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.

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.

#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;
}

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.

/* 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;
}


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

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