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

• Difficulty Level : Basic
• Last Updated : 07 Aug, 2022

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.  ```

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``?>`

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(logn)

Space Complexity: O(1)
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