Start by recognizing that pow(x, n) requires handling negative exponents and can be optimized with recursion to reduce repeated multiplication. Recursively halve the exponent while multiplying results carefully, ensuring that large negative or positive n values do not overflow. This approach directly applies the Math plus Recursion pattern to achieve logarithmic time complexity while keeping the solution clean and interview-ready.
Problem Statement
Implement a function pow(x, n) that calculates x raised to the power n (x^n). The function must handle negative exponents and large n efficiently while avoiding floating-point errors.
For example, pow(2.00000, 10) should return 1024.00000, pow(2.10000, 3) should return 9.26100, and pow(2.00000, -2) should return 0.25000. Constraints include -100.0 < x < 100.0, -2^31 <= n <= 2^31-1, n being an integer, and either x not zero or n > 0.
Examples
Example 1
Input: x = 2.00000, n = 10
Output: 1024.00000
Example details omitted.
Example 2
Input: x = 2.10000, n = 3
Output: 9.26100
Example details omitted.
Example 3
Input: x = 2.00000, n = -2
Output: 0.25000
2-2 = 1/22 = 1/4 = 0.25
Constraints
- -100.0 < x < 100.0
- -231 <= n <= 231-1
- n is an integer.
- Either x is not zero or n > 0.
- -104 <= xn <= 104
Solution Approach
Recursive Exponentiation by Squaring
Use recursion to halve the exponent at each step. If n is negative, compute pow(x, -n) and return 1/result. Multiply intermediate results for even and odd exponents separately to maintain correctness.
Handle Negative Powers Carefully
Check if n is negative before recursion. Convert n to positive safely, considering integer overflow limits, then apply the recursive pattern. Return 1/result for the final output to handle negative exponents.
Optimize Base Cases and Multiplications
Define base cases for n = 0 and n = 1 to stop recursion early. Reduce redundant multiplications by reusing the result of pow(x, n//2) when n is even, ensuring logarithmic recursion depth.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(log n) due to halving the exponent at each recursive call. Space complexity is O(log n) from recursion stack usage.
What Interviewers Usually Probe
- Recognizes negative exponent handling is required.
- Uses recursion to optimize repeated multiplications.
- Considers integer overflow and floating-point precision issues.
Common Pitfalls or Variants
Common pitfalls
- Failing to handle n = INT_MIN correctly when negating.
- Recomputing pow(x, n//2) multiple times instead of storing intermediate results.
- Ignoring base cases for n = 0 or n = 1, causing unnecessary recursion.
Follow-up variants
- Implement iterative version to reduce recursion stack usage.
- Extend to handle fractional exponents with floating-point math.
- Support modular exponentiation for large integer powers under modulo constraints.
How GhostInterview Helps
- Highlights the Math plus Recursion pattern directly in solution steps.
- Guides correct handling of negative powers and edge cases automatically.
- Ensures optimized recursion to avoid timeouts and stack overflow in interviews.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the main pattern used in Pow(x, n)?
The problem uses Math plus Recursion pattern, optimizing exponentiation by halving the exponent recursively.
How do I handle negative exponents in pow(x, n)?
Convert n to positive and return 1/pow(x, -n), carefully avoiding integer overflow for n = INT_MIN.
What is the time complexity of the recursive solution?
Time complexity is O(log n) since the exponent is halved at each recursive step.
Why should pow(x, n//2) results be reused?
Reusing avoids redundant calculations and ensures logarithmic recursion depth, preventing unnecessary multiplications.
What are the base cases in Pow(x, n)?
Base cases are n = 0 returning 1 and n = 1 returning x, which stop recursion safely.
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