Leetcode Variant

Problem Statement

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 <= 2311
• 104 <= xn <= 104

Problem Link

Pow(x, n) - LeetCode

Code

class Solution {
public:
    double myPow(double x, int n) {
        
        if(n==0)
            return 1;
        
       
        if(n<0)
            return 1/x * (myPow(1/x,-(n+1)));//condition to avoid overflow
        
        if(n%2==0)
            return myPow(x*x,n/2);
        
        return x*myPow(x,n-1);
        
    }
};

Interview Bit Variant

Problem Statement

Implement pow(x, n) % d.