# Day 01 - Report Repair

## Puzzle

After saving Christmas five years in a row, you've decided to take a vacation at a nice resort on a tropical island. Surely, Christmas will go on without you.

The tropical island has its own currency and is entirely cash-only. The gold coins used there have a little picture of a starfish; the locals just call them stars. None of the currency exchanges seem to have heard of them, but somehow, you'll need to find fifty of these coins by the time you arrive so you can pay the deposit on your room.

To save your vacation, you need to get all fifty stars by December 25th.

Collect stars by solving puzzles. Two puzzles will be made available on each day in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!

Before you leave, the Elves in accounting just need you to fix your expense report (your puzzle input); apparently, something isn't quite adding up.

Specifically, they need you to find the two entries that sum to 2020 and then multiply those two numbers together.

For example, suppose your expense report contained the following:

```1721
979
366
299
675
1456
```

In this list, the two entries that sum to 2020 are 1721 and 299. Multiplying them together produces 1721 * 299 = 514579, so the correct answer is 514579.

Of course, your expense report is much larger. Find the two entries that sum to 2020; what do you get if you multiply them together?

### Part 2

The Elves in accounting are thankful for your help; one of them even offers you a starfish coin they had left over from a past vacation. They offer you a second one if you can find three numbers in your expense report that meet the same criteria.

Using the above example again, the three entries that sum to 2020 are 979, 366, and 675. Multiplying them together produces the answer, 241861950.

In your expense report, what is the product of the three entries that sum to 2020?

## Solution

Slurping the input file to `@inputs`. `@inputs` will contain an array of integers from the input file which were seperated by a newline.

```my @inputs;
@inputs = "input".IO.lines>>.Int;
```

Iterate over all possible combinations of `@inputs`, so 2 nested `for` loop would've solved it. But a better solution would be to remove elements from `@inputs` while iterating over it.

Consider this list: (`1, 2`). A nested `for` loop on it means we have to check for (`1, 1`, `1, 2`, `2, 1`, `2, 2`). But `1 + 2` is the same as `2 + 1` so we could've dropped either the 1st or the 2nd index from that array.

We do just that by `pop`'ing the value from `@inputs` while iterating. That way we just check for (`1, 1`, `1, 2`, `2, 2`). (`1, 2`) is skipped because `1` was `pop`'ed out in `while` loop.

```while pop @inputs -> \$num_1 {
my Int \$diff = 2020 - \$num_1;
for @inputs -> \$num_2 {
say "Part 1: ", \$num_2 * \$num_1 if \$diff == \$num_2;
}
}
```

Also, `\$diff` is calculated to be `2020 - \$num_1` & checked for `\$diff` being equal to `\$num_2`. The check should be `\$num_1 + \$num_2` being equal to `2020`. But that would mean this addition would be run the number of times `for` loop is run. It is optimized by calculating `\$diff` once for each `for` iteration.

For example, consider this list: (`1, 2`). Without this optimization we go through these steps:

```## first iteration through while loop
if 1 + 1 == 2020 then print 1 * 1 # for loop 1
if 1 + 2 == 2020 then print 1 * 2 # for loop 2

## second iteration through while loop
if 2 + 2 == 2020 then print 2 * 2 # for loop 1
```

But with the optimization we go through these steps:

```## first iteration through while loop
diff = 2020 - 1
if 1 == diff then print 1 * 1 # for loop 1
if 2 == diff then print 1 * 2 # for loop 2

## second iteration through while loop
diff = 2020 - 2
if 2 == diff then print 2 * 2 # for loop 2
```

You see what changes? We replaced all that addition in each `for` iteration with a single subtraction. In our example list, instead of 3 addition operations we just perform 2 subtraction operations but the optimization will better work on longer list.

Now this is not the best solution, we could do some more optimization but I'll leave it at this.

### Part 2

Input is again slurped to `@inputs`. This time 3 nested for loops are implemented instead. This was the simplest solution that came to my mind.

```@inputs = "input".IO.lines>>.Int;
for @inputs -> \$num_1 {
for @inputs -> \$num_2 {
for @inputs -> \$num_3 {
if 2020 == \$num_1 + \$num_2 + \$num_3 {
say "Part 2: ", (\$num_1 * \$num_2 * \$num_3);
exit;
}
}
}
}
```

Andinus / / Modified: 2020-12-04 Fri 20:42 Emacs 27.2 (Org mode 9.4.4)