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# Program to find whether a given number is power of 2

• Difficulty Level : Easy
• Last Updated : 18 Nov, 2022

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 = 4

Input : n = 32
Output : Yes
Explanation: 25 = 32

Recommended Practice

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 ` `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 ` `#include ` `#include `   `/* 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

 ``

## Javascript

 ``

Output

```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 ` `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 ` `#include `   `/* 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

 ``

## Javascript

 ``

Output

```No
Yes```

Time Complexity: O(log N)
Auxiliary Space: O(1)

## C++

 `// C++ program for above approach` `#include ` `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 ` `#include `   `// 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

 ``

Output

```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 ` `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

 ``

Output

```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 –> 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 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).

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `/* 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) ? 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 ` `#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

 ``

## Javascript

 ``

Output

```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

Below is the implementation of the above approach:

## C++

 `// C++ program of the above approach`   `#include ` `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

 ``

Output

```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)

Below is the implementation of the above approach:

## C++

 `// C++ program of the above approach`   `#include ` `using` `namespace` `std;`   `/* Function to check if x is power of 2*/` `bool` `isPowerofTwo(``long` `long` `n)` `{` `    ``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

 ``

Output

```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 2^4

Biased Exponent 1023+4=1027
1027 = 10000000011
Normalised Mantissa = 01110000
We will add 0’s to complete the 52 bits

The IEEE 754 Double Precision is:
= 0 10000000011 0111000000000000000000000000000000000000000000000000

Notice that the mantissa is not 0.

——————————————————————————————

Now let’s take a power of 2, say 16
16 = 00010000
=1.0000000 x 2^4

Biased Exponent 1023+4=1027
1027 = 10000000011
Normalised Mantissa = 00000000
We will add 0’s to complete the 52 bits

The IEEE 754 Double Precision is:
= 0 10000000011 0000000000000000000000000000000000000000000000000000

Now the mantissa is strictly 0.

——————————————————————————————

Below is the implementation of the above approach:

## C

 `#include ` `#include ` ` `  `// 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 `   `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.`

Output

```No
Yes
No
Yes```

Time Complexity: O(1)
Auxiliary Space: O(1)