# Check if a number is power of k using base changing method

This program checks whether a number n can be expressed as power of k and if yes, then to what power should k be raised to make it n. Following example will clarify :

Examples:

```Input :   n = 16, k = 2
Output :  yes : 4
Explanation : Answer is yes because 16 can
be expressed as power of 2.

Input :   n = 27, k = 3
Output :  yes : 3
Explanation : Answer is yes as 27 can be
expressed as power of 3.

Input :  n = 20, k = 5
Output : No
Explanation : Answer is No as 20 cannot
be expressed as power of 5.
```

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

We have discussed two methods in below post
:Check if a number is a power of another number

In this post, a new Base Changing method is discussed.

In Base Changing Method, we simply change the base of number n to k and check if the first digit of Changed number is 1 and remaining all are zero.

Example for this : Let’s take n = 16 and k = 2.
Change 16 to base 2. i.e. (10000)2. Since first digit is 1 and remaining are zero. Hence 16 can be expressed as power of 2. Count the length of (10000)2 and subtract 1 from it, that’ll be the number to which 2 must be raised to make 16. In this case 5 – 1 = 4.

Another example : Let’s take n = 20 and k = 3.
20 in base 3 is (202)3. Since there are two non-zero digit, hence 20 cannot be expressed as power of 3.

## C++

 `// CPP program to check if a number can be ` `// raised to k ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `bool` `isPowerOfK(unsigned ``int` `n, unsigned ``int` `k) ` `{ ` `    ``// loop to change base n to base = k ` `    ``bool` `oneSeen = ``false``; ` `    ``while` `(n > 0) { ` ` `  `        ``// Find current digit in base k ` `        ``int` `digit = n % k; ` ` `  `        ``// If digit is neither 0 nor 1  ` `        ``if` `(digit > 1) ` `            ``return` `false``; ` ` `  `        ``// Make sure that only one 1 ` `        ``// is present.  ` `        ``if` `(digit == 1) ` `        ``{ ` `            ``if` `(oneSeen) ` `            ``return` `false``; ` `            ``oneSeen = ``true``; ` `        ``}      ` ` `  `        ``n /= k; ` `    ``} ` `     `  `    ``return` `true``;  ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 64, k = 4; ` ` `  `    ``if` `(isPowerOfK(n ,k)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` `} `

## Java

 `// Java program to check if a number can be ` `// raised to k ` ` `  `class` `GFG ` `{ ` `    ``static` `boolean` `isPowerOfK(``int` `n,``int` `k) ` `    ``{ ` `        ``// loop to change base n to base = k ` `        ``boolean` `oneSeen = ``false``; ` `        ``while` `(n > ``0``)  ` `        ``{ ` `     `  `            ``// Find current digit in base k ` `            ``int` `digit = n % k; ` `     `  `            ``// If digit is neither 0 nor 1  ` `            ``if` `(digit > ``1``) ` `                ``return` `false``; ` `     `  `            ``// Make sure that only one 1 ` `            ``// is present.  ` `            ``if` `(digit == ``1``) ` `            ``{ ` `                ``if` `(oneSeen) ` `                ``return` `false``; ` `                ``oneSeen = ``true``; ` `            ``}      ` `     `  `            ``n /= k; ` `        ``} ` `         `  `        ``return` `true``;  ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `n = ``64``, k = ``4``; ` `     `  `        ``if` `(isPowerOfK(n ,k)) ` `            ``System.out.print(``"Yes"``); ` `        ``else` `            ``System.out.print(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python program to ` `# check if a number can be ` `# raised to k ` ` `  `def` `isPowerOfK(n, k): ` ` `  `    ``# loop to change base ` `    ``# n to base = k ` `    ``oneSeen ``=` `False` `    ``while` `(n > ``0``): ` `  `  `        ``# Find current digit in base k ` `        ``digit ``=` `n ``%` `k ` `  `  `        ``# If digit is neither 0 nor 1  ` `        ``if` `(digit > ``1``): ` `            ``return` `False` `  `  `        ``# Make sure that only one 1 ` `        ``# is present.  ` `        ``if` `(digit ``=``=` `1``): ` `         `  `            ``if` `(oneSeen): ` `                ``return` `False` `            ``oneSeen ``=` `True` `  `  `        ``n ``/``/``=` `k ` `     `  `    ``return` `True` `     `  `# Driver code ` ` `  `n ``=` `64` `k ``=` `4` `  `  `if` `(isPowerOfK(n , k)): ` `    ``print``(``"Yes"``) ` `else``: ` `    ``print``(``"No"``) ` ` `  `# This code is contributed ` `# by Anant Agarwal. `

## C#

 `// C# program to check if a number can be ` `// raised to k ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `bool` `isPowerOfK(``int` `n, ``int` `k) ` `    ``{ ` `         `  `        ``// loop to change base n to base = k ` `        ``bool` `oneSeen = ``false``; ` `        ``while` `(n > 0)  ` `        ``{ ` `     `  `            ``// Find current digit in base k ` `            ``int` `digit = n % k; ` `     `  `            ``// If digit is neither 0 nor 1  ` `            ``if` `(digit > 1) ` `                ``return` `false``; ` `     `  `            ``// Make sure that only one 1 ` `            ``// is present.  ` `            ``if` `(digit == 1) ` `            ``{ ` `                ``if` `(oneSeen) ` `                    ``return` `false``; ` `                     `  `                ``oneSeen = ``true``; ` `            ``}  ` `     `  `            ``n /= k; ` `        ``} ` `         `  `        ``return` `true``;  ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `n = 64, k = 4; ` `     `  `        ``if` `(isPowerOfK(n ,k)) ` `            ``Console.WriteLine(``"Yes"``); ` `        ``else` `            ``Console.WriteLine(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` 0)  ` `    ``{ ` ` `  `        ``// Find current  ` `        ``// digit in base k ` `        ``\$digit` `= ``\$n` `% ``\$k``; ` ` `  `        ``// If digit is  ` `        ``// neither 0 nor 1  ` `        ``if` `(``\$digit` `> 1) ` `            ``return` `false; ` ` `  `        ``// Make sure that ` `        ``// only one 1 ` `        ``// is present.  ` `        ``if` `(``\$digit` `== 1) ` `        ``{ ` `            ``if` `(``\$oneSeen``) ` `            ``return` `false; ` `            ``\$oneSeen` `= true; ` `        ``}  ` ` `  `        ``\$n` `= (int)``\$n` `/ ``\$k``; ` `    ``} ` `     `  `    ``return` `true;  ` `} ` ` `  `// Driver code ` `\$n` `= 64; ` `\$k` `= 4; ` ` `  `if` `(isPowerOfK(``\$n``, ``\$k``)) ` `    ``echo` `"Yes"``; ` `else` `    ``echo` `"No"``; ` ` `  `// This code is contributed  ` `// by ajit ` `?> `

Output:

```Yes
```

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Improved By : vt_m, jit_t

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