# Pow(x, n) LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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 ProblemPow(x, n)– LeetCode Problem

`Pow(x, n)– LeetCode Problem`

### Problem:

Implement pow(x, n), which calculates `x` raised to the power `n` (i.e., `xn`).

Example 1:

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

Example 2:

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

Example 3:

```Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
```

Constraints:

• `-100.0 < x < 100.0`
• `-231 <= n <= 231-1`
• `-104 <= xn <= 104`
`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)``` #### Ads Blocker Detected!!!

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