readme.md

50. Pow(x, n)

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

Example 1:

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

Example 2:

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

Example 3:

Input: x = 2.00000, n = -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 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

class Solution:
    def myPow(self, x: float, n: int) -> float:
        # Helper function to recursively compute power
        def fast_pow(base, exp):
            if exp == 0:
                return 1  # Base case: any number to the power of 0 is 1

            if exp < 0:
                # If exponent is negative, invert the base and make exponent positive
                return fast_pow(1 / base, -exp)

            half = fast_pow(base, exp // 2)

            if exp % 2 == 0:
                return half * half  # If exponent is even, use the square of half-exponent
            else:
                return half * half * base  # If odd, multiply an additional base to the result

        # Call the helper function with initial parameters
        return fast_pow(x, n)

Converting to an iterative solution makes this problem faster but still has the same time complexity

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n == 0:
            return 1
        if n < 0:
            x = 1 / x
            n = -n

        result = 1
        current_product = x
        while n > 0:
            if n % 2 == 1:  # If n is odd, multiply the current product to the result
                result *= current_product
            current_product *= current_product  # Square the base
            n //= 2  # Reduce n by half

        return result

Thoughts

Not exactly tricky, doable but edges cases are difficult

Time Complexity

O(log n) for the recursion tree / loop

Space Complexity

O(log n) for the recursion tree