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# 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 = 4

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

Recommended Practice

## Finding whether a given number is a power of 2 using Log operator:

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)`

## Javascript

 ``

## PHP

 ``

Output

```No
Yes
```

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

## Finding whether a given number is a power of 2 using the modulo & division operator:

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`

## Javascript

 ``

## PHP

 ``

Output

```No
Yes
```

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

## Finding 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.`

## 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`

## 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`

## Javascript

 ``

Output

```No
Yes
```

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

## Finding 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).

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 ``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 ``#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`

## Javascript

 ``

## PHP

 ``

Output

```No
Yes
```

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

## Finding 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 ``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)