "False\n" ); } // This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program for // the above approach import java.util.*; class GFG { // Function which checks // whether a number is a // power of 2 static boolean powerOf2( int n) { // base cases // '1' is the only odd number // which is a power of 2(2^0) if (n == 1 ) return true ; // all other odd numbers are // not powers of 2 else if (n % 2 != 0 || n == 0 ) return false ; // recursive function call return powerOf2(n / 2 ); } // Driver Code public static void main(String[] args) { // True int n = 64 ; // False int m = 12 ; // Function call 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" ); } } // This code is contributed by Princi Singh |
Python3
# Python program for above approach # function which checks whether a # number is a power of 2 def powerof2(n): # base cases # '1' is the only odd number # which is a power of 2(2^0) if n = = 1 : return True # all other odd numbers are not powers of 2 elif n % 2 ! = 0 or n = = 0 : return False # recursive function call return powerof2(n / 2 ) # Driver Code if __name__ = = "__main__" : # Function call print (powerof2( 64 )) # True print (powerof2( 12 )) # False # code contributed by Moukthik a.k.a rowdyninja |
C#
// C# program for above approach using System; class GFG { // Function which checks whether a // number is a power of 2 static bool powerOf2( int n) { // Base cases // '1' is the only odd number // which is a power of 2(2^0) if (n == 1) return true ; // All other odd numbers // are not powers of 2 else if (n % 2 != 0 || n == 0) return false ; // Recursive function call return powerOf2(n / 2); } // Driver code static void Main() { int n = 64; // True int m = 12; // False // Function call if (powerOf2(n)) { Console.Write( "True" + "\n" ); } else { Console.Write( "False" + "\n" ); } if (powerOf2(m)) { Console.Write( "True" ); } else { Console.Write( "False" ); } } } // This code is contributed by rutvik_56 |
Javascript
<script> // javascript program for // the above approach // Function which checks // whether a number is a // power of 2 function powerOf2(n) { // base cases // '1' is the only odd number // which is a power of 2(2^0) if (n == 1) return true ; // all other odd numbers are // not powers of 2 else if (n % 2 != 0 || n ==0) return false ; // recursive function call return powerOf2(n / 2); } // Driver Code //True var n = 64; //False 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" ); // This code contributed by shikhasingrajput </script> |
True False
Time Complexity: O(log N)
Auxiliary Space: O(log N)
Find 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++
// 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
// 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. |
C#
// 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 |
Python3
# 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 |
Javascript
<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)
Find 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++
// 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
// 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
// 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
# 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#
// 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 |
PHP
<?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 ?> |
Javascript
<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> |
No Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
Find 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++
// 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
// 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. |
Python3
# 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#
// 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 |
Javascript
<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)
Find whether a given number is a power of 2 using Brian Kernighan’s algorithm:
To solve the problem follow the below idea:
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)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++
// 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 false ; return (n != 0) && ((n & (n - 1)) == 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 Suruchi Kumari |
Java
// Java program of the above approach import java.io.*; class GFG { /* Function to check if x is power of 2*/ public static boolean isPowerofTwo( long n) { return (n != 0 ) && ((n & (n - 1 )) == 0 ); } // Driver code public static void main(String[] args) { // Function call if (isPowerofTwo( 30 )) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } if (isPowerofTwo( 128 )) { System.out.println( "Yes" ); } else { System.out.println( "No" ); } } } // This code is contributed by akashish__ |
Python3
# Python3 program of the above approach # Function to check if x is power of 2 def isPowerofTwo(n): return (n ! = 0 ) and ((n & (n - 1 )) = = 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 aditya942003patil |
C#
// C# program of the above approach using System; public class GFG { /* Function to check if x is power of 2*/ static public bool isPowerofTwo( ulong n) { return (n != 0) && ((n & (n - 1)) == 0); } // Driver code static public void Main() { // Function call if (isPowerofTwo(30)) { System.Console.WriteLine( "Yes" ); } else { System.Console.WriteLine( "No" ); } if (isPowerofTwo(128)) { System.Console.WriteLine( "Yes" ); } else { System.Console.WriteLine( "No" ); } } } // This code is contributed by akashish__ |
Javascript
<script> function isPowerofTwo(n){ return (n != 0) && ((n & (n - 1)) == 0); } /* Function to check if x is power of 2*/ if (isPowerofTwo(30)) { document.write( "Yes" ); } else { document.write( "No" ); } if (isPowerofTwo(128)) { document.write( "Yes" ); } else { document.write( "No" ); } </script> |
No Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
Find whether a given number is a power of 2 using a floating point bit hack:
We can also harness a unique property of IEEE Standard 754 to infer if the given integer is a power of 2 using the following bit hack that only works in a few languages that allow pointer casting.
This works because we know that a power of 2 only has 1 bit high and all the other bits low. Therefore if we represent such a number in scientific notation, we’ll always be left with a mantissa of 1. But in IEEE Standard 754 the 1 is discarded from the mantissa as it is redundant. Now we should be left with 0, if not, then the number must not be a power of 2. We will be using double precision.
Example:
Let’s take 23 first –
23 = 00010111
=1.0111000 x 2^4Biased Exponent 1023+4=1027
1027 = 10000000011
Normalised Mantissa = 01110000
We will add 0’s to complete the 52 bitsThe IEEE 754 Double Precision is:
= 0 10000000011 0111000000000000000000000000000000000000000000000000Notice that the mantissa is not 0.
——————————————————————————————
Now let’s take a power of 2, say 16 –
16 = 00010000
=1.0000000 x 2^4Biased Exponent 1023+4=1027
1027 = 10000000011
Normalised Mantissa = 00000000
We will add 0’s to complete the 52 bitsThe IEEE 754 Double Precision is:
= 0 10000000011 0000000000000000000000000000000000000000000000000000Now the mantissa is strictly 0.
——————————————————————————————
Below is the implementation of the above approach:
C
#include <stdio.h> #include <stdint.h> // Function to check if x is power of 2. int isPowerOfTwo( int x) { // Power of 2 can't be less than 1. if (x < 1) return 0; // Converting the number to double precision floating point. double n = x; /* * Storing the binary representation of double to an unsigned 64 bit integer. * This is not the same as direct casting to integer, as the binary representation changes. */ uint64_t r = *((uint64_t*)(&n)); // Discarding 1 sign bit and 11 exponent bits. r = (r << 12); return r == 0; } // Driver code int main() { // Function calls isPowerOfTwo(31) ? printf ( "Yes\n" ) : printf ( "No\n" ); isPowerOfTwo(32) ? printf ( "Yes\n" ) : printf ( "No\n" ); isPowerOfTwo(33) ? printf ( "Yes\n" ) : printf ( "No\n" ); isPowerOfTwo(64) ? printf ( "Yes\n" ) : printf ( "No\n" ); return 0; } // This code is contributed by Aryan Rai. |
C++
#include <iostream> using namespace std; // Function to check if x is power of 2. bool isPowerOfTwo( int x) { // Power of 2 can't be less than 1. if (x < 1) return false ; // Converting the number to double precision floating point. double n = x; /* * Storing the binary representation of double to an unsigned 64 bit integer. * This is not the same as direct casting to integer, as the binary representation changes. */ uint64_t r = *((uint64_t*)(&n)); // Discarding 1 sign bit and 11 exponent bits. r = (r << 12); return r == 0; } // Driver code int main() { // Function calls cout << ( isPowerOfTwo(31) ? "Yes" : "No" ) << endl; cout << ( isPowerOfTwo(32) ? "Yes" : "No" ) << endl; cout << ( isPowerOfTwo(33) ? "Yes" : "No" ) << endl; cout << ( isPowerOfTwo(64) ? "Yes" : "No" ) << endl; return 0; } // This code is contributed by Aryan Rai. |
Java
import java.math.BigInteger; public class Main { static boolean isPowerOfTwo( int x) { // Power of 2 can't be less than 1. if (x < 1 ) return false ; // Converting the number to a BigInteger. BigInteger n = BigInteger.valueOf(x); /* * Storing the binary representation of BigInteger as * a string and checking if there is only 1 * '1' bit in the string representation. */ return n.bitCount() == 1 ; } public static void main(String[] args) { // Function calls System.out.println(isPowerOfTwo( 31 ) ? "Yes" : "No" ); System.out.println(isPowerOfTwo( 32 ) ? "Yes" : "No" ); System.out.println(isPowerOfTwo( 33 ) ? "Yes" : "No" ); System.out.println(isPowerOfTwo( 64 ) ? "Yes" : "No" ); } } |
Python3
import math def isPowerOfTwo(x): # Power of 2 can't be less than 1. if x < 1 : return False # check if log2 is an integer using modulo return math.log2(x) % 1 = = 0 # Driver code if __name__ = = "__main__" : # Function calls print ( "Yes" if isPowerOfTwo( 31 ) else "No" ) print ( "Yes" if isPowerOfTwo( 32 ) else "No" ) print ( "Yes" if isPowerOfTwo( 33 ) else "No" ) print ( "Yes" if isPowerOfTwo( 64 ) else "No" ) |
C#
// C# code using System; public class GFG { static bool IsPowerOfTwo( int x) { // Power of 2 can't be less than 1. if (x < 1) return false ; // check if log2 is an integer using modulo return Math.Log(x,2) % 1 == 0; } // Driver code public static void Main( string [] args) { // Function calls System.Console.WriteLine(IsPowerOfTwo(31) ? "Yes" : "No" ); System.Console.WriteLine(IsPowerOfTwo(32) ? "Yes" : "No" ); System.Console.WriteLine(IsPowerOfTwo(33) ? "Yes" : "No" ); System.Console.WriteLine(IsPowerOfTwo(64) ? "Yes" : "No" ); } } // This code is contributed by Aman Kumar. |
Javascript
function isPowerOfTwo(x) { // Power of 2 can't be less than 1. if (x < 1) { return false ; } // check if log2 is an integer using modulo return Math.log2(x) % 1 === 0; } // Driver code console.log(isPowerOfTwo(31) ? "Yes" : "No" ); console.log(isPowerOfTwo(32) ? "Yes" : "No" ); console.log(isPowerOfTwo(33) ? "Yes" : "No" ); console.log(isPowerOfTwo(64) ? "Yes" : "No" ); |
No Yes No Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
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