# Java 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 2^{2}= 4 Input : n = 7 Output : No Input : n = 32 Output : Yes 2^{5}= 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.

## Java

`// Java Program to find whether a` `// no is power of two` `class` `GFG {` ` ` `/* Function to check if x is power of 2*/` ` ` `static` `boolean` `isPowerOfTwo(` `int` `n)` ` ` `{` ` ` `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)` ` ` `{` ` ` `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` |

**Output:**

No Yes

**Time Complexity: **O(log_{2}n)

**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.

## 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 program` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `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.` |

**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.

## Java

`// Java program to efficiently` `// check for power for 2` `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 method` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `System.out.println(isPowerOfTwo(` `31` `) ? ` `"Yes"` `: ` `"No"` `);` ` ` `System.out.println(isPowerOfTwo(` `64` `) ? ` `"Yes"` `: ` `"No"` `);` ` ` `}` `}` `// This program is contributed by Gaurav Miglani` |

**Output:**

No Yes

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