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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

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).

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 
?> 

Output:

64 is a power of 8

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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