Go solution to Leetcode 1006 Clumsy Factorial

In this Leetcode problem, we are asked to compute clumsy(N) = N * (N-1) / (N-2) + (N-3) — (N-4) * (N-5) / (N-6) + (N-7) ... (stopping when we reach 1), respecting the operations priorities (* and / before + and -).

Problem statement

One thing to notice for the recursion is that it is curious that N doesn’t have a minus sing in front, and the series goes (+) * / + — * / + - instead of the more natural — * / + — * / + -.

So I used the trick to actually compute- N* (N-1) / (N-2) + (N-3) — (N-4) * (N-5) / (N-6) + (N-7) ... and at the end I added the required:
2*(N(N-1)/(N-2)).

There isn’t much more than that for the recursion. Here is the code. It runs in 0ms beating 100% of answers.

func clumsy(N int) int {
    if N == 1 {
        return 1
    }
    if N == 2 {
        return 2
    }
    if N == 3 {
        return 6
    }
    if N == 4 {
        return 7
    }
    return 2*(N*(N-1)/(N-2)) + clumsyRec(N)
}
func clumsyRec(N int) int {
    if N == 1 {
        return -1
    }
    if N == 2 {
        return -2
    }
    if N == 3 {
        return -6
    }
    if N == 4 {
        return -5
    }
    return - N*(N-1)/(N-2) + (N-3) + clumsyRec(N-4)
}

Runs test in 0ms

Hope it helped, happy coding!