Pow(x, n)

Problem Overview

The problem "Pow(x, n)" on LeetCode (Problem Number: 50) requires us to implement a function that calculates x raised to the power n, i.e., x^n. Given two numbers, a double x and an integer n, we need to find the result efficiently.

The function signature:

double myPow(double x, int n);

public double myPow(double x, int n);

Understanding the Problem

The task is to compute the power function without using direct power computations like pow(x, n). Additionally, the solution must handle both positive and negative exponents due to integer n, including the edge cases such as n being zero or extremely large.

Examples:

• myPow(2.0, 10) returns 1024.0
• myPow(2.1, 3) returns 9.261
• myPow(2.0, -2) returns 0.25

The challenge lies in efficiently handling both large values of n and negative powers, ensuring robust handling of possible integer overflow cases.

Key DSA Concepts and Theory

Exponentiation by Squaring

This problem revolves around the concept of exponentiation by squaring, which is an optimization technique to compute power of a number in logarithmic time, i.e., O(log n).

Theory:

• Basic case: x^0 = 1
• Recursive case for even n: x^n = (x^(n/2)) * (x^(n/2))
• Recursive case for odd n: x^n = x * (x^(n//2)) * (x^(n//2))

For negative exponents (n < 0), the result x^n is computed as 1/x^(-n).

Edge Cases Management

1. n = 0: Any number raised to the power of zero is 1.
2. x = 0 and n > 0: The answer is 0.
3. Negative Powers: Compute the reciprocal of the positive power.

Solution Approach

Given the understanding of exponentiation by squaring, we can break down the solution into manageable steps:

C++ Implementation

double myPow(double x, int n) {
    if (n == 0) return 1;
    double half = myPow(x, n / 2);
    if (n % 2 == 0) return half * half;
    else return n > 0 ? half * half * x : half * half / x;
}

Java Implementation

public double myPow(double x, int n) {
    if (n == 0) return 1;
    double half = myPow(x, n / 2);
    if (n % 2 == 0) return half * half;
    else return n > 0 ? half * half * x : half * half / x;
}

Step-by-Step Explanation

1. Base Case: If n is zero, return 1 as x^0 is always 1.
2. Recursive Approach: Compute x^(n/2) only once and store in half.
3. Even n: If n is even, simply return half * half.
4. Odd n: If n is odd:
   • For positive n, return half * half * x.
   • For negative n, account for the negative exponent by returning half * half / x.

Time and Space Complexity Analysis

• Time Complexity: O(log n): The power is divided by two recursively, resulting in logarithmic time complexity.
• Space Complexity: O(log n): Due to the recursive stack space used during the computation.

Common Mistakes to Avoid

1. Handling Negative Powers Incorrectly: Forgetting to account for negative powers directly using x^n = 1/x^(-n).
2. Integer Overflow: The edge case when n is Integer.MIN_VALUE in Java, due to negative bounds.

Similar Problems on LeetCode

1. Problem 372: Super Pow
2. Problem 258: Add Digits

Additional Resources and References

• Exponentiation by Squaring - Wikipedia
• LeetCode Discuss section for insights and alternative approaches from the community.
• Algorithm textbooks that discuss various power algorithms and their optimizations.

By understanding these concepts and leveraging exponentiation by squaring, we can efficiently solve the "Pow(x, n)" problem using either recursive or iterative approaches.