I usually skim a page before reading everything closely.
Today, I saw:
And I feared this would be another shortest path challenge.
Then I read it.
And breathed a sigh of relief...at least for Part 1.
I need to find all valid paths.
That...I can do!
I have to find all the 0s:
input = input .split('\n') .map( line => line.split('').map(char => +char) ) let zeros = [] for (let r = 0; r < input.length; r++) { for (let c = 0; c < input[0].length; c++) { if (input[r][c] == 0) { zeros.push([r,c]) } } }
0s: found!
From each 0, a valid path is nine steps where each number is one greater than the last, ending at 9.
This sounds like a job for recursion.
I need a base case:
Here's how my algorithm should work:
Input: 1. Originating number 2. Current coordinates Get current number If it is not exactly one greater than the originating number Return false Else If it is 9 Return the current coordinates If it is not 9 Continue with the coordinates in each orthogonal direction
Having now written it, the part I missed was tracking the ledger of valid end coordinates.
I struggled with this for some time.
I kept getting an error that wrongly made me think I couldn't pass in a Set or even an array.
But, thankfully, I just forgot to pass it into further calls of the recursive function.
Here's my working recursive algorithm:
let dirs = [[-1,0],[0,-1],[0,1],[1,0]] function pathFinder(num, coord, memo) { let current = input[coord[0]][coord[1]] if (current - num !== 1) { return false } else if (current == 9) { memo.add(coord.join(',')) return } else { dirs.forEach(dir => { if ( coord[0] + dir[0] >= 0 && coord[0] + dir[0] < input.length && coord[1] + dir[1] >= 0 && coord[1] + dir[1] < input[0].length ) { pathFinder( current, [coord[0] + dir[0], coord[1] + dir[1]], memo ) } }) } }
Since I must start with the coordinates of 0s, my first call uses -1:
pathFinder(-1, zeroCoordinate, matches)
Lastly, to get the correct score, I iterate through each zero, generating the unique set of destination 9s, keep and sum up the sizes of the sets:
let part1 = zeros.map(z => { let matches = new Set() pathFinder(-1, z, matches) return matches.size }).reduce((a, c) => a + c)
It generated the correct answer for the small example input.
And for the larger example input.
And...
...for my puzzle input!!!
Woohoo!!!
What will Part 2 challenge me with?
Is it possible that the way I wrote my algorithm in Part 1 means this will only require a few small changes to get the correct answer?
Right now, I add each valid 9 to a set.
For Part 2, I think I just need to increment a counter for each valid 9.
Worth a try!
Correct answer for the example.
Correct answer for my puzzle input.
Wow. Wow. Wow.
On to the next day...which is likely to be much harder.
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