Pow(x, n)

Problem Statement: Implement pow(x, n), which calculates x raised to the power n (i.e., x^n).

1. Clarify the problem:

• Can the input values be negative?
• Can the input values be floating-point numbers?

2. Analyze the problem:

• Input: Floating-point number x, integer n
• Output: Floating-point number representing x raised to the power n
• Constraints:
  • -100.0 < x < 100.0
  • -2^31 <= n <= 2^31-1

3. Design an algorithm:

• We can solve this problem using a recursive approach.
• If the exponent n is 0, we return 1.
• If the exponent n is negative, we calculate 1 divided by x raised to the power -n.
• If the exponent n is positive, we divide the problem into subproblems:
  • If n is even, we calculate x raised to the power n/2 and square the result.
  • If n is odd, we calculate x raised to the power n-1 recursively and multiply it by x.

4. Explain your approach:

• We will use a recursive approach to calculate the power of x.
• We divide the problem into smaller subproblems by recursively calculating x raised to the power n/2.
• Based on the parity of n, we perform different operations to obtain the final result.
• This approach reduces the number of multiplications required to calculate the power of x.

5. Write clean and readable code:

python

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n == 0:
            return 1.0
        
        if n < 0:
            x = 1.0 / x
            n = -n
        
        return self.powHelper(x, n)
    
    def powHelper(self, x: float, n: int) -> float:
        if n == 1:
            return x
        
        if n % 2 == 0:
            subproblem = self.powHelper(x, n // 2)
            return subproblem * subproblem
        else:
            subproblem = self.powHelper(x, n - 1)
            return x * subproblem

6. Test your code: I will test the code with the following test cases:

• Test case 1: x = 2.0, n = 10
• Test case 2: x = 2.1, n = 3
• Test case 3: x = 2.0, n = -2

python

solution = Solution()
print(solution.myPow(2.0, 10))  # Expected output: 1024.0
print(solution.myPow(2.1, 3))  # Expected output: 9.261
print(solution.myPow(2.0, -2))  # Expected output: 0.25

7. Optimize if necessary: The provided algorithm is already quite efficient with a time complexity of O(log n) since we divide the problem into subproblems of size n/2 at each step.

8. Handle error cases: The code handles the following error cases:

• If the exponent n is 0, it returns 1.0.
• If the exponent n is negative, it calculates 1 divided by x raised to the power -n.
• The code does not handle invalid inputs for x since the problem statement specifies the input range.

9. Discuss complexity analysis:

• The time complexity of the algorithm is O(log n) since we divide the problem into subproblems of size n/2 at each step.
• The space complexity is O(log n) due to the recursive calls, as the maximum depth of the recursion is log n.
• The algorithm efficiently calculates the power of x by dividing the problem into smaller subproblems, resulting in a relatively low time complexity.