Write a program to calculate pow(x,n)

Given two integers x and n, write a function to compute xⁿ. We may assume that x and n are small and overflow doesn’t happen.

Examples :

Input : x = 2, n = 3
Output : 8

Input : x = 7, n = 2
Output : 49

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Below solution divides the problem into subproblems of size y/2 and call the subproblems recursively.

C++

// C++ program to calculate pow(x,n) 
#include<iostream> 
using namespace std; 
class gfg 
{ 
      
/* Function to calculate x raised to the power y */
public: 
int power(int x, unsigned int y) 
{ 
    if (y == 0) 
        return 1; 
    else if (y % 2 == 0) 
        return power(x, y / 2) * power(x, y / 2); 
    else
        return x * power(x, y / 2) * power(x, y / 2); 
} 
}; 
  
/* Driver code */
int main() 
{ 
    gfg g; 
    int x = 2; 
    unsigned int y = 3; 
  
    cout << g.power(x, y); 
    return 0; 
} 
  
// This code is contributed by SoM15242 

C

#include<stdio.h> 
  
/* Function to calculate x raised to the power y */
int power(int x, unsigned int y) 
{ 
    if (y == 0) 
        return 1; 
    else if (y%2 == 0) 
        return power(x, y/2)*power(x, y/2); 
    else
        return x*power(x, y/2)*power(x, y/2); 
} 
  
/* Program to test function power */
int main() 
{ 
    int x = 2; 
    unsigned int y = 3; 
  
    printf("%d", power(x, y)); 
    return 0; 
} 

Java

class GFG { 
    /* Function to calculate x raised to the power y */
    static int power(int x, int y) 
    { 
        if (y == 0) 
            return 1; 
        else if (y % 2 == 0) 
            return power(x, y / 2) * power(x, y / 2); 
        else
            return x * power(x, y / 2) * power(x, y / 2); 
    } 
  
    /* Program to test function power */
    public static void main(String[] args) 
    { 
        int x = 2; 
        int y = 3; 
  
        System.out.printf("%d", power(x, y)); 
    } 
} 
  
// This code is contributed by Smitha Dinesh Semwal 

Python3

# Python3 program to calculate pow(x,n) 
  
# Function to calculate x 
# raised to the power y  
def power(x, y): 
  
    if (y == 0): return 1
    elif (int(y % 2) == 0): 
        return (power(x, int(y / 2)) *
               power(x, int(y / 2))) 
    else: 
        return (x * power(x, int(y / 2)) *
                   power(x, int(y / 2))) 
  
# Driver Code 
x = 2; y = 3
print(power(x, y)) 
  
# This code is contributed by Smitha Dinesh Semwal. 

C#

using System; 
  
public class GFG { 
      
    // Function to calculate x raised to the power y 
    static int power(int x, int y) 
    { 
        if (y == 0) 
            return 1; 
        else if (y % 2 == 0) 
            return power(x, y / 2) * power(x, y / 2); 
        else
            return x * power(x, y / 2) * power(x, y / 2); 
    } 
  
    // Program to test function power 
    public static void Main() 
    { 
        int x = 2; 
        int y = 3; 
  
        Console.Write(power(x, y)); 
    } 
} 
  
// This code is contributed by shiv_bhakt. 

PHP

<?php 
// Function to calculate x  
// raised to the power y 
function power($x, $y) 
{ 
    if ($y == 0) 
        return 1; 
    else if ($y % 2 == 0) 
        return power($x, (int)$y / 2) *  
               power($x, (int)$y / 2); 
    else
        return $x * power($x, (int)$y / 2) *  
                    power($x, (int)$y / 2); 
} 
  
// Driver Code 
$x = 2; 
$y = 3; 
  
echo power($x, $y); 
      
// This code is contributed by ajit 
?> 

Output :

Time Complexity: O(n)
Space Complexity: O(1)
Algorithmic Paradigm: Divide and conquer.

Above function can be optimized to O(logn) by calculating power(x, y/2) only once and storing it.

/* Function to calculate x raised to the power y in O(logn)*/
int power(int x, unsigned int y) 
{ 
    int temp; 
    if( y == 0) 
        return 1; 
    temp = power(x, y/2); 
    if (y%2 == 0) 
        return temp*temp; 
    else
        return x*temp*temp; 
} 

Time Complexity of optimized solution: O(logn)
Let us extend the pow function to work for negative y and float x.

C++

/* Extended version of power function  
that can work for float x and negative y*/
#include <bits/stdc++.h> 
using namespace std; 
  
float power(float x, int y)  
{  
    float temp;  
    if(y == 0)  
        return 1;  
    temp = power(x, y / 2);  
    if (y % 2 == 0)  
        return temp * temp;  
    else
    {  
        if(y > 0)  
            return x * temp * temp;  
        else
            return (temp * temp) / x;  
    }  
}  
  
// Driver Code 
int main()  
{  
    float x = 2;  
    int y = -3;  
    cout << power(x, y);  
    return 0;  
}  
  
// This is code is contributed  
// by rathbhupendra 

C

/* Extended version of power function that can work 
 for float x and negative y*/
#include<stdio.h> 
  
float power(float x, int y) 
{ 
    float temp; 
    if( y == 0) 
       return 1; 
    temp = power(x, y/2);        
    if (y%2 == 0) 
        return temp*temp; 
    else
    { 
        if(y > 0) 
            return x*temp*temp; 
        else
            return (temp*temp)/x; 
    } 
}   
  
/* Program to test function power */
int main() 
{ 
    float x = 2; 
    int y = -3; 
    printf("%f", power(x, y)); 
    return 0; 
} 

Java

/* Java code for extended version of power function 
that can work for float x and negative y */
class GFG { 
      
    static float power(float x, int y) 
    { 
        float temp; 
        if( y == 0) 
            return 1; 
        temp = power(x, y/2);  
          
        if (y%2 == 0) 
            return temp*temp; 
        else
        { 
            if(y > 0) 
                return x * temp * temp; 
            else
                return (temp * temp) / x; 
        } 
    }  
      
    /* Program to test function power */
    public static void main(String[] args) 
    { 
        float x = 2; 
        int y = -3; 
        System.out.printf("%f", power(x, y)); 
    } 
} 
  
// This code is contributed by  Smitha Dinesh Semwal. 

Python3

# Python3 code for extended version 
# of power function that can work 
# for float x and negative y 
  
def power(x, y): 
  
    if(y == 0): return 1
    temp = power(x, int(y / 2))  
      
    if (y % 2 == 0): 
        return temp * temp 
    else: 
        if(y > 0): return x * temp * temp 
        else: return (temp * temp) / x 
      
# Driver Code 
x, y = 2, -3
print('%.6f' %(power(x, y))) 
  
# This code is contributed by Smitha Dinesh Semwal.  

C#

// C# code for extended version of power function 
// that can work for float x and negative y 
  
using System; 
  
public class GFG{ 
      
    static float power(float x, int y) 
    { 
        float temp; 
          
        if( y == 0) 
            return 1; 
        temp = power(x, y/2);  
          
        if (y % 2 == 0) 
            return temp * temp; 
        else
        { 
            if(y > 0) 
                return x * temp * temp; 
            else
                return (temp * temp) / x; 
        } 
    }  
      
    // Program to test function power  
    public static void Main() 
    { 
        float x = 2; 
        int y = -3; 
          
        Console.Write(power(x, y)); 
    } 
} 
  
// This code is contributed by shiv_bhakt. 

PHP

<?php 
// Extended version of power 
// function that can work 
// for float x and negative y 
  
function power($x, $y) 
{ 
    $temp; 
    if( $y == 0) 
    return 1; 
    $temp = power($x, $y / 2);      
    if ($y % 2 == 0) 
        return $temp * $temp; 
    else
    { 
        if($y > 0) 
            return $x *  
                   $temp * $temp; 
        else
            return ($temp *  
                    $temp) / $x; 
    } 
}  
  
// Driver Code 
$x = 2; 
$y = -3; 
echo power($x, $y); 
  
// This code is contributed by ajit 
?> 

Output :

0.125000

Write an iterative O(Log y) function for pow(x, y)
Modular Exponentiation (Power in Modular Arithmetic)

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

C++

C

Java

Python3

C#

PHP

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