Given a positive integer n, find if it can be expressed as xy where y > 1 and x > 0. x and y both are integers.
Examples :
Input: n = 8 Output: true 8 can be expressed as 23 Input: n = 49 Output: true 49 can be expressed as 72 Input: n = 48 Output: false 48 can't be expressed as xy
The idea is simple to try all numbers x starting from 2 to square root of n (given number). For every x, try x^y where y starts from 2 and increases one by one until either x^y becomes n or greater than n.
Below is the implementation of the above idea.
Try It Yourself
// C++ program to check if a given number can be expressed
// as power
#include <bits/stdc++.h>
using namespace std;
// Returns true if n can be written as x^y
bool isPower(unsigned n)
{
if (n==1) return true;
// Try all numbers from 2 to sqrt(n) as base
for (int x=2; x<=sqrt(n); x++)
{
unsigned y = 2;
unsigned p = pow(x, y);
// Keep increasing y while power 'p' is smaller
// than n.
while (p<=n && p>0)
{
if (p==n)
return true;
y++;
p = pow(x, y);
}
}
return false;
}
// Driver Program
int main()
{
for (int i =2; i<100; i++)
if (isPower(i))
cout << i << " ";
return 0;
}
// Java code to check if a number can be expressed
// as x^y (x raised to power y)
class GFG {
// Returns true if n can be written as x^y
static boolean isPower(int n)
{
for (int x = 2; x <= Math.sqrt(n); x++) {
int y = 2;
double p = Math.pow(x, y);
while (p <= n && p > 0) {
if (p == n)
return true;
y++;
p = Math.pow(x, y);
}
}
return false;
}
// Driver function
public static void main(String[] args)
{
for (int i = 2; i < 100; i++)
if (isPower(i))
System.out.print(i + " ");
}
}
// This code is submitted by Kamal Rawal
# Python3 program to check if
# a given number can be expressed
# as power
import math
# Returns true if n can be written as x^y
def isPower(n) :
if (n==1) :
return True
# Try all numbers from 2 to sqrt(n) as base
for x in range(2,(int)(math.sqrt(n))+1) :
y = 2
p = (int)(math.pow(x, y))
# Keep increasing y while power 'p' is smaller
# than n.
while (p<=n and p>0) :
if (p==n) :
return True
y = y + 1
p = math.pow(x, y)
return False
# Driver Program
for i in range(2,100 ) :
if (isPower(i)) :
print(i,end=" ")
# This code is contributed by Nikita Tiwari.
// C# code to check if a number
// can be expressed as x^y
// (x raised to power y)
using System;
class GFG
{
// Returns true if n can
// be written as x^y
static bool isPower(int n)
{
for (int x = 2; x <= Math.Sqrt(n); x++)
{
int y = 2;
double p = Math.Pow(x, y);
while (p <= n && p > 0)
{
if (p == n)
return true;
y++;
p = Math.Pow(x, y);
}
}
return false;
}
// Driver Code
static public void Main ()
{
for (int i = 2; i < 100; i++)
if (isPower(i))
Console.Write(i + " ");
}
}
// This code is submitted by ajit.
<?php
// PHP program to check if a given
// number can be expressed as power
// Returns true if n can
// be written as x^y
function isPower($n)
{
if ($n == 1) return true;
// Try all numbers from 2
// to sqrt(n) as base
for ($x = 2; $x <= sqrt($n); $x++)
{
$y = 2;
$p = pow($x, $y);
// Keep increasing y while
// power 'p' is smaller than n.
while ($p <= $n && $p > 0)
{
if ($p == $n)
return true;
$y++;
$p = pow($x, $y);
}
}
return false;
}
// Driver Code
for ($i = 2; $i < 100; $i++)
if (isPower($i))
echo $i , " ";
// This code is contributed by aj_36
?>
<script>
// javascript code to check if a number can be expressed
// as x^y (x raised to power y)
// Returns true if n can be written as x^y
function isPower(n)
{
for (x = 2; x <= Math.sqrt(n); x++) {
var y = 2;
var p = Math.pow(x, y);
while (p <= n && p > 0) {
if (p == n)
return true;
y++;
p = Math.pow(x, y);
}
}
return false;
}
// Driver function
for (i = 2; i < 100; i++)
if (isPower(i))
document.write(i + " ");
// This code is contributed by gauravrajput1
</script>
Output :
4 8 9 16 25 27 32 36 49 64 81
Time Complexity: O(100 * n*log n) //as inbuilt sqrt function takes log n time to execute
Auxiliary Space: O(1)
One optimization in the above solution is to avoid the call to pow() by multiplying p with x one by one.
// C++ program to check if a given number can be expressed
// as power
#include <bits/stdc++.h>
using namespace std;
// Returns true if n can be written as x^y
bool isPower(unsigned int n)
{
// Base case
if (n <= 1) return true;
// Try all numbers from 2 to sqrt(n) as base
for (int x=2; x<=sqrt(n); x++)
{
unsigned p = x;
// Keep multiplying p with x while is smaller
// than or equal to x
while (p <= n)
{
p *= x;
if (p == n)
return true;
}
}
return false;
}
// Driver Program
int main()
{
for (int i =2; i<100; i++)
if (isPower(i))
cout << i << " ";
return 0;
}
// Java code to check if a number can be expressed
// as x^y (x raised to power y)
class GFG {
// Returns true if n can be written as x^y
static boolean isPower(int n)
{
for (int x = 2; x <= Math.sqrt(n); x++) {
int p = x;
while (p <= n) {
p = p * x;
if (p == n)
return true;
}
}
return false;
}
// Driver function
public static void main(String[] args)
{
for (int i = 2; i < 100; i++)
if (isPower(i))
System.out.print(i + " ");
}
}
// This code is submitted by Kamal Rawal
# Python3 program to check if
# a given number can be expressed
# as power
import math
# Returns true if n can be written
# as x ^ y
def isPower(n) :
# Base case
if (n <= 1) :
return True
# Try all numbers from 2 to sqrt(n)
# as base
for x in range(2, (int)(math.sqrt(n)) + 1) :
p = x
# Keep multiplying p with x while
# is smaller than or equal to x
while (p <= n) :
p = p * x
if (p == n) :
return True
return False
# Driver Program
for i in range(2, 100) :
if (isPower(i)) :
print( i, end =" ")
# This code is contributed by Nikita Tiwari.
// C# code to check if a number
// can be expressed as x^y (x
// raised to power y)
using System;
class GFG
{
// Returns true if n can
// be written as x^y
static bool isPower(int n)
{
for (int x = 2;
x <= Math.Sqrt(n); x++)
{
int p = x;
while (p <= n)
{
p = p * x;
if (p == n)
return true;
}
}
return false;
}
// Driver Code
public static void Main()
{
for (int i = 2; i < 100; i++)
if (isPower(i))
Console.Write(i + " ");
}
}
// This code is submitted by nitin mittal.
<?php
// PHP program to check if a
// given number can be expressed
// as power
// Returns true if n can
// be written as x^y
function isPower($n)
{
// Base case
if ($n <= 1) return true;
// Try all numbers from 2
// to sqrt(n) as base
for ($x = 2; $x <= sqrt($n); $x++)
{
$p = $x;
// Keep multiplying p with
// x while is smaller
// than or equal to x
while ($p <= $n)
{
$p *= $x;
if ($p == $n)
return true;
}
}
return false;
}
// Driver Code
for ($i = 2; $i < 100; $i++)
if (isPower($i))
echo $i , " ";
// This code is contributed by ajit
?>
<script>
// Javascript code to check if a number
// can be expressed as x^y (x
// raised to power y)
// Returns true if n can
// be written as x^y
function isPower(n)
{
for (let x = 2; x <= parseInt(Math.sqrt(n), 10); x++)
{
let p = x;
while (p <= n)
{
p = p * x;
if (p == n)
return true;
}
}
return false;
}
for (let i = 2; i < 100; i++)
if (isPower(i))
document.write(i + " ");
// This code is contributed by divyeshrabadiya07.
</script>
Output:
4 8 9 16 25 27 32 36 49 64 81
Time Complexity: O(100 * n*log(n)), where n is the number ranging from 1 to 100
Auxiliary Space: O(1), as no extra space is required
Alternate Implementation :
// C++ program to check if a given number can be expressed
// as power
#include <bits/stdc++.h>
using namespace std;
// Returns true if n can be written as x^y
bool isPower(unsigned n)
{
float p;
if (n <= 1)
return 1;
for (int i = 2; i <= sqrt(n); i++) {
p = log2(n) / log2(i);
if ((ceil(p) == floor(p)) && p > 1)
return true;
}
return false;
}
// Driver Program
int main()
{
for (int i = 2; i < 100; i++)
if (isPower(i))
cout << i << " ";
return 0;
}
// Java program to check if a given
// number can be expressed as power
import java.io.*;
import java.lang.Math;
class GFG {
// Returns true if n can be written as x^y
static boolean isPower(int n)
{
double p;
if (n <= 1)
{
return true;
}
for(int i = 2; i <= Math.sqrt(n); i++)
{
p = Math.log(n) / Math.log(i);
if ((Math.ceil(p) == Math.floor(p)) && p > 1)
{
return true;
}
}
return false;
}
// Driver Code
public static void main (String[] args)
{
for(int i = 2; i < 100; i++)
{
if (isPower(i))
System.out.print(i + " ");
}
}
}
// This code is contributed by shubhamsingh10
# Python3 program to check if a given
# number can be expressed as power
import math
# Returns true if n can be
# written as x^y
def isPower(n):
p = 0
if (n <= 1):
return 1
for i in range(2, (int)(math.sqrt(n)) + 1):
p = math.log2(n) / math.log2(i)
if ((math.ceil(p) ==
math.floor(p)) and p > 1):
return 1
return 0
# Driver code
for i in range(2, 100):
if isPower(i):
print(i, end = " ")
# This code is contributed by sallagondaavinashreddy7
// C# program to check if a given
// number can be expressed as power
using System;
class GFG{
// Returns true if n can be written as x^y
static bool isPower(int n)
{
double p;
if (n <= 1)
{
return true;
}
for(int i = 2; i <= Math.Sqrt(n); i++)
{
p = Math.Log(n) / Math.Log(i);
if ((Math.Ceiling(p) == Math.Floor(p)) && p > 1)
{
return true;
}
}
return false;
}
// Driver code
static public void Main ()
{
for(int i = 2; i < 100; i++)
{
if (isPower(i))
Console.Write(i + " ");
}
}
}
// This code is contributed by shubhamsingh10
<script>
// Javascript program to check if a given
// number can be expressed as power
// Returns true if n can be written as x^y
function isPower(n)
{
let p;
if (n <= 1)
{
return true;
}
for(let i = 2; i <= parseInt(Math.sqrt(n), 10); i++)
{
p = (Math.log(n) / Math.log(i)).toFixed(14);
if ((Math.ceil(p) == Math.floor(p)) && (p > 1))
{
return true;
}
}
return false;
}
for(let i = 2; i < 100; i++)
{
if (isPower(i))
document.write(i + " ");
}
</script>
Output:
4 8 9 16 25 27 32 36 49 64 81
Time Complexity: O(100 * log n), where n is the number ranging from 1 to 100
Auxiliary Space: O(1), as no extra space is required
Efficient Solution: Check if a number can be expressed as a^b | Set 2
