The provided code is an implementation of the `pow(x, n)`...

The provided code is an implementation of the pow(x, n) function, which raises a number x to the power of n. Let me clarify what the code does step by step:

1. Base Cases:

   • If the base x is 1 or the exponent n is 0, the result is 1 (because any number to the power of 0 is 1, and 1 raised to any power remains 1).
   • If the exponent n is 1, the result is x (base case for recursion).

2. Handling Negative Exponents:
   If n is negative, it calculates the reciprocal by inverting x and flipping the sign of n:

   return 1 / myPow(x, -n);
   

   However, a special condition is added for n == INT_MIN because -INT_MIN cannot be directly represented as a positive integer due to integer overflow. In this case, it calculates 1 / (myPow(x, INT_MAX) * x).

3. Optimization for Even and Odd Exponents:

   • If n is even, the problem is broken into two smaller subproblems by halving the exponent:

     return myPow(x, n / 2) * myPow(x, n / 2);
     

     (This is simplified further in the improved version below.)
   • If n is odd, the function calculates the result recursively by reducing n by 1 and multiplying by x:

     return x * myPow(x, n - 1);
     

Problem with the Code:

The issue in your implementation is Time Limit Exceeded (TLE) for the case where x = 0.00001 and n = 2147483647. The reason is that the recursive implementation has a time complexity of O(n) in the worst case, especially for large n, as it computes the result repeatedly for subproblems.

Solution to Optimize the Code:

Instead of solving the problem in O(n) time, we can use Exponentiation by Squaring, which reduces the time complexity to O(log n). By doing so, you avoid redundant recursive calls and massively speed up the computation.

Here is your updated and optimized code:

class Solution {
public:
    double myPow(double x, int n) {
        if (x == 1 || n == 0) {
            return 1; // Base case: x^0 = 1, or 1^n = 1
        }

        if (n < 0) {
            if (n == INT_MIN) {
                return 1 / (myPow(x, INT_MAX) * x); // Handle INT_MIN edge case
            }
            return 1 / myPow(x, -n); // Convert to positive exponent
        }

        double result = 1;
        while (n > 0) {
            if (n % 2 == 1) { // If n is odd
                result *= x;
            }
            x *= x; // Square the base
            n /= 2; // Halve the exponent
        }
        return result;
    }
};

What Changed?

1. Removed Redundant Recursion and Used Iteration:
   The recursive calls were replaced with a loop to reduce function call overhead and directly implement Exponentiation by Squaring.

2. Exponentiation by Squaring Logic:

   • If n is odd, multiply the current result by x.
   • Square the base (x *= x) and halve the exponent (n /= 2).

3. Time Complexity:
   The new implementation has O(log n) time complexity, as the exponent is halved in each iteration.

4. Space Complexity:
   The space complexity is now O(1) because all computations are done iteratively without additional stack space.

This optimization ensures that cases like x = 0.00001 and n = 2147483647 run efficiently without TLE.