Advent of Code 2020 Day 1 – Our Solutions

James

I decided to solve the problem using both APL and Python.

Part 1 (APL):

q1_1←2020∘{×/⍵[⍸⍵∊⍺-⍵]}

My solution for part one is fairly straight forward.

The only issues I had related not to the problem itself, but to the use of ]LINK, an APL source code management utility. This Advent of Code was a fresh opportunity for me, and I saw it as a good time to try LINK for the first time.

I did have a bit of a battle getting used to it as I would often create a variable in the session and forget to run ]LINK.add to track it. I also encountered an issue early on as this solution uses composition; I join 2020 as a left argument to my function when assigning it to a name. LINK doesn’t handle this, but Adám from Dyalog helped me find a way around it:

∇ q1_1←q1_1
  q1_1←2020∘{×/⍵[⍸⍵∊⍺-⍵]}
∇

2020 ∘ ∇{×/⍵[∪⊃¨⍸⍺=⍵∘.+,∘.+⍨⍵⌿⍨⍵≤⍺-+/2↑⍵]} 
∇ ⊂∘⍋⌷⊢

∇ r←q1_2 input;goal;sortedInput;sumOfTwo;indLargestThird;largestThirds;totals;filtered

 goal←2020                                                  ⍝ Set target
 sortedInput←(⊂∘⍋⌷⊢)input                                   ⍝ Sort input
 filtered←goal{⍵⌿⍨⍵≤⍺-+/2↑⍵}sortedInput                     ⍝ Remove any numbers which are less than the sum of the first two
 sumOfTwo←∘.+⍨filtered                                      ⍝ Find the sum of remaining numbers
 indLargestThird←(⌈/⍸)⍤1⊢(goal-⊃sortedInput)>sumOfTwo       ⍝ Find the index of the largest possible third number
 largestThirds←filtered[indLargestThird]                    ⍝ Lookup those numbers
 totals←largestThirds+⍤1⊢sumOfTwo                           ⍝ and sum with the previous two
 r←goal{×/⍵,⍺-+/⍵}sortedInput[∊⍸totals∊goal]                ⍝ Find the index where this sums is the goal, and return the product of it's three elements
∇

def q1_1(target, data):
    # Find the two entries that sum to 2020; what do you get if you multiply them together?

        # Comparing every number
        for num1 in data:

            # Generate the necessary second number to reach the target
            # and if it exists, return the product of the two
            num2 = target - num1
            if num2 in data:
                return num1 * num2

    return -1   # return not found

def q1_2(target, data):
    # what is the product of the three entries that sum to 2020?

        # Comparing every number
        for num1 in data:
       
            # With every number
            for num2 in data:

                # Generate the necessary third number to reach the target
                # and if it exists, return the product of the three numbers
                num3 = target - num1 - num2
                if num3 in data:
                    return num1 * num2 * num3

    return -1   # return not found

Side note: I realise none of my solutions factor in *distinct* number solutions. I wonder if the question was asking for non-duplicate values? In my case, all the numbers were unique, but I wonder if this is always the case...