Check if given number is a power of d where d is a power of 2

Given an integer n, find whether it is a power of d or not, where d is itself a power of 2.
Examples:

Input : n = 256, d = 16
Output : Yes

Input : n = 32, d = 16
Output : No

Method 1 Take log of the given number on base d, and if we get an integer then number is power of d.

C++

// CPP program to find if a number is power
// of d where d is power of 2.
#include<bits/stdc++.h>


bool isPowerOfd(int n, int d)
{
  //Take log of the given number on base d,
  //and if we get an integer then number is power of d.

    return ((int)(log(n) /log(d)) == (int)log(n) / (int)log(d));

}

/* Driver program to test above function*/
int main()
{
int n = 64, d = 8;
if (isPowerOfd(n, d))
    printf("%d is a power of %d", n, d);
else
    printf("%d is not a power of %d", n, d);
return 0;
}

Python3

#Python3 program to find if a number is power
# of d where d is power of 2.

from math import log

def isPowerOfd(n, d):

  #Take log of the given number on base d,
  #and if we get an integer then number is power of d.
  return log(n) / log(d) == log(n) // log(d)


# Driver program to test above function*
n = 64
d = 8
if (isPowerOfd(n, d)):
    print(n, "is a power of", d)
else:
    print(n, "is not a power of", d)


#This code is contributed by phasing17

Java

public class Main {

  public static boolean isPowerOfd(int n, int d) {
    //Take log of the given number on base d,
    //and if we get an integer then number is power of d.
    return (Math.log(n) / Math.log(d) == (int) (Math.log(n) / Math.log(d)));
  }

  public static void main(String[] args) {
    int n = 64, d = 8;
    if (isPowerOfd(n, d))
      System.out.println(n + " is a power of " + d);
    else
      System.out.println(n + " is not a power of " + d);
  }
}

C#

using System;

public class Mainn {
    public static bool IsPowerOfd(int n, int d) {
        //Take log of the given number on base d,
        //and if we get an integer then number is power of d.
        return (Math.Log(n) / Math.Log(d) == (int) (Math.Log(n) / Math.Log(d)));
    }

    public static void Main(string[] args) {
        int n = 64, d = 8;
        if (IsPowerOfd(n, d))
            Console.WriteLine(n + " is a power of " + d);
        else
            Console.WriteLine(n + " is not a power of " + d);
    }
}

Javascript

// JavaScript program to find if a number is power
// of d where d is power of 2.

function isPowerOfd(n, d)
{
  //Take log of the given number on base d,
  //and if we get an integer then number is power of d.

    return (Math.floor((Math.log(n) / Math.log(d)))) == (Math.log(n) /Math.log(d));

}

/* Driver program to test above function*/
let n = 64, d = 8;
if (isPowerOfd(n, d))
    console.log(n + " is a power of " + d);
else
    console.log(n + " is not a power of " + d);


//This code is contributed by phasing17

Output

64 is a power of 8

Method 2 Keep dividing the number by d, i.e, do n = n/d iteratively. In any iteration, if n%d becomes non-zero and n is not 1 then n is not a power of d, otherwise n is a power of d.
Method 3(Bitwise)
A number n is a power of d if following conditions are met. 
a) There is only one bit set in the binary representation of n (Note : d is a power of 2) 
b) The count of zero bits before the (only) set bit is a multiple of log₂(d).
For example: For n = 16 (10000) and d = 4, 16 is a power of 4 because there is only one bit set and count of 0s before the set bit is 4 which is a multiple of log₂(4).

Steps to solve this problem:

1. declare variable count=0.

2. check if n&&!(n&(n-1)) is true than :

          *while n is greater than 1:

                   *n>>=1.

                   *update count to count+1.

          *return count%(log2n(d))==0.

3. return false.

C++

// CPP program to find if a number is power
// of d where d is power of 2.
#include<stdio.h>

unsigned int Log2n(unsigned int n)
{
return (n > 1)? 1 + Log2n(n/2): 0;
}

bool isPowerOfd(unsigned int n, unsigned int d)
{
int count = 0;

/* Check if there is only one bit set in n*/
if (n && !(n&(n-1)) )
{
    /* count 0 bits before set bit */
    while (n > 1)
    {
    n >>= 1;
    count += 1;
    }

    /* If count is a multiple of log2(d)
    then return true else false*/
    return (count%(Log2n(d)) == 0);
}

/* If there are more than 1 bit set
    then n is not a power of 4*/
return false;
}

/* Driver program to test above function*/
int main()
{
int n = 64, d = 8;
if (isPowerOfd(n, d))
    printf("%d is a power of %d", n, d);
else
    printf("%d is not a power of %d", n, d);
return 0;
}

Java

// Java program to find if
// a number is power of d
// where d is power of 2.

class GFG
{
static int Log2n(int n)
{
    return (n > 1)? 1 +
    Log2n(n / 2): 0;
}

static boolean isPowerOfd(int n,
                        int d)
{
int count = 0;

/* Check if there is
only one bit set in n*/
if (n > 0 && (n &
(n - 1)) == 0)
{
    /* count 0 bits
    before set bit */
    while (n > 1)
    {
        n >>= 1;
        count += 1;
    }

    /* If count is a multiple
    of log2(d) then return
    true else false*/
    return (count %
        (Log2n(d)) == 0);
}

/* If there are more
than 1 bit set then
n is not a power of 4*/
return false;
}

// Driver Code
public static void main(String[] args)
{
    int n = 64, d = 8;
    if (isPowerOfd(n, d))
        System.out.println(n +
                    " is a power of " + d);
    else
        System.out.println(n +
                    " is not a power of " + d);
}
}

// This code is contributed by mits

Python3

# Python3 program to find if a number
# is power of d where d is power of 2.

def Log2n(n):
    return (1 + Log2n(n / 2)) if (n > 1) else 0;

def isPowerOfd(n, d):
    count = 0;

    # Check if there is only
    # one bit set in n

    if (n and (n & (n - 1))==0):
        # count 0 bits
        # before set bit
        while (n > 1):
            n >>= 1;
            count += 1;
        # If count is a multiple of log2(d)
        # then return true else false
        return (count%(Log2n(d)) == 0);

    # If there are more than 1 bit set
    # then n is not a power of 4
    return False;

# Driver Code
n = 64;
d = 8;
if (isPowerOfd(n, d)):
    print(n,"is a power of",d);
else:
    print(n,"is not a power of",d);

# This code is contributed by mits

C#

// C# program to find if
// a number is power of d
// where d is power of 2.
using System;

class GFG
{
static int Log2n(int n)
{
    return (n > 1)? 1 +
    Log2n(n / 2): 0;
}

static bool isPowerOfd(int n,
                    int d)
{
int count = 0;

/* Check if there is
only one bit set in n*/
if (n > 0 && (n & (n - 1)) == 0)
{
    /* count 0 bits
    before set bit */
    while (n > 1)
    {
    n >>= 1;
    count += 1;
    }

    /* If count is a multiple
    of log2(d) then return
    true else false*/
    return (count % (Log2n(d)) == 0);
}

/* If there are more than
1 bit set then n is not
a power of 4*/
return false;
}

// Driver Code
static void Main()
{
int n = 64, d = 8;
if (isPowerOfd(n, d))
    Console.WriteLine("{0} is a " +
                    "power of {1}",
                            n, d);
else
    Console.WriteLine("{0} is not a"+
                    " power of {1}",
                            n, d);
}

// This code is contributed by mits
}

PHP

<?php
// PHP program to find if a number
// is power of d where d is power of 2.

function Log2n($n)
{
    return ($n > 1)? 1 +
    Log2n($n / 2): 0;
}

function isPowerOfd($n, $d)
{
    $count = 0;

// Check if there is only
// one bit set in n
if ($n && !($n & ($n - 1)))
{

    // count 0 bits
    // before set bit
    while ($n > 1)
    {
        $n >>= 1;
        $count += 1;
    }

    /* If count is a multiple of log2(d)
    then return true else false*/
    return ($count%(Log2n($d)) == 0);
}

    /* If there are more than 1 bit set
    then n is not a power of 4*/
    return false;
}

// Driver Code
$n = 64;
$d = 8;
if (isPowerOfd($n, $d))
    echo $n," ","is a power of ", $d;
else
    echo $n," ","is not a power of ", $d;

// This code is contributed by m_kit
?>

Javascript

<script>

// Javascript program to find if
// a number is power of d
// where d is power of 2
function Log2n(n)
{
    return (n > 1) ? 1 +
      Log2n(n / 2) : 0;
}

// Function to count the number
// of ways to paint  N * 3 grid
// based on given conditions
function isPowerOfd(n, d)
{
    var count = 0;

    /* Check if there is
    only one bit set in n*/
    if (n > 0 && (n & (n - 1)) == 0)
    {

        /* count 0 bits
        before set bit */
        while (n > 1)
        {
            n >>= 1;
            count += 1;
        }

        /* If count is a multiple
        of log2(d) then return
        true else false*/
        return (count % (Log2n(d)) == 0);
    }

    /* If there are more
    than 1 bit set then
    n is not a power of 4*/
    return false;
}

// Driver code
var n = 64, d = 8;

if (isPowerOfd(n, d))
    document.write(n +
                " is a power of " + d);
else
    document.write(n +
                " is not a power of " + d);

// This code is contributed by Khushboogoyal499

</script>

Output

64 is a power of 8

Time Complexity: O(log₂n)
Auxiliary Space: O(1)