# 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
```

## 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.

```#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)

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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# GATE CS Corner    Company Wise Coding Practice

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