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# Python Program to find whether a no is power of two

• Difficulty Level : Hard
• Last Updated : 06 Jan, 2022

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```

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```

Please refer complete article on Program to find whether a no is power of two for more details!

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