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.
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Examples :
Input : x = 2, n = 3 Output : 8 Input : x = 7, n = 2 Output : 49
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 = 3print(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 ?> |
Javascript
<script> // Javascript program to calculate pow(x,n) // 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, parseInt(y / 2, 10)) * power(x, parseInt(y / 2, 10)); else return x * power(x, parseInt(y / 2, 10)) * power(x, parseInt(y / 2, 10)); } // Driver code let x = 2; let y = 3; document.write(power(x, y)); // This code is contributed by mukesh07 </script> |
Output :
8
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.
C++
/* 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; } // This code is contributed by Shubhamsingh10 |
C
/* 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; } |
Java
/* Function to calculate x raised to the power y in O(logn)*/static int power(int x, 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; } // This code is contributed by divyeshrabadiya07. |
Python3
# Function to calculate x raised to the power y in O(logn) def power(x,y): temp = 0 if( y == 0): return 1 temp = power(x, int(y / 2)) if (y % 2 == 0) return temp * temp; else return x * temp * temp; # This code is contributed by avanitrachhadiya2155 |
C#
/* Function to calculate x raised to the power y in O(logn)*/static int power(int x, 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; } // This code is contributed by divyesh072019. |
Javascript
<script> /* Function to calculate x raised to the power y in O(logn)*/function power(x , y) { var temp; if( y == 0) return 1; temp = power(x, y / 2); if (y % 2 == 0) return temp*temp; else return x*temp*temp; } // This code is contributed by todaysgaurav </script> |
Time Complexity of optimized solution: O(logn)
Auxiliary Space: O(1)
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, -3print('%.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 ?> |
Javascript
<script> // Javascript code for extended // version of power function that // can work for var x and negative y function power(x, y) { var temp; if (y == 0) return 1; temp = power(x, parseInt(y / 2)); if (y % 2 == 0) return temp * temp; else { if (y > 0) return x * temp * temp; else return (temp * temp) / x; } } // Driver code var x = 2; var y = -3; document.write( power(x, y).toFixed(6)); // This code is contributed by aashish1995 </script> |
Output :
0.125000
Time Complexity: O(log|n|)
Auxiliary Space: O(1)
Using recursion:
This approach is almost similar to iterative solution
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; int power(int x, int y) { // If x^0 return 1 if (y == 0) return 1; // If we need to find of 0^y if (x == 0) return 0; // For all other cases return x * power(x, y - 1); } // Driver Code int main() { int x = 2; int y = 3; cout << (power(x, y)); } // This code is contributed by ukasp. |
Java
// Java program for the above approach import java.io.*; class GFG { public static int power(int x, int y) { // If x^0 return 1 if (y == 0) return 1; // If we need to find of 0^y if (x == 0) return 0; // For all other cases return x * power(x, y - 1); } // Driver Code public static void main(String[] args) { int x = 2; int y = 3; System.out.println(power(x, y)); } } |
Python3
# Python3 program for the above approach def power(x, y): # If x^0 return 1 if (y == 0): return 1 # If we need to find of 0^y if (x == 0): return 0 # For all other cases return x * power(x, y - 1) # Driver Code x = 2y = 3 print(power(x, y)) # This code is contributed by shivani. |
C#
// C# program for the above approach using System; class GFG{ public static int power(int x, int y) { // If x^0 return 1 if (y == 0) return 1; // If we need to find of 0^y if (x == 0) return 0; // For all other cases return x * power(x, y - 1); } // Driver Code public static void Main(String[] args) { int x = 2; int y = 3; Console.WriteLine(power(x, y)); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // javascript program for the above approach function power(x , y) { // If x^0 return 1 if (y == 0) return 1; // If we need to find of 0^y if (x == 0) return 0; // For all other cases return x * power(x, y - 1); } // Driver Code var x = 2; var y = 3; document.write(power(x, y)); // This code is contributed by Rajput-Ji </script> |
Output:
8
Time Complexity: O(n)
Auxiliary Space: O(1)
Using Math.pow() function of java:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; int power(int x, int y) { // return type of pow() // function is double return (int)pow(x, y); } // Driver Code int main() { int x = 2; int y = 3; cout << (power(x, y)); } // This code is contributed by hemantraj712. |
Java
// Java program for the above approach import java.io.*; class GFG { public static int power(int x, int y) { // Math.pow() is a function that // return floating number return (int)Math.pow(x, y); } // Driver Code public static void main(String[] args) { int x = 2; int y = 3; System.out.println(power(x, y)); } } |
Python3
# Python3 program for the above approach def power(x, y): # Return type of pow() # function is double return pow(x, y) # Driver Code x = 2y = 3 print(power(x, y)) # This code is contributed by susmitakundugoaldanga |
C#
// C# program for the above approach using System; public class GFG { public static int power(int x, int y) { // Math.pow() is a function that // return floating number return (int)Math.Pow(x, y); } // Driver code static public void Main() { int x = 2; int y = 3; Console.WriteLine(power(x, y)); } } |
Javascript
<script> // Javascript program for the above approach function power( x, y) { // Math.pow() is a function that // return floating number return parseInt(Math.pow(x, y)); } // Driver Code let x = 2; let y = 3; document.write(power(x, y)); // This code is contributed by sravan kumar </script> |
Output:
8
Time Complexity: O(1)
Auxiliary Space: O(1)
Non-Recursive Method: This is a very efficient method for novices, who feel hard to understand the iterative recursive calls. This method also takes T(n)= O(log(n)) complexity.
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; int power(int x, int y) { int result = 1; while (y > 0) { if (y % 2 == 0) // y is even { x = x * x; y = y / 2; } else // y isn't even { result = result * x; y = y - 1; } } return result; } // Driver Code int main() { int x = 2; int y = 3; cout << (power(x, y)); return 0; } // This code is contributed by Madhav Mohan |
Java
// Java program for above approach class GFG{ static int power(int x, int y) { int result = 1; while (y > 0) { // y is even if (y % 2 == 0) { x = x * x; y = y / 2; } // y isn't even else { result = result * x; y = y - 1; } } return result; } // Driver Code public static void main(String[] args) { int x = 2; int y = 3; System.out.println(power(x, y)); } } // This code is contributed by hritikrommie |
Python3
# Python program for the above approach def power( x, y): result = 1 while (y > 0): if (y % 2 == 0): # y is even x = x * x y = y / 2 else: # y isn't even result = result * x y = y - 1 return result # Driver Code x = 2y = 3print((power(x, y))) # This code is contributed by shivanisinghss2110 |
C#
// C# program for above approach using System; class GFG{ static int power(int x, int y) { int result = 1; while (y > 0) { // y is even if (y % 2 == 0) { x = x * x; y = y / 2; } // y isn't even else { result = result * x; y = y - 1; } } return result; } // Driver Code public static void Main(String[] args) { int x = 2; int y = 3; Console.Write(power(x, y)); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // Javascript program for the above approach function power(x,y) { let result = 1; while (y > 0) { if (y % 2 == 0) // y is even { x = x * x; y = Math.floor(y / 2); } else // y isn't even { result = result * x; y = y - 1; } } return result; } // Driver Code let x = 2; let y = 3; document.write(power(x, y)) // This code is contributed by rag2127 </script> |
Output
8
Time Complexity: O(log2y)
Auxiliary Space: O(1)
Write an iterative O(Log y) function for pow(x, y)
Modular Exponentiation (Power in Modular Arithmetic)
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