Write an iterative O(Log y) function for pow(x, y)

Given an integer x and a positive number y, write a function that computes x^y under following conditions. 
a) Time complexity of the function should be O(Log y) 
b) Extra Space is O(1) 

Examples:

Input: x = 3, y = 5
Output: 243

Input: x = 2, y = 5
Output: 32
 

We strongly recommend that you click here and practice it, before moving on to the solution.

We have discussed recursive O(Log y) solution for power. The recursive solutions are generally not preferred as they require space on call stack and they involve function call overhead. 

Following is implementation to compute x^y.  

C++

// Iterative C program to implement pow(x, n)
#include <iostream>
using namespace std;

/* Iterative Function to calculate (x^y) in O(logy) */
int power(int x, unsigned int y)
{
    int res = 1; // Initialize result

    while (y > 0) {
        // If y is odd, multiply x with result
        if (y & 1)
            res = res * x;

        // y must be even now
        y = y >> 1; // y = y/2
        x = x * x; // Change x to x^2
    }
    return res;
}

// Driver program to test above functions
int main()
{
    int x = 3;
    unsigned int y = 5;

    cout<<"Power is "<<power(x, y);

    return 0;
}

// this code is contributed by shivanisinghss2110

C

// Iterative C++ program to implement pow(x, n)
#include <stdio.h>

/* Iterative Function to calculate (x^y) in O(logy) */
int power(int x, unsigned int y)
{
    int res = 1; // Initialize result

    while (y > 0) {
        // If y is odd, multiply x with result
        if (y & 1)
            res = res * x;

        // y must be even now
        y = y >> 1; // y = y/2
        x = x * x; // Change x to x^2
    }
    return res;
}

// Driver program to test above functions
int main()
{
    int x = 3;
    unsigned int y = 5;

    printf("Power is %d", power(x, y));

    return 0;
}

Java

// Iterative Java program
// to implement pow(x, n)
import java.io.*;

class GFG
{

/* Iterative Function to
calculate (x^y) in O(logy) */
static int power(int x, int y)
{
    // Initialize result
    int res = 1;

    while (y > 0)
    {
        // If y is odd,
        // multiply
        // x with result
        if ((y & 1) == 1)
            res = res * x;

        // y must be even now
        y = y >> 1; // y = y/2
        x = x * x; // Change x to x^2
    }
    return res;
}

// Driver Code
public static void main (String[] args)
{
    int x = 3;
    int y = 5;

    System.out.println("Power is " +
                        power(x, y));
}
}

// This code is contributed
// by aj_36

Python3

# Iterative Python3 program
# to implement pow(x, n)

# Iterative Function to
# calculate (x^y) in O(logy)
def power(x, y):

    # Initialize result
    res = 1

    while (y > 0):

        # If y is odd, multiply
        # x with result
        if ((y & 1) == 1) :
            res = res * x

        # y must be even
        # now y = y/2
        y = y >> 1

        # Change x to x^2
        x = x * x

    return res


# Driver Code
x = 3
y = 5

print("Power is ",
       power(x, y))

# This code is contributed
# by ihritik

C#

// Iterative C# program
// to implement pow(x, n)
using System;

class GFG
{

/* Iterative Function to
calculate (x^y) in O(logy) */
static int power(int x, int y)
{
    int res = 1; // Initialize result

    while (y > 0)
    {
        // If y is odd, multiply
        // x with result
        if ((y & 1) == 1)
            res = res * x;

        // y must be even now
        y = y >> 1; // y = y/2
        x = x * x; // Change x to x^2
    }
    return res;
}

// Driver Code
static public void Main ()
{
int x = 3;
int y = 5;

Console.WriteLine("Power is "+
                   power(x, y));
}
}

// This code is contributed
// by aj_36

PHP

<?php
// Iterative php program
// to implement pow(x, n)>

// Iterative Function to
// calculate (x^y) in O(logy)

function power($x, $y)
{

    // Initialize result
    $res = 1;

    while ($y > 0)
    {

        // If y is odd, multiply
        // x with result
        if ($y & 1)
            $res = $res * $x;

        // y must be even now

        // y = y/2
        $y = $y >> 1;

        // Change x to x^2
        $x = $x * $x;
    }
    return $res;
}

    // Driver Code
    $x = 3;
    $y = 5;

    echo "Power is ", power($x, $y);

// This code is contributed by ajit
?>

Javascript

<script>

// Iterative Javascript program to implement pow(x, n)

/* Iterative Function to calculate (x^y) in O(logy) */
function power(x, y)
{
// Initialize result
    let res = 1;

    while (y > 0) {
        // If y is odd, multiply x with result
        if (y & 1)
            res = res * x;

        // y must be even now
        y = y >> 1; // y = y/2
        x = x * x; // Change x to x^2
    }
    return res;
}

// Driver program to test above functions

    let x = 3;
    y = 5;

    document.write("Power is " + power(x, y));


// This code is contributed by Mayank Tyagi

</script>

Output

Power is 243

Time Complexity: O(log y), since in loop each time the value of y decreases by half it’s current value.
Auxiliary Space: O(1), since no extra space has been taken.

Another approach:

Step 1: Start the function with the base and exponent as input parameters.
Step 2: Check if the exponent is equal to zero, return 1.
Step 3: Recursively call the function with the base and the exponent divided by 2.
Step 4: If the exponent is even, return the square of the result obtained from the recursive call.
Step 5: If the exponent is odd, return the product of the base, the square of the result obtained from the recursive call, and the base again.

C++

#include <iostream>
using namespace std;

// Recursive Function to calculate (x^y) in O(logy)
int power(int x, int y)
{
    if (y == 0)
        return 1;

    int temp = power(x, y / 2);
    if (y % 2 == 0)
        return temp * temp;
    else
        return x * temp * temp;
}

// Driver program to test above functions
int main()
{
    int x = 3;
    int y = 5;

    cout << "Power is " << power(x, y);

    return 0;
}

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
           int x=3;
          int y=5;

          System.out.println("Power is: "+power(x,y));

    }
    public static int power(int x, int y)
    {
        if (y == 0)
        {
            return 1;
        }

        int temp = power(x, y / 2);
          // if y is even
        if (y % 2 == 0)
        {
            return temp * temp;
        }
        else
        {
            return x * temp * temp;
        }
    }
}
//This code is contributed by aeroabrar_31

Python3

#python code to demonstrate the above approach
def power(x, y):
    if y==0:
        return 1

    temp = power(x, y // 2)
    # if y is even
    if y % 2 == 0:
        return temp * temp
    else:
        return x * temp * temp


def main():
    x = 3
    y = 5

    print("Power is : ", power(x,y))


main()

C#

using System;

class GFG {
    // Recursive Function to calculate (x^y) in O(logy)
    static int power(int x, int y)
    {
        if (y == 0)
            return 1;

        int temp = power(x, y / 2);
        if (y % 2 == 0)
            return temp * temp;
        else
            return x * temp * temp;
    }

    // Driver program to test above functions
    static void Main()
    {
        int x = 3;
        int y = 5;

        Console.WriteLine("Power is " + power(x, y));
    }
}

Javascript

function power(x, y) {
    if (y == 0)
        return 1;

    let temp = power(x, Math.floor(y / 2));
    if (y % 2 == 0)
        return temp * temp;
    else
        return x * temp * temp;
}

let x = 3;
let y = 5;

console.log("Power is " + power(x, y));

Output

Power is 243 

Complexity Analysis:

The time complexity of the power function implemented using recursion is O(log n), where n is the value of the exponent.

In this implementation, the function repeatedly divides the exponent by 2 and squares the base until the exponent becomes zero. The number of iterations required to reach the base case is log₂(n), where log₂ denotes the logarithm to the base 2. Therefore, the time complexity of the function is proportional to the number of iterations, which is log₂(n).

This approach reduces the number of multiplications needed to calculate the result and improves the performance of the function compared to the naive iterative approach, which has a time complexity of O(n).

However, the space complexity of the recursive implementation is also O(log n), since each recursive call adds a new stack frame to the call stack until the base case is reached. Therefore, the space required by the function grows logarithmically with the size of the input. This can become a concern for very large values of n, as it may lead to a stack overflow error.

In summary, the recursive implementation of the power function has a time complexity of O(log n) and a space complexity of O(log n).

This article is contributed by Udit Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above