50. Pow(x, n) - Detailed Explanation
Problem Statement
Description:
Implement a function to calculate x raised to the power n, i.e., compute xⁿ. Note that n can be negative, and the result should be returned as a double.
Example 1:
- Input: x = 2.00000, n = 10
- Output: 1024.00000
- Explanation: 2¹⁰ = 1024
Example 2:
- Input: x = 2.10000, n = 3
- Output: 9.26100
- Explanation: 2.1³ ≈ 9.261
Example 3:
- Input: x = 2.00000, n = -2
- Output: 0.25000
- Explanation: 2⁻² = 1/(2²) = 0.25
Constraints:
- -100.0 < x < 100.0
- -2³¹ ≤ n ≤ 2³¹ - 1
- n is an integer
- xⁿ will be in the range of a double
Hints and Key Ideas
-
Handling Negative Exponents:
If n is negative, compute the power for the positive exponent and then take the reciprocal (i.e., result = 1 / (x^(-n))). -
Brute Force vs. Fast Exponentiation:
A simple solution might multiply x by itself n times, but when n is large, this approach becomes inefficient. Instead, using the divide-and-conquer technique (also known as exponentiation by squaring) reduces the number of multiplications significantly.
Approaches
Approach 1: Brute Force Multiplication
Explanation:
The brute force approach simply multiplies x by itself n times.
- For Positive n: Loop n times and multiply.
- For Negative n: Compute x^|n| and return its reciprocal.
- Time Complexity: O(n)
- Space Complexity: O(1)
Python Code (Brute Force):
Java Code (Brute Force):
Approach 2: Fast Exponentiation (Divide and Conquer)
Explanation:
The fast exponentiation method (or exponentiation by squaring) reduces the number of multiplications by using the following logic:
- If n is 0: Return 1.
- Recursively compute:
- Let half = x^(n//2).
- If n is even, then xⁿ = half * half.
- If n is odd, then xⁿ = half * half * x.
- For negative n: Convert the problem into computing x^|n| and then return the reciprocal.
- Time Complexity: O(log n)
- Space Complexity: O(log n) due to recursion stack (or O(1) in an iterative version)
Python Code (Fast Exponentiation):
Java Code (Fast Exponentiation):
Final Thoughts
Edge Cases:
- n = 0: Any number raised to the power 0 is 1.
- Negative Exponents: The method should correctly handle negative exponents by computing the positive power and taking the reciprocal.
- Large Exponents: The fast exponentiation approach efficiently handles large values of n using O(log n) multiplications.
Common Pitfalls:
- Overflow: In languages like Java, converting n to a long when taking the absolute value can avoid overflow issues when n is Integer.MIN_VALUE.
- Precision: Floating-point arithmetic might lead to precision issues, so be cautious in scenarios requiring very high precision.
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