Coding

Pow(x,n)

Asked at

Goldman Sachs

Zoho

1Give yourself 5 minutes

2Answer out loud, not in your head

3Then compare with the answer below

The problem

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

**Note **: In output print 4 digits places after decimal point.

Input : x = 2.0000 , n = 10 Output : 1024.0000 Explanation : Answer = 2^10 => 1024.

Input : x = 2.0000 , n = -2 Output : 0.2500 Explanation : Answer = 2^(-2) = 1/4 => 0.25.

Input : x = 2.5000 , n = 2

-100.0 <= x <= 100.0
-231 <= n <= 231 - 1
-104 <= xn <= 104
Either x is not zero or n>0.
n is an integer.

cpp

class Solution {
public:
    double myPow(double x, int n) {
        //your code goes here
    }
};

java

class Solution {
    public double myPow(double x, int n) {
          //your code goes here
    }
}

python

class Solution:    
    def myPow(self, x, n):
        #your code goes here

javascript

class Solution {
    myPow(x, n) {
         //your code goes here
    }
}

csharp

public class Solution
{
    public double MyPow(double x, int n)
    {
        //your code goes here
    }
}

func myPow(x float64, n int) float64 {
    //your code goes here
}

Stuck? Show a way to structure it+

01Promote n to a wider signed type.
02Invert x and negate exponent when n is negative.
03Maintain result and current base.
04Square the base and halve the exponent.
05Multiply result whenever the current exponent bit is one.

Reference answer

Use iterative binary exponentiation, not n repeated multiplications. Promote n to a wide signed integer first because negating the minimum 32-bit integer overflows in its own type. If n is negative, replace x with 1 divided by x and negate the promoted exponent. Initialize result to 1. While exponent is positive, if its low bit is one multiply result by the current base; square the base, then divide exponent by two. Each bit of the exponent is processed once, giving O(log absolute(n)) time and O(1) extra space. Explain the invariant: result times base raised to exponent always equals the original requested power. This method naturally handles exponent zero, x equal to one, and negative powers. Floating point rounding is expected, so return the computed double and let presentation formatting handle the requested decimal places.

Then expect these follow-ups

Can you write the recursive equivalent?
Tests: follow-up reasoning
What numerical issues arise for very large magnitudes?
Tests: constraint adaptation

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