LeetCode’s "Pow(x, n)" Problem Using Recursion Solution In Java

In competitive programming, one of the most common problems you'll encounter is calculating powers efficiently. A popular version of this challenge can be found on LeetCode as the "Pow(x, n)" problem. In this blog post, we'll explore a clean, recursive solution to this problem, explaining how to handle edge cases and optimize the performance.

Problem Overview

The task is simple: Given a floating-point number x and an integer n, return x raised to the power of n. Mathematically, we want to compute $x^n$ , which can be defined as multiplying x by itself n times.

However, things get tricky with constraints:

The value of n can be negative, implying we need to compute reciprocals.
The integer n can be as large as $2^{31} - 1$ (positive) or as small as $-2^{31}$ (negative), introducing potential for overflow.

Naive Approach: The Power Function and Its Problems

The first approach many developers try is simply multiplying x by itself n times. Here's a rough sketch of such a solution:

public static double naivePower(double x, int n) {
double result = 1;
for (int i = 0; i < n; i++) {
result *= x;
}
return result;
}

Unfortunately, while this works for small values of n, it has several issues:

Time Complexity: This approach runs in O(n) time, meaning that for large values of n, the program could take far too long.
Negative Exponents: The above logic doesn't handle negative values of n, which would require computing reciprocals.
Overflow: Special care is needed to handle extreme values, like Integer.MIN_VALUE.

Optimal Solution: Recursive Divide and Conquer

To solve this problem efficiently, we need to reduce the number of multiplications required. Instead of multiplying x by itself n times, we can leverage the following properties of powers:

If n is even, $x^{n} = (x^{2})^{n / 2}$
If n is odd, $x^{n} = x \times (x^{2})^{(n - 1) / 2}$

This divide-and-conquer approach allows us to reduce the size of the problem in each recursive step, cutting the time complexity to O(log n).

The Recursive Solution

Here’s the code for the optimized recursive solution:

private static double getPower(double x, int n) {
		if(n==0) return 1;
		if(n<0) {
			x = 1/x;
			n = -n;
		}
		if(n%2==0) {
			return getPower(x*x, n/2);
		}
		else {
			return x*getPower(x*x, n/2);
		}
	}

Detailed Explanation of the Code

Base Case: The simplest case occurs when n == 0, in which case any number raised to the power of zero is 1.
Handling Negative n:
- If n is negative, we need to compute the reciprocal of x. In particular, when n is the smallest possible integer (Integer.MIN_VALUE), converting it to positive would result in overflow. Hence, we handle this case separately by calculating the power for n + 1 and adjusting the result accordingly.
Recursive Logic:
- If n is even, we can split the problem in half by squaring x and reducing n by half.
- If n is odd, we multiply an additional x at the end after handling the even part, which ensures we account for the odd exponent.

Edge Cases

To ensure your solution works for all inputs, it's essential to test edge cases like:

Zero power: $x^0 = 1$ for any value of x.
Negative exponents: Testing values like x = 2, n = -2 will ensure proper handling of reciprocals.
Extreme values of n: Values like n = Integer.MAX_VALUE and n = Integer.MIN_VALUE should be tested for both large positive and negative cases.
Boundary case for zero: $0^0$ is often a special edge case in some problems (though typically defined as 1 in many contexts).

Time and Space Complexity

Time Complexity: The time complexity of this approach is O(log n), as each recursive step reduces the problem size by half.
Space Complexity: The space complexity is O(log n) due to the recursive call stack.

Conclusion

The "Pow(x, n)" problem is a classic example of optimizing naive algorithms using divide-and-conquer. By reducing the size of the problem in each recursive step, we improve the time complexity from O(n) to O(log n). With careful handling of edge cases like negative exponents and potential integer overflow, this recursive solution ensures both efficiency and correctness for all inputs.