This code defines a class `Solution` with a method `myPow`...

This code defines a class Solution with a method myPow that calculates the result of raising a number x to the power of an integer n. Here's a detailed breakdown of its functionality:

Code Functionality:

1. Base Cases:

   • If x is 1 or n is 0, the result is 1 (since any number to the power of 0 or 1 raised to any power is 1).

   if(x==1 || n==0){
       return 1;
   }
   

2. Handle the Case When n == 1:

   • If n == 1, the result is simply x.

   if(n==1){
       return x;
   }
   

3. Handle Negative Powers (n < 0):

   • If n is negative, the power can be rewritten as ( \frac{1}{x^{-n}} ). To prevent infinite recursion when n is equal to INT_MIN (the smallest possible integer value), the code handles this special case:
     • ( \text{myPow}(x, \text{INT_MIN}) = \frac{1}{\text{myPow}(x, \text{INT_MAX}) \cdot x} ).

   if(n<0){
       if (n == INT_MIN) {
           return 1 / (myPow(x, INT_MAX) * x);
       }
       return 1 / myPow(x, -n);
   }
   

4. Efficient Exponentiation Using Recursion:

   • The code uses exponentiation by squaring:
     1. If n is even, it reduces the problem by calculating myPow(x, n/2) recursively.

     if(n%2==0){
         return myPow(x,n/2);
     }
     

     2. If n is odd, it reduces the problem by calculating ( x \cdot \text{myPow}(x, n-1) ).

     else{
         return x*myPow(x,n-1);
     }
     

The Problem with x = 0.00001, n = 2147483647 (TLE Issue):

The code becomes inefficient for very large values of n (like 2147483647), leading to a Time Limit Exceeded (TLE) error. This happens because the recursive calls stack up and are not optimized for such cases. Specifically:

1. The logic to compute the result involves excessive recursion due to the naive handling of n.
2. Although the logic divides n in half for even powers, it still incurs a large recursion depth proportional to the size of n.

Optimizations to Fix TLE:

To resolve the inefficiency, you can use an iterative version or ensure the recursive algorithm handles exponentiation by squaring more effectively:

• Combine the calculations for even and odd powers more explicitly.
• Use an iterative approach to avoid recursion overhead.

For example, an iterative approach might look like this:

double myPowIterative(double x, int n) {
    long long power = n;
    if (power < 0) {
        power = -power;
        x = 1 / x;
    }
    double result = 1;
    while (power > 0) {
        if (power % 2 == 1) {
            result *= x;
        }
        x *= x;
        power /= 2;
    }
    return result;
}

This approach works efficiently for very large values of n, avoiding excessive recursion depth and preventing TLE.