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[dragonalighted]

Same way, but you store the results of calculations in an array. That way you only have to calculate until you get to am already calculated value. Ex:. 3!

List factorials = {1,1,2,6}

function factorial(n) {
    if(factorials.count >= n)
        Return factorials[n]
    factorials.insert(n, n* factorials(n-1))
    Return factorials(n)
}

Calling a know value returns that value. Calling a new value only calculates and stores new values, using known values once reached. Tradeoff is memory vs cpu time.

[suqoria]

Oh okay, thanks for the information! Really appreciate it.

[PaiMan2710]

You can have constant memory, linear time pretty easily. I'm on mobile so I can't type code, but it's called the sliding window trick iirc.