Friday, December 27, 2024
Google search engine
HomeData Modelling & AIProgram to find whether a given number is power of 2

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 <bits/stdc++.h>
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 <math.h>
#include <stdbool.h>
#include <stdio.h>
 
/* 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




<script>
// javascript Program to find whether a
// no is power of two
 
    /* Function to check if x is power of 2 */
    function isPowerOfTwo(n)
    {
        if (n == 0)
            return false;
 
        return parseInt( (Math.ceil((Math.log(n) / Math.log(2))))) == parseInt( (Math.floor(((Math.log(n) / Math.log(2))))));
    }
 
    // Driver Code
     
    if (isPowerOfTwo(31))
        document.write("Yes<br/>");
    else
        document.write("No<br/>");
 
    if (isPowerOfTwo(64))
        document.write("Yes<br/>");
    else
        document.write("No<br/>");
 
// This code is contributed by shikhasingrajput.
</script>


PHP




<?php
// PHP Program to find
// whether a no is
// power of two
 
// Function to check
// Log base 2
function Log2($x)
{
    return (log10($x) /
            log10(2));
}
 
 
// Function to check
// if x is power of 2
function isPowerOfTwo($n)
{
    return (ceil(Log2($n)) ==
            floor(Log2($n)));
}
 
// Driver Code
 
// Function call
if(isPowerOfTwo(31))
echo "Yes\n";
else
echo "No\n";
 
if(isPowerOfTwo(64))
echo "Yes\n";
else
echo "No\n";
     
// This code is contributed
// by Sam007
?>


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 <bits/stdc++.h>
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 <stdbool.h>
#include <stdio.h>
 
/* 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




<script>
 
    /* Function to check if x is power of 2*/
    function isPowerOfTwo(n)
    {
        if (n == 0)
            return 0;
        while (n != 1)
        {
            if (n%2 != 0)
                return 0;
            n = n/2;
        }
        return 1;
    
     
    isPowerOfTwo(31)? document.write("Yes" + "</br>"): document.write("No" + "</br>");
    isPowerOfTwo(64)? document.write("Yes"): document.write("No");
 
</script>


PHP




<?php
// php program for the above approach
 
// Function to check if
// x is power of 2
function isPowerOfTwo($n)
{
if ($n == 0)
    return 0;
while ($n != 1)
{
    if ($n % 2 != 0)
        return 0;
    $n = $n / 2;
}
return 1;
}
 
// Driver Code
 
// Function call
if(isPowerOfTwo(31))
    echo "Yes\n";
else
    echo "No\n";
 
if(isPowerOfTwo(64))
    echo "Yes\n";
else
    echo "No\n";
 
// This code is contributed
// by Sam007
?>


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 <bits/stdc++.h>
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




<script>
    // JavaScript code for the above approach
 
    // Function to check if x is power of 2
    function isPowerofTwo(n)
    {
        let 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
 
    if (isPowerofTwo(30) == true)
            document.write("Yes" + "<br/>");
        else
            document.write("No" + "<br/>");
 
        if (isPowerofTwo(128) == true)
            document.write("Yes" + "<br/>");
        else
            document.write("No" + "<br/>");
             
            // This code is contributed by sanjoy_62.
  </script>


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 <bits/stdc++.h>
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 <stdio.h>
#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




<script>
 
// JavaScript program to efficiently
// check for power for 2  
 
/* Method to check if x is power of 2*/
    function isPowerOfTwo (x)
    {
      /* First x in the below expression is
        for the case when x is 0 */
        return x!=0 && ((x&(x-1)) == 0);
         
    }
     
// Driver method
document.write(isPowerOfTwo(31) ? "Yes" : "No");
document.write("<br>"+(isPowerOfTwo(64) ? "Yes" : "No"));
 
// This code is contributed by 29AjayKumar
 
</script>


PHP




<?php
// php program for the above approach
 
// Function to check if
// x is power of 2
function isPowerOfTwo ($x)
{
// First x in the below expression
// is for the case when x is 0
return $x && (!($x & ($x - 1)));
}
 
// Driver Code
 
// Function call
if(isPowerOfTwo(31))
    echo "Yes\n" ;
else
    echo "No\n";
 
if(isPowerOfTwo(64))
    echo "Yes\n" ;
else
    echo "No\n";
         
// This code is contributed by Sam007
?>


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 <bits/stdc++.h>
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




<script>
// javascript program of the above approach
  
    // Function to check if x is power of 2
    function isPowerofTwo(n)
    {
        if (n == 0)
            return false;
        if ((n & (~(n - 1))) == n)
            return true;
        return false;
    }
 
        if (isPowerofTwo(30) == true)
            document.write("Yes<br/>");
        else
            document.write("No<br/>");
       
          if (isPowerofTwo(128) == true)
            document.write("Yes<br/>");
        else
            document.write("No<br/>");
                 
// This code is contributed by umadevi9616
</script>


Output

No
Yes

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

Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above. 

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

RELATED ARTICLES

Most Popular

Recent Comments