Pow(x, n) Python Solution with Tests

LeetCode 50, Medium. Topics: Math, Recursion. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 16 parametrized pytest cases, and a playground notebook:

lcpy gen -n 50   # by problem number
lcpy gen -s powx_n   # by problem name

Problem

Implement pow(x, n), which calculates x raised to the power n (i.e., x<sup>n</sup>).

Examples

Input: x = 2.00000, n = 10
Output: 1024.00000

Input: x = 2.10000, n = 3
Output: 9.26100

Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2^-2 = 1/2^2 = 1/4 = 0.25

Constraints

-100.0 < x < 100.0
-2^31 <= n <= 2^31 - 1
n is an integer.
Either x is not zero or n > 0.
-10^4 <= x^n <= 10^4

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(log n)
    # Space: O(log n)
    def my_pow(self, x: float, n: int) -> float:
        if n == 0:
            return 1.0
        if n < 0:
            return 1.0 / self.my_pow(x, -n)
        if n % 2 == 0:
            half = self.my_pow(x, n // 2)
            return half * half
        return x * self.my_pow(x, n - 1)

Complexity

Time	Space
O(log n)	O(log n)

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Pow(x, n) Python Solution with Tests

Problem

Examples

Constraints

Solution

Complexity

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