Monday, July 24th

 Can the power function be implemented without using exponents?

That is the subject of today's problem


The most obvious way was described as brute-force, namely iterating over a value, multiplying a variable any number of times;

e.g. for a positive exponent:

    double ans=1.0;
    if(n>0) {
        for(int i=0; i<n; i++) {
            ans*=x;
        }
    } 

However, this times out for large inputs

To perform quicker, we will need to reduce the number of multiplications.  For brute force there are n multiplies.  

To do better we observe the fact that:
    
    x^n = (x^2) ^(n/2)

which means inserting a multiplication to square x requires 1 multiplication, but we're saving n/2 multiplications on the outer exponent.  When x is odd, the equation is:

    x^n = x * (x^2)^((n-1)/2)

We can use this information to solve the problem recursively.  We start with an answer of 1.  Then, while n is greater than 0, we multiply x by x (squaring) then divide n by 2.

    double ans=1.0;
    while(n) {
        if(n%2==1) {
            ans*=x;
            n-=1;
        }
         x*=x;
         n/=2;
    }
    
    return ans;

If n is odd, we multiply the answer by x and reduce n by 1.  This occurs during the final iteration because n will equal 1

Also, if n is negative, we multiply n by -1 and invert x

    if(n<0) {
        n*=(-1);
        x=1.0/x;
    }

And don't forget the base case

    if(n==0) return 1;

Also, since n-1 may be executed for an n at the negative boundary of integer values, we change the type of n to long

    double myPow(double x, long n) {

While solving, I tried a variation of binary exponentiation, and got through the time exceeded test case, but then failed a test case where x=1.000000001 because of an inaccurate decimal place

Tricky problem, thanks to the Editorial for filling in on the time exceeded issue

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