Check if a number is a power of another number
Last Updated :
05 Jun, 2025
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
[Naive Approach] Repeated Multiplication Method
This approach checks whether a number y is a power of another number x by repeatedly multiplying x until it either matches or exceeds y. It starts with an initial value of 1 and keeps multiplying it by x in a loop. If the resulting value becomes equal to y, then y is a power of x; otherwise, if it exceeds y without a match, the function returns false. This method efficiently verifies powers by iteratively computing exponential values instead of using logarithms or recursion.
C++
#include <bits/stdc++.h>
using namespace std;
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);
}
int main()
{
cout << boolalpha;
cout << isPower(10, 1) << endl;
cout << isPower(1, 20) << endl;
cout << isPower(2, 128) << endl;
cout << isPower(2, 30) << endl;
return 0;
}
#include <stdio.h>
int 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);
}
int main()
{
printf("%d\n", isPower(10, 1));
printf("%d\n", isPower(1, 20));
printf("%d\n", isPower(2, 128));
printf("%d\n", isPower(2, 30));
return 0;
}
public class Main {
public static boolean isPower(int x, long y) {
// The only power of 1 is 1 itself
if (x == 1)
return (y == 1);
// Repeatedly compute power of x
long pow = 1;
while (pow < y)
pow *= x;
// Check if power of x becomes y
return (pow == y);
}
public static void main(String[] args) {
System.out.println(isPower(10, 1));
System.out.println(isPower(1, 20));
System.out.println(isPower(2, 128));
System.out.println(isPower(2, 30));
}
}
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 *= x
# Check if power of x becomes y
return pow == y
if __name__ == '__main__':
print(isPower(10, 1))
print(isPower(1, 20))
print(isPower(2, 128))
print(isPower(2, 30))
using System;
class Program {
static bool isPower(int x, long y) {
// The only power of 1 is 1 itself
if (x == 1)
return (y == 1);
// Repeatedly compute power of x
long pow = 1;
while (pow < y)
pow *= x;
// Check if power of x becomes y
return (pow == y);
}
static void Main() {
Console.WriteLine(isPower(10, 1));
Console.WriteLine(isPower(1, 20));
Console.WriteLine(isPower(2, 128));
Console.WriteLine(isPower(2, 30));
}
}
function isPower(x, y) {
// The only power of 1 is 1 itself
if (x === 1)
return y === 1;
// Repeatedly compute power of x
let pow = 1;
while (pow < y)
pow *= x;
// Check if power of x becomes y
return pow === y;
}
console.log(isPower(10, 1));
console.log(isPower(1, 20));
console.log(isPower(2, 128));
console.log(isPower(2, 30));
Outputtrue
false
true
false
Time complexity: O(Logxy)
Auxiliary space: O(1)
[Better Approach] Exponentiation and Binary Search Method
This approach efficiently checks if a number y is a power of another number x by leveraging exponential growth and binary search. It first attempts to reach or exceed y by repeatedly squaring x, which significantly reduces the number of multiplications compared to simple iteration. If the squared value matches y, the function returns true. If it overshoots y, a binary search is performed between the last two computed powers of x to find an exact match.
C++
#include <bits/stdc++.h>
using namespace std;
bool isPower(int x, long int y) {
if (x == 1) return (y == 1);
if(y==1) return true;
long int pow = x, i = 1;
while (pow < y) {
pow *= pow;
i *= 2;
}
if (pow == y) return true;
long int low = x, high = pow;
while (low <= high) {
long int mid = low + (high - low) / 2;
long int result = powl(x, int(log2(mid) / log2(x)));
if (result == y) return true;
if (result < y) low = mid + 1;
else high = mid - 1;
}
return false;
}
int main() {
cout << boolalpha;
cout << isPower(10, 1) << endl;
cout << isPower(1, 20) << endl;
cout << isPower(2, 128) << endl;
cout << isPower(2, 30) << endl;
return 0;
}
public class PowerCheck {
public static boolean isPower(int x, long y) {
if (x == 1) return (y == 1);
if(y==1) return true;
long pow = x;
while (pow < y) {
pow *= pow;
}
if (pow == y) return true;
long low = x, high = pow;
while (low <= high) {
long mid = low + (high - low) / 2;
long result = (long) Math.pow(x, (int)(Math.log(mid) / Math.log(x)));
if (result == y) return true;
if (result < y) low = mid + 1;
else high = mid - 1;
}
return false;
}
public static void main(String[] args) {
System.out.println(isPower(10, 1));
System.out.println(isPower(1, 20));
System.out.println(isPower(2, 128));
System.out.println(isPower(2, 30));
}
}
import math
def isPower(x, y):
if x == 1:
return y == 1
if y == 1:
return 1
pow = x
while pow < y:
pow *= x
if pow == y:
return True
low, high = x, pow
while low <= high:
mid = low + (high - low) // 2
result = int(x ** int(math.log(mid) / math.log(x)))
if result == y:
return True
if result < y:
low = mid + 1
else:
high = mid - 1
return False
print("true" if isPower(10, 1) else "false")
print("true" if isPower(1, 20) else "false")
print("true" if isPower(2, 128) else "false")
print("true" if isPower(2, 30) else "false")
using System;
class Program {
public static bool IsPower(int x, long y) {
if (x == 1) return (y == 1);
if(y==1) return true;
long pow = x;
while (pow < y) {
pow *= pow;
}
if (pow == y) return true;
long low = x, high = pow;
while (low <= high) {
long mid = low + (high - low) / 2;
long result = (long)Math.Pow(x, (int)(Math.Log(mid) / Math.Log(x)));
if (result == y) return true;
if (result < y) low = mid + 1;
else high = mid - 1;
}
return false;
}
static void Main() {
Console.WriteLine(IsPower(10, 1));
Console.WriteLine(IsPower(1, 20));
Console.WriteLine(IsPower(2, 128));
Console.WriteLine(IsPower(2, 30));
}
}
function isPower(x, y) {
if (x === 1) return (y === 1);
if (y==1) return true;
let pow = x;
while (pow < y) {
pow *= pow;
}
if (pow === y) return true;
let low = x, high = pow;
while (low <= high) {
let mid = Math.floor((low + high) / 2);
let result = Math.pow(x, Math.floor(Math.log(mid) / Math.log(x)));
if (result === y) return true;
if (result < y) low = mid + 1;
else high = mid - 1;
}
return false;
}
console.log(isPower(10, 1));
console.log(isPower(1, 20));
console.log(isPower(2, 128));
console.log(isPower(2, 30));
Outputtrue
false
true
false
Time Complexity - O(log log y)
Auxiliary Space - O(1)
[Expected Approach] Logarithmic Method
This approach checks if y is a power of x using logarithms. By applying the logarithm change of base formula, it computes logy(x)=log(y)/log(x). If the result is an integer, it means that y is an exact power of x, and the function returns true. Otherwise, it returns false.
C++
#include <iostream>
#include <math.h>
using namespace std;
bool isPower(int x, int y)
{
float res1 = log(y) / log(x);
return res1 == floor(res1);
}
int main()
{
cout << boolalpha;
cout << isPower(2, 128) << endl;
return 0;
}
#include <stdio.h>
#include <math.h>
#include <stdbool.h>
bool isPower(int x, int y)
{
double res1 = log(y) / log(x);
return res1 == floor(res1);
}
int main()
{
printf("%s\n", isPower(2, 128) ? "true" : "false");
return 0;
}
import java.lang.Math;
public class Main {
public static boolean isPower(int x, int y) {
double res1 = Math.log(y) / Math.log(x);
return res1 == Math.floor(res1);
}
public static void main(String[] args) {
System.out.println(isPower(2, 128));
}
}
import math
def isPower(x, y):
res1 = math.log(y) / math.log(x)
return res1 == math.floor(res1)
print(isPower(2, 128))
using System;
class Program {
static bool IsPower(int x, int y) {
double res1 = Math.Log(y) / Math.Log(x);
return res1 == Math.Floor(res1);
}
static void Main() {
Console.WriteLine(IsPower(2, 128));
}
}
function isPower(x, y) {
let res1 = Math.log(y) / Math.log(x);
return res1 === Math.floor(res1);
}
console.log(isPower(2, 128));
Time complexity: O(1)
Auxiliary space: O(1)
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