Python Program to find whether a no is power of two
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.
Python3
# Python3 Program to find # whether a no is # power of two import math # Function to check # Log base 2 def Log2(x): 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 (isPowerOfTwo( 31 )): print ( "Yes" ); else : print ( "No" ); if (isPowerOfTwo( 64 )): print ( "Yes" ); else : print ( "No" ); # This code is contributed # by mits |
No Yes
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.
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 (isPowerOfTwo( 31 )): print ( 'Yes' ) else : print ( 'No' ) if (isPowerOfTwo( 64 )): print ( 'Yes' ) else : print ( 'No' ) # This code is contributed by Danish Raza |
No Yes
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.w3wiki.net/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).
Below is the implementation of this method.
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 (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 (isPowerOfTwo( 31 )): print ( 'Yes' ) else : print ( 'No' ) if (isPowerOfTwo( 64 )): print ( 'Yes' ) else : print ( 'No' ) # This code is contributed by Danish Raza |
No Yes
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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