LeetCode Problem: Implement pow(x, n) leetcode:- 50

Understanding the Fast Exponentiation Algorithm: Exponentiation by Squaring

In this blog post, we will explore the implementation of the myPow function, a common problem found on LeetCode. The goal of this problem is to compute \(x^n\) efficiently, even when \(n\) is a very large number. Instead of using a naive approach, we employ a method called Exponentiation by Squaring to significantly reduce the number of multiplications required.

Problem Overview

Given a floating-point number x and an integer n, our task is to implement a function that computes \(x^n\). A straightforward solution might involve multiplying x by itself n times, but this approach is too slow when n is large.

Intuition Behind Exponentiation by Squaring

The main idea is to reduce the problem size by halving the exponent at each recursive step. Consider the following:

  • If n is even, then:
    \(x^n = x^{n/2} \times x^{n/2}\)
  • If n is odd, then:
    \(x^n = x^{n//2} \times x^{n//2} \times x\)

By breaking down the exponent in this manner, we drastically reduce the number of multiplications from \(O(n)\) in the naive approach to \(O(\log n)\) in this optimized version.

Detailed Code Explanation

Below is the complete Python solution: 

class Solution:
    def myPow(self, x: float, n: int) -> float:
        def power(x, n):
            if x == 0:
                return 0
            if n == 0:
                return 1
            res = power(x, n // 2)
            res = res * res
            if n % 2 == 1:
                return res * x
            return res
        
        ans = power(x, abs(n))
        if n >= 0:
            return ans
        else:
            return 1 / ans

Let's break down the code:

  1. The Recursive Helper Function: The power function handles the main logic. It has two base cases:
    • If x is 0, it returns 0 immediately since \(0^n = 0\) for \(n > 0\).
    • If n is 0, it returns 1 because any number raised to the power 0 is 1.
    Then, the function computes the power recursively by halving the exponent.
  2. Combining the Results: After obtaining the result from the recursive call (res = power(x, n // 2)), the code squares the result. If the exponent n is odd, it multiplies the result by x once more.
  3. Handling Negative Exponents: Since the recursion only handles non-negative exponents efficiently, the main function calls power(x, abs(n)) and then adjusts the final result. If n is negative, it returns the reciprocal of the computed power.

Complexity Analysis

  • Time Complexity: \(O(\log n)\). The algorithm reduces the problem size by half at each recursive call, leading to a logarithmic number of steps.
  • Space Complexity: \(O(\log n)\). The recursion depth is proportional to the logarithm of n, which is used in the call stack.

Conclusion

Exponentiation by Squaring is a powerful technique for computing large exponents efficiently. The solution provided not only meets the problem requirements but also demonstrates an elegant and optimal way of thinking about recursive solutions. This approach is widely used in many areas of computer science and mathematics due to its efficiency and simplicity.

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