50 Pow(x, n)

Here, we define a few base-cases that pertain to the rule of math. But the brute force approach to reduce n by 1 every function call and calculate the power gives a TLE.

In this case, the optimization is to use math where ever we can. So, once we reach an even power, instead of decrementing n by 1, we half the value of n, and multiply x by itself. This way, by halving n at every function call, this linear input set is reduced to logarithmic input space.

Fun fact: just returning x**n works, lol

Code:

Python3 (brute force)

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

Python3 (optimized)

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n == 0: return 1
 
				if n < 0: return 1 / self.myPow(x, -n)
 
				if n % 2 != 0: return self.myPow(x*x, n/2)  # half the input space in half
 
        return x * self.myPow(x, n-1)

Big O Analysis

• Runtime The runtime complexity here is $O(log(N))$. This is because once we reach even power ($n\%2==0$), we decrement $n$ by half and multiply x by itself in that particular function call. Since, we half the input n every call, it’s logarithmic optimization.
• Memory The memory usage is $O(1)$ , constant space.

— A

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