Implement power function using binary exponentiation

[aryaman0406]

PR Title Format: 50.Pow(x, n).cpp💡 IntuitionThe naive approach of multiplying $x$ by itself $n$ times would be $O(n)$, which is too slow for large values of $n$. We can leverage the mathematical property that $x^n = x^{n/2} \cdot x^{n/2}$. This allows us to reduce the number of multiplications significantly by halving the exponent in each step. This approach is $O(\log n)$. Special care must be taken to handle negative exponents and the edge case where the exponent $n$ is the minimum integer value (INT_MIN).✍️ ApproachThe solution uses a recursive helper function, binaryExp(x, n), which handles the exponentiation:Base Case: If the exponent $n$ is $0$, return $1$ (since $x^0 = 1$).Negative Exponent Handling: If $n$ is negative, we use the property $x^{-n} = 1 / x^n$. The initial function myPow casts $n$ to a long to prevent overflow when negating the INT_MIN value, as -INT_MIN is not representable in a standard int.Odd Exponent: If $n$ is odd, we have $x^n = x \cdot x^{n-1}$. Since $n-1$ is even, we can rewrite this as $x \cdot (x^2)^{(n-1)/2}$. We multiply by $x$ once, square the base $x$ to get $x^2$, and halve the new exponent to $(n-1)/2$.Even Exponent: If $n$ is even, we have $x^n = (x^2)^{n/2}$. We square the base $x$ to get $x^2$ and halve the exponent to $n/2$.Efficiency: The total number of recursive calls (multiplications) is proportional to $\log n$, resulting in an $O(\log n)$ time complexity.Code Solution (C++)C++class Solution {
public:
/**
* @brief Computes x raised to the power n using Binary Exponentiation.
* * @param x The base.
* @param n The exponent (integer).
* @return double The result x^n.
*/
double myPow(double x, int n) {
// Cast n to long to safely handle the INT_MIN case, where -n would overflow int.
return binaryExp(x, static_cast(n));
}

private:
double binaryExp(double x, long n) {
// Base case: x^0 = 1
if (n == 0) {
return 1.0;
}

    // Negative exponent: x^-n = 1 / x^n
    if (n < 0) {
        // Note: -n is safe because n was cast to long
        return 1.0 / binaryExp(x, -n);
    }
   
    // Odd exponent: x^n = x * (x^2)^((n-1)/2)
    if (n % 2 == 1) {
        return x * binaryExp(x * x, (n - 1) / 2);
    } 
    // Even exponent: x^n = (x^2)^(n/2)
    else {
        return binaryExp(x * x, n / 2);
    }
}

};
Related IssuesBy submitting this PR, I confirm that:[x] This is my original work not totally AI generated[x] I have tested the solution thoroughly on leetcode[x] I have maintained proper PR description format[x] This is a meaningful contribution, not spam

Summary by Sourcery

New Features:

• Add recursive binary exponentiation implementation for computing x^n in O(log n) time with negative exponent and INT_MIN handling

[sourcery-ai]

Reviewer's guide (collapsed on small PRs)

Reviewer's Guide

This PR adds a recursive binary exponentiation implementation (myPow and binaryExp) in C++, handling negative exponents and the INT_MIN edge case by casting the exponent to a long, and achieves O(log n) time complexity.

File-Level Changes

Change	Details	Files
Introduced public myPow wrapper	Cast int exponent to long to avoid overflow Invoke binaryExp with long exponent	50.Pow(x,n).cpp
Implemented binaryExp recursive logic	Return 1 for exponent zero Handle negative exponents via reciprocal recursion Process odd exponents by multiplying once and halving Process even exponents by squaring base and halving	50.Pow(x,n).cpp

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[github-actions]

Thanks for raising the PR, the owner will be review it soon' keep patience, keep contributing>>>!!! make sure you have star ⭐ the repo

[sourcery-ai]

Hey there - I've reviewed your changes and they look great!

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[SjxSubham]

this solution is already exixts