# 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.

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 <iostream>` `#include <algorithm>` `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

`<?php` `// PHP program to check ` `// if a number can be` `// raised to k` `function` `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` `= (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

`<script>` `// JavaScript program to check if a number can be` `// raised to k` ` ` `function` `isPowerOfK(n,k)` ` ` `{` ` ` `// loop to change base n to base = k` ` ` `let oneSeen = ` `false` `;` ` ` `while` `(n > 0) ` ` ` `{` ` ` ` ` `// Find current digit in base k` ` ` `let 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 = Math.floor(n / k);` ` ` `}` ` ` ` ` `return` `true` `; ` ` ` `}` `// Driver Code` ` ` `let n = 64, k = 4;` ` ` ` ` `if` `(isPowerOfK(n ,k))` ` ` `document.write(` `"Yes"` `);` ` ` `else` ` ` `document.write(` `"No"` `);` ` ` `</script>` |

**Output:**

Yes

**Time Complexity: **O(logn)

**Space Complexity:** O(1)

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