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In this post, you will find the solution for the **Pow(x, n)** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Pow(x, n)– LeetCode Problem

Pow(x, n)– LeetCode Problem

**Problem:**

Implement pow(x, n), which calculates `x`

raised to the power `n`

(i.e., `x`

).^{n}

**Example 1:**

Input:x = 2.00000, n = 10Output:1024.00000

**Example 2:**

Input:x = 2.10000, n = 3Output:9.26100

**Example 3:**

Input:x = 2.00000, n = -2Output:0.25000Explanation:2^{-2}= 1/2^{2}= 1/4 = 0.25

**Constraints:**

`-100.0 < x < 100.0`

`-2`

^{31}<= n <= 2^{31}-1`-10`

^{4}<= x^{n}<= 10^{4}

Pow(x, n)– LeetCode Solutions

Pow(x, n) in C++:

class Solution { public: double myPow(double x, long n) { if (n == 0) return 1; if (n < 0) return 1 / myPow(x, -n); if (n & 1) return x * myPow(x, n - 1); return myPow(x * x, n / 2); } };

Pow(x, n) in Java:

class Solution { public double myPow(double x, long n) { if (n == 0) return 1; if (n < 0) return 1 / myPow(x, -n); if (n % 2 == 1) return x * myPow(x, n - 1); return myPow(x * x, n / 2); } }

Pow(x, n) in Python:

class Solution: def myPow(self, x: float, n: int) -> float: if n == 0: return 1 if n < 0: return 1 / self.myPow(x, -n) if n % 2: return x * self.myPow(x, n - 1) return self.myPow(x * x, n / 2)