In this guide we will provide 50. Pow(x, n)LeetCode Solution with best time and space complexity. The solution to Pow(x, n) problem is provided in various programming languages like C++, Java and python. This will be helpful for you if you are preparing for placements, hackathon, interviews or practice purposes. The solutions provided here are very easy to follow and with detailed explanations.
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 0.
-104 <= xn <= 104
Complexity Analysis
Time Complexity: O(\log n)
Space Complexity: O(1)
50. Pow(x, n)LeetCode Solution 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 % 2 == 1)
return x * myPow(x, n - 1);
return myPow(x * x, n / 2);
}
}; /* code provided by PROGIEZ */
50. Pow(x, n)LeetCode Solution 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);
}
} // code provided by PROGIEZ
50. Pow(x, n)LeetCode Solution 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 == 1:
return x * self.myPow(x, n - 1)
return self.myPow(x * x, n // 2) #code by PROGIEZ
