The given code implements the power function `myPow(x, n)`, which...

The given code implements the power function myPow(x, n), which calculates (x^n) (x raised to the power of n) using recursion. However, the issue arises because of an edge case when n is a very large negative number. Let’s break down the problem and understand the issue in detail.

What the code does:

1. If (x = 1) or (n = 0), the result is 1. This is because any number raised to the power of 0 is 1, and (1^n = 1) for any (n).
2. If (n = 1), the function simply returns (x) because (x^1 = x).
3. If (n < 0), it calculates the power as (1 / myPow(x, -n)). This handles the case where the exponent is negative (e.g., (x^{-n} = 1 / x^n)).
4. When (n) is even, the function recursively calls itself with (myPow(x, n / 2)). This is based on the property ((x^n = (x^{n/2})^2)) to optimize the calculation.
5. If (n) is odd, the function recursively calls itself with (x \times myPow(x, n - 1)), which handles the property (x^n = x \times x^{n-1}).

The problem:

The issue occurs when (n < 0), specifically when (n) is a very large negative number. In that case, the statement return 1 / myPow(x, -n) causes an integer overflow because (-n) exceeds the range of permissible integers. For example, if (n = -2^{31} (-2147483648)), then (-n = 2147483648), which is out of range for a 32-bit signed integer. This overflow results in undefined behavior and produces an incorrect output.

Why your specific test case fails:

When x = 2.00000 and n = 10, the output should be (2^{10} = 1024.00000), which is correct. However, when testing edge cases (e.g., when n is a very large negative integer), the function will fail to compute correctly due to integer overflow.

The fix:

To fix the issue, you should handle the case where (n < 0) more carefully. Instead of converting (-n) directly (which causes overflow), you can use long to store the value of (|n|) to prevent overflow. Here’s the corrected code:

class Solution {
public:
    double myPow(double x, int n) {
        if (x == 1 || n == 0) {
            return 1.0;
        }

        long long N = n; // Convert n to long long to avoid overflow
        if (N < 0) {
            x = 1 / x;
            N = -N;
        }

        double result = 1.0;
        while (N > 0) {
            if (N % 2 == 1) {
                result *= x;
            }
            x *= x;
            N /= 2;
        }

        return result;
    }
};

What this fix does:

1. Converts n to a long long type (N) to prevent overflow when handling large negative values.
2. Handles the negative exponent by taking the reciprocal of x ((1 / x)) and converting n to its absolute value using N = -N.
3. Implements an iterative approach (binary exponentiation) for better efficiency and to avoid potential stack overflow issues with deep recursion.

Now, the code will correctly handle large negative exponents and give the expected output without encountering overflow errors. For your specific test case x = 2.00000 and n = 10, the output will correctly be 1024.00000.