Pow(x n) LeetCode – Python Solutions

LeetCode has a Medium coding Problem in Its’ Algorithm Section “Pow(x n) LeetCode”. Today We are going to solve this problem in Python. LeetCode Link of the Problem is HERE

Question

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

Examples

```Input: x = 2.00000, n = 10
Output: 1024.00000
```
```Input: x = 2.10000, n = 3
Output: 9.26100
```
```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`

Solution to Pow(x n) LeetCode

``````class Solution(object):
def myPow(self, x, n):
"""
:type x: float
:type n: int
:rtype: float
"""
pow = 1
if n < 0:
n = (-n)
x = 1/x
while n:

# if `n` is odd, multiply the result by `x`
if n & 1:
pow *= x

# divide `n` by 2
n = n >> 1

# multiply `x` by itself
x = x * x

# return result
return pow``````

Success

Runtime: 21 ms, faster than 63.45% of Python online submissions for Pow(x, n).

Memory Usage: 13.3 MB, less than 68.21% of Python online submissions for Pow(x, n).

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