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Python Program to find whether a no is power of two

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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


Output: 

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


Output: 

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.geeksforgeeks.org/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   


Output: 

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!
 



Last Updated : 06 Apr, 2023
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