 Open in App
Not now

# Check if a number is a power of another number

• Difficulty Level : Easy
• Last Updated : 11 Jan, 2023

Given two positive numbers x and y, check if y is a power of x or not.
Examples :

Input:  x = 10, y = 1
Output: True
x^0 = 1

Input:  x = 10, y = 1000
Output: True
x^3 = 1

Input:  x = 10, y = 1001
Output: False

A simple solution is to repeatedly compute the powers of x. If a power becomes equal to y, then y is a power, else not.

## C++

 `// C++ program to check if a number is power of``// another number``#include ``using` `namespace` `std;` `/* Returns 1 if y is a power of x */``bool` `isPower(``int` `x, ``long` `int` `y)``{``    ``// The only power of 1 is 1 itself``    ``if` `(x == 1)``        ``return` `(y == 1);` `    ``// Repeatedly compute power of x``    ``long` `int` `pow` `= 1;``    ``while` `(``pow` `< y)``        ``pow` `*= x;` `    ``// Check if power of x becomes y``    ``return` `(``pow` `== y);``}` `/* Driver program to test above function */``int` `main()``{``    ``cout << isPower(10, 1) << endl;``    ``cout << isPower(1, 20) << endl;``    ``cout << isPower(2, 128) << endl;``    ``cout << isPower(2, 30) << endl;``    ``return` `0;``}`

## Java

 `// Java program to check if a number is power of``// another number``public` `class` `Test {``    ``// driver method to test power method``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// check the result for true/false and print.``        ``System.out.println(isPower(``10``, ``1``) ? ``1` `: ``0``);``        ``System.out.println(isPower(``1``, ``20``) ? ``1` `: ``0``);``        ``System.out.println(isPower(``2``, ``128``) ? ``1` `: ``0``);``        ``System.out.println(isPower(``2``, ``30``) ? ``1` `: ``0``);``    ``}``    ``/* Returns true if y is a power of x */``    ``public` `static` `boolean` `isPower(``int` `x, ``int` `y)``    ``{``        ``// The only power of 1 is 1 itself``        ``if` `(x == ``1``)``            ``return` `(y == ``1``);` `        ``// Repeatedly compute power of x``        ``int` `pow = ``1``;``        ``while` `(pow < y)``            ``pow = pow * x;` `        ``// Check if power of x becomes y``        ``return` `(pow == y);``    ``}``}` `// This code is contributed by Jyotsna.`

## Python3

 `# python program to check``# if a number is power of``# another number` `# Returns true if y is a``# power of x``def` `isPower (x, y):``    ` `    ``# The only power of 1``    ``# is 1 itself``    ``if` `(x ``=``=` `1``):``        ``return` `(y ``=``=` `1``)``        ` `    ``# Repeatedly compute``    ``# power of x``    ``pow` `=` `1``    ``while` `(``pow` `< y):``        ``pow` `=` `pow` `*` `x` `    ``# Check if power of x``    ``# becomes y``    ``return` `(``pow` `=``=` `y)``    ` `    ` `# Driver Code``# check the result for``# true/false and print.``if``(isPower(``10``, ``1``)):``    ``print``(``1``)``else``:``    ``print``(``0``)` `if``(isPower(``1``, ``20``)):``    ``print``(``1``)``else``:``    ``print``(``0``)``if``(isPower(``2``, ``128``)):``    ``print``(``1``)``else``:``    ``print``(``0``)``if``(isPower(``2``, ``30``)):``    ``print``(``1``)``else``:``    ``print``(``0``)``    ` `# This code is contributed``# by Sam007.`

## C#

 `// C# program to check if a number``// is power of another number``using` `System;` `class` `GFG``{``    ` `    ``// Returns true if y is a power of x``    ``public` `static` `bool` `isPower (``int` `x, ``int` `y)``    ``{``        ``// The only power of 1 is 1 itself``        ``if` `(x == 1)``        ``return` `(y == 1);` `        ``// Repeatedly compute power of x``        ``int` `pow = 1;``        ``while` `(pow < y)``        ``pow = pow * x;` `        ``// Check if power of x becomes y``        ``return` `(pow == y);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``        ``//check the result for true/false and print.``        ``Console.WriteLine(isPower(10, 1) ? 1 : 0);``        ``Console.WriteLine(isPower(1, 20) ? 1 : 0);``        ``Console.WriteLine(isPower(2, 128) ? 1 : 0);``        ``Console.WriteLine(isPower(2, 30) ? 1 : 0);``    ``}``    ` `}` `// This code is contributed by Sam007`

## PHP

 ``

## Javascript

 ``

Output

```1
0
1
0```

Time complexity: O(Logxy)
Auxiliary space: O(1)

Optimization:
We can optimize above solution to work in O(Log Log y). The idea is to do squaring of power instead of multiplying it with x, i.e., compare y with x^2, x^4, x^8, …etc. If x becomes equal to y, return true. If x becomes more than y, then we do binary search for power of x between previous power and current power, i.e., between x^i and x^(i/2).
Following are detailed step.

```1) Initialize pow = x, i = 1
2) while (pow < y)
{
pow = pow*pow
i *= 2
}
3) If pow == y
return true;
4) Else construct an array of powers
from x^i to x^(i/2)
5) Binary Search for y in array constructed
Else return true.```

Alternate Solution :
The idea is to take log of y in base x. If it turns out to be an integer, we return true. Else false.

## C++

 `// CPP program to check given number y``// is power of x``#include ``#include ``using` `namespace` `std;` `bool` `isPower(``int` `x, ``int` `y)``{``    ``// logarithm function to calculate value``    ``double` `res1 = ``log``(y) / ``log``(x);``    ``double` `res2 = ``log``(y) / ``log``(x); ``// Note : this is double` `    ``// compare to the result1 or result2 both are equal``    ``return` `(res1 == res2);``}` `// Driven program``int` `main()``{``    ``cout << isPower(2, 128) << endl;``    ``return` `0;``}`

## Java

 `// Java program to check given``// number y is power of x` `class` `GFG``{``    ``static` `boolean` `isPower(``int` `x,``                           ``int` `y)``    ``{``        ``// logarithm function to``        ``// calculate value``        ``double` `res1 = Math.log(y) /``                   ``Math.log(x);``                   ` `         ``// Note : this is double         ``        ``double` `res2 = Math.log(y) /``                      ``Math.log(x);``    ` `        ``// compare to the result1 or``        ``// result2 both are equal``        ``return` `(res1 == res2);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``if``(isPower(``2``, ``128``))``            ``System.out.println(``"1"``);``        ``else``            ``System.out.println(``"0"``);``    ``}``}` `// This code is contributed by Sam007`

## PHP

 ``

Output

`1`

Time complexity: O(1)
Auxiliary space: O(1)

Thanks to Gyayak Jain for suggesting this solution.