Pow(x, n)

LeetCode: https://leetcode.com/problems/powx-n/

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

Use fast exponentiation to achieve O(log n).

1def myPow(x: float, n: int) -> float:
2    if n == 0:
3        return 1.0
4    if n < 0:
5        return 1.0 / myPow(x, -n)
6        
7    half = myPow(x, n // 2)
8    if n % 2 == 0:
9        return half * half
10    return half * half * x

Fibonacci Number

Reverse String

idx

Step 1 / 3

Step 1:

Compute myPow(2, 10). Split into half exponent: myPow(2, 5).

Pointers: idx=0

Focus: select @ [0]