I've been going through the Advent of Code, and it's been pretty straight forward day to day. I got to day 15, and it took me a little bit to think of how to get an array of 4 digits that also adds up to 100. After some thought, I then gave into the easy, brute force solution. There was a previous puzzle for "incrementing" a password and I thought I could solve this one the same way.

In other words, I'd start with [0,0,0,0] and always increment the right-most digit (array with index = len(array)-1) and when it hit 100, it would increment the digit at position len(array)-2 and reset the current one back to 0. This would take F@#%inG FOREVER. But it would loop through all possible combinations. Before calculating the "cookie recipe" based on the array, I would first check if they added up to 100, if not I would just increment until it did add up to 100. One quick brilliant optimization I made was to start with the initial array of [0,0,0,99] :) (I kid on the brilliance of it... it saved 99 iterations in what could be a hundred million).

So then I thought again about it. There were two other previous puzzles where you could brute force the solution by first getting all possible permutations of the elements, and looping over all permutations, applying the logic, and sorting them to tell you the highest / lowest combined value. I thought, this seems similar in a way to that, moreso than it did to the password incrementing.

So you'd start with inserting 100 into a blank array, then recursively call a method that would give you permutations of the remaining value, in this case would be 0. So in the end it would return [100,0,0,0]

If I inserted 50 into the array first, it would then give me permutations of the remaining 50.

So, [50,25,25,0], [50,25,0,25], [50,26,0,24] etc, all the way to [50,0,0,50], [50,50,0,0], [50,0,50,0]. Totally exhaustive but also, 100% correct and I never have to check if the sum of the array adds up to 100, because it's *guaranteed to*! :)

Here's that method in Go

`func Perms(total, num int) [][]int {`

ret := [][]int{}

if num == 2 {

for i := 0; i < total/2 + 1; i++ {

ret = append(ret, []int{ total-i, i })

if i != total - i {

ret = append(ret, []int{ i, total-i })

}

}

} else {

for i := 0; i <= total; i++ {

perms := Perms(total-i, num-1)

for _, p := range perms {

q := append([]int{ i }, p...)

ret = append(ret, q)

}

}

}

return ret

}

`perms := Perms(5, 3)`

fmt.Println(perms)

`And the output`

`[[0 5 0] [0 0 5] [0 4 1] [0 1 4] [0 3 2] [0 2 3] [1 4 0] [1 0 4] [1 3 1] [1 1 3]`

[1 2 2] [2 3 0] [2 0 3] [2 2 1] [2 1 2] [3 2 0] [3 0 2] [3 1 1] [4 1 0] [4 0 1]

[5 0 0]]

YES!!! I thought about that one for a while. I knew there had to be a way to do it :)

More importantly, for the Perms(100,4) input in the puzzle, it returns only 176,851 possible permutations, and my original brute force method could have looped 100,000,000 (one hundred million) times!! Quite the improvement.