C# 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 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.
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) { 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() { 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) |
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.
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 program public static void Main() { Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" ); Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" ); } } // This code is contributed by Sam007 |
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.w3wiki.net/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).
Below is the implementation of this method.
C#
// C# program to efficiently // check for power for 2 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 method public static void Main() { Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No" ); Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No" ); } } // This code is contributed by Sam007 |
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!
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