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++
#include<bits/stdc++.h>
using namespace std;
bool isPowerOfTwo(int n)
{
if(n==0)
return false;
return (ceil(log2(n)) == floor(log2(n)));
}
int main()
{
isPowerOfTwo(31)? cout<<"Yes"<<endl: cout<<"No"<<endl;
isPowerOfTwo(64)? cout<<"Yes"<<endl: cout<<"No"<<endl;
return 0;
}
|
C
#include<stdio.h>
#include<stdbool.h>
#include<math.h>
bool isPowerOfTwo(int n)
{
if(n==0)
return false;
return (ceil(log2(n)) == floor(log2(n)));
}
int main()
{
isPowerOfTwo(31)? printf("Yes
"): printf("No
");
isPowerOfTwo(64)? printf("Yes
"): printf("No
");
return 0;
}
|
Java
class GFG
{
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)))));
}
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");
}
}
|
Python3
import math
def Log2(x):
if x == 0:
return false;
return (math.log10(x) /
math.log10(2));
def isPowerOfTwo(n):
return (math.ceil(Log2(n)) ==
math.floor(Log2(n)));
if(isPowerOfTwo(31)):
print("Yes");
else:
print("No");
if(isPowerOfTwo(64)):
print("Yes");
else:
print("No");
|
C#
using System;
class GFG
{
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)))));
}
public static void Main()
{
if(isPowerOfTwo(31))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
if(isPowerOfTwo(64))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
|
PHP
<?php
function Log2($x)
{
return (log10($x) /
log10(2));
}
function isPowerOfTwo($n)
{
return (ceil(Log2($n)) ==
floor(Log2($n)));
}
if(isPowerOfTwo(31))
echo "Yes
";
else
echo "No
";
if(isPowerOfTwo(64))
echo "Yes
";
else
echo "No
";
?>
|
Output:
No
Yes
2. Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2.
C
#include<stdio.h>
#include<stdbool.h>
bool isPowerOfTwo(int n)
{
if (n == 0)
return 0;
while (n != 1)
{
if (n%2 != 0)
return 0;
n = n/2;
}
return 1;
}
int main()
{
isPowerOfTwo(31)? printf("Yes
"): printf("No
");
isPowerOfTwo(64)? printf("Yes
"): printf("No
");
return 0;
}
|
Java
import java.io.*;
class GFG {
static boolean isPowerOfTwo(int n)
{
if (n == 0)
return false;
while (n != 1)
{
if (n % 2 != 0)
return false;
n = n / 2;
}
return true;
}
public static void main(String args[])
{
if (isPowerOfTwo(31))
System.out.println("Yes");
else
System.out.println("No");
if (isPowerOfTwo(64))
System.out.println("Yes");
else
System.out.println("No");
}
}
|
Python3
def isPowerOfTwo(n):
if (n == 0):
return False
while (n != 1):
if (n % 2 != 0):
return False
n = n // 2
return True
if(isPowerOfTwo(31)):
print('Yes')
else:
print('No')
if(isPowerOfTwo(64)):
print('Yes')
else:
print('No')
|
C#
using System;
class GFG
{
static bool isPowerOfTwo(int n)
{
if (n == 0)
return false;
while (n != 1) {
if (n % 2 != 0)
return false;
n = n / 2;
}
return true;
}
public static void Main()
{
Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No");
Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No");
}
}
|
PHP
<?php
function isPowerOfTwo($n)
{
if ($n == 0)
return 0;
while ($n != 1)
{
if ($n % 2 != 0)
return 0;
$n = $n / 2;
}
return 1;
}
if(isPowerOfTwo(31))
echo "Yes
";
else
echo "No
";
if(isPowerOfTwo(64))
echo "Yes
";
else
echo "No
";
?>
|
Output :
No
Yes
3. All power of two numbers have only one bit set. So count the no. of set bits and if you get 1 then number is a power of 2. Please see Count set bits in an integer for counting set bits.
4. If we subtract a power of 2 numbers by 1 then all unset bits after the only set bit become set; and the set bit become unset.
For example for 4 ( 100) and 16(10000), we get following after subtracting 1
3 –> 011
15 –> 01111
So, if a number n is a power of 2 then bitwise & of n and n-1 will be zero. We can say n is a power of 2 or not based on value of n&(n-1). The expression n&(n-1) will not work when n is 0. To handle this case also, our expression will become n& (!n&(n-1)) (thanks to https://tutorialspoint.dev/slugresolver/program-to-find-whether-a-no-is-power-of-two/Mohammad for adding this case).
Below is the implementation of this method.
C
#include<stdio.h>
#define bool int
bool isPowerOfTwo (int x)
{
return x && (!(x&(x-1)));
}
int main()
{
isPowerOfTwo(31)? printf("Yes
"): printf("No
");
isPowerOfTwo(64)? printf("Yes
"): printf("No
");
return 0;
}
|
Java
class Test
{
static boolean isPowerOfTwo (int x)
{
return x!=0 && ((x&(x-1)) == 0);
}
public static void main(String[] args)
{
System.out.println(isPowerOfTwo(31) ? "Yes" : "No");
System.out.println(isPowerOfTwo(64) ? "Yes" : "No");
}
}
|
Python
def isPowerOfTwo (x):
return (x and (not(x & (x - 1))) )
if(isPowerOfTwo(31)):
print('Yes')
else:
print('No')
if(isPowerOfTwo(64)):
print('Yes')
else:
print('No')
|
C#
using System;
class GFG
{
static bool isPowerOfTwo (int x)
{
return x != 0 && ((x & (x - 1)) == 0);
}
public static void Main()
{
Console.WriteLine(isPowerOfTwo(31) ? "Yes" : "No");
Console.WriteLine(isPowerOfTwo(64) ? "Yes" : "No");
}
}
|
PHP
<?php
function isPowerOfTwo ($x)
{
return $x && (!($x & ($x - 1)));
}
if(isPowerOfTwo(31))
echo "Yes
" ;
else
echo "No
";
if(isPowerOfTwo(64))
echo "Yes
" ;
else
echo "No
";
?>
|
Output :
No
Yes
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