This code defines a function within a `Solution` class that...

This code defines a function within a Solution class that calculates ( x^n ) (x raised to the power of n) using a recursive approach. Here's a step-by-step explanation of what it does:

Breakdown:

1. Base cases:

   • If x == 1 or n == 0, the result is 1 because any number raised to the power of 0 is 1, and 1 raised to any power is always 1.
   • If n == 1, the result is x because any number raised to the power of 1 is itself.

2. Handling negative powers:

   • When n < 0, the function recognizes this and calculates ( x^n ) as ( 1 / (x^n) ). This corresponds to the mathematical property that ( x^{-n} = \frac{1}{x^n} ).

3. Handling even powers:

   • If n % 2 == 0 (i.e., n is even), the function exploits the property of exponents ( x^n = (x^{n/2})^2 ). Instead of directly computing ( x^n ), the function recursively calls itself with n / 2, which reduces the computational complexity.

4. Handling odd powers:

   • If none of the other conditions apply, it means n is odd. The function uses the property ( x^n = x \times x^{n-1} ) to recursively compute the result by reducing the problem to calculating ( x^{n-1} ).

Concerns:

• Infinite recursion for negative powers: The code for handling negative powers (return 1 / myPow(x, n);) is incorrect and causes infinite recursion because it doesn't update the n value in the recursive call (it keeps passing the negative n). It should instead convert n to positive by passing -n.

Fixed Code:

You can fix the infinite recursion issue for negative n like this:

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

Key Properties Used:

1. ( x^0 = 1 )
2. ( x^n = x \times x^{n-1} ) (for odd n)
3. ( x^n = (x^{n/2}) \times (x^{n/2}) ) (for even n)

Computational Efficiency:

This implementation uses recursion and halves the problem size for even powers, leading to a time complexity of ( O(\log n) ), making it efficient for computing powers.