- The DSA Woodshed
- Algorithms
- Recursion
- Pow X N
Pow X N
Problem
Implement pow(x, n), which calculates x raised to the power n (i.e., x^n). Handle negative exponents.
Approach
Fast exponentiation (exponentiation by squaring). If n is even, pow(x, n) = pow(x*x, n//2). If n is odd, pow(x, n) = x * pow(x, n-1). For negative n, compute pow(1/x, -n).
When to Use
Any "repeated operation" that can be halved — exponentiation, matrix power, modular arithmetic. Foundation of divide-and-conquer thinking.
Complexity
| Time | O(log n) |
| Space | O(log n) — recursion stack |
Source
"""Pow(x, n) — compute x raised to the power n using fast exponentiation.
Problem:
Implement pow(x, n), which calculates x raised to the power n
(i.e., x^n). Handle negative exponents.
Approach:
Fast exponentiation (exponentiation by squaring). If n is even,
pow(x, n) = pow(x*x, n//2). If n is odd, pow(x, n) = x * pow(x, n-1).
For negative n, compute pow(1/x, -n).
When to use:
Any "repeated operation" that can be halved — exponentiation, matrix
power, modular arithmetic. Foundation of divide-and-conquer thinking.
Complexity:
Time: O(log n)
Space: O(log n) — recursion stack
"""
def my_pow(x: float, n: int) -> float:
"""Compute x raised to the power n via fast exponentiation.
>>> my_pow(2.0, 10)
1024.0
>>> my_pow(2.0, -2)
0.25
>>> my_pow(0.0, 0)
1.0
"""
if n < 0:
return my_pow(1.0 / x, -n)
if n == 0:
return 1.0
if n % 2 == 0:
return my_pow(x * x, n // 2)
return x * my_pow(x, n - 1)This page lives in git. Anyone can propose an edit. Edit this page View source