Day 6: Trash Compactor

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FAQ

  • Jayjader@jlai.lu
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    5 hours ago

    (Browser-based) Javascript

    I got lazy and lucky writing the first part; hard-coded the number of lines / operands and it worked on the first try! I didn’t factor in the final block not being padded by spaces on both sides for part 2, and so needed to test on the example code – which has 3 lines of operands instead of the problem input’s 4 lines, so I had to properly solve the problem across all possible amounts of lines of input 😩.

    I am very jealous of all y’all who have a transpose function available in your language of choice.

    Code
    function part1(inputText) {
      const [firstArgs, secondArgs, thirdArgs, fourthArgs, operators] = inputText.trim().split('\n').map(line => line.trim().split(/\s+/));
      let totalSum = 0;
      for (let i = 0; i < firstArgs.length; i++) {
        if (operators[i] === '+') {
          totalSum += Number.parseInt(firstArgs[i], 10) + Number.parseInt(secondArgs[i], 10) + Number.parseInt(thirdArgs[i], 10) + Number.parseInt(fourthArgs[i], 10);
        } else if (operators[i] === '*') {
          totalSum += Number.parseInt(firstArgs[i], 10) * Number.parseInt(secondArgs[i], 10) * Number.parseInt(thirdArgs[i], 10) * Number.parseInt(fourthArgs[i], 10);
        }
      }
      return totalSum
    }
    
    {
      const start = performance.now();
      const result = part1(document.body.textContent);
      const end = performance.now();
      console.info({
        day: 6,
        part: 1,
        time: end - start,
        result
      })
    }
    
    function part2(inputText) {
      const lines = inputText.trimEnd().split('\n');
      const interBlockIndices = [];
      for (let i = 0; i < lines[0].length; i++) {
        let allEmpty = true;
        for (let j = 0; j < lines.length; j++) {
          if (lines[j][i] !== ' ') {
            allEmpty = false;
            break;
          }
        }
        if (allEmpty) {
          interBlockIndices.push(i);
        }
      }
      interBlockIndices.push(lines[0].length);
    
      let totalSum = 0;
      let blockStart = 0;
      for (const interBlockIndex of interBlockIndices) {
        // compute calculation of block [blockStart, interBlockIndex - 1]
        const operands = [];
        for (let i = interBlockIndex - 1; i >= blockStart; i--) {
          // parse operands
          let operand = 0;
          for (let j = 0; j < lines.length - 1; j++) {
            if (lines[j][i] !== ' ') {
              operand *= 10;
              operand += Number.parseInt(lines[j][i], 10);
            }
          }
          operands.push(operand)
        }
        if (lines.at(-1)[blockStart] === '+') {
          totalSum += operands.reduce((accu, next) => accu + next, 0);
        } else if (lines.at(-1)[blockStart] === '*') {
          totalSum += operands.reduce((accu, next) => accu * next, 1);
        }
        //  console.debug({ totalSum, operands, blockStart, interBlockIndex });
        blockStart = interBlockIndex + 1;
      }
      return totalSum;
    }
    
    {
      const example = `123 328  51 64 
     45 64  387 23 
      6 98  215 314
    *   +   *   +  
    `;
      const start = performance.now();
      const result = part2(document.body.textContent);
      //   const result = part2(example);
      const end = performance.now();
      console.info({
        day: 6,
        part: 2,
        time: end - start,
        result
      })
    }
    
    
  • GiantTree@feddit.org
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    1 day ago

    Kotlin

    More grid stuff and two-dimensional problem solving, I like it!
    The first part just requires extracting the numbers and operators, transposing the grid and summing/multiplying the numbers.
    The second part is also not too hard. I just search for the numbers in the transposed grid, making sure to leave out the last column. That one might contain an operator (“+” or “*”). Remember it for later. If the entire row is made of spaces, we have finished parsing a math problem. Just remember to account for the last one! 😅
    As with part one, just reduce the found problems and we’re done!

    This solution requires the trailing spaces to be present in the input files. I had to disable an option in my IDE to prevent it from breaking my nice solution.

    Code on Github

    Code
    class Day06 : AOCSolution {
        override val year = 2025
        override val day = 6
    
        override fun part1(inputFile: String): String {
            val worksheet = readResourceLines(inputFile)
            // Arrange the problem in a grid and transpose it, so that the operation is the last element of each row
            val problems = worksheet.map { line -> line.trim().split(spaceSplitRegex) }.toGrid().transposed()
    
            val grandTotal = problems.rows().sumOf { line ->
                // Map the all but the last element to a number
                val numbers = line.mapToLongArray(0, line.lastIndex - 1, String::toLong)
                // Extract the operation
                val operation = line.last()[0]
    
                // Call the correct reduction
                // The "else" branch is needed for the compiler
                when (operation) {
                    ADD -> numbers.sum()
                    MULTIPLY -> numbers.reduce { acc, value -> acc * value }
                    else -> 0
                }
            }
            return grandTotal.toString()
        }
    
        override fun part2(inputFile: String): String {
            val worksheet = readResourceLines(inputFile)
    
            // In this part the problem is more complicated and dependent on the individual characters.
            val charGrid = worksheet.map(CharSequence::toList).toCharGrid().transposed()
    
            val numbers = mutableListOf<Long>()
            val sb = StringBuilder(charGrid.width)
    
            val problems = buildList {
                // Begin with an empty operation
                // Assume the operation will be set to a valid value
                var operation = SPACE
                for (y in 0 until charGrid.height) {
                    // Extract each row (transposed column)
                    sb.clear().append(charGrid[y])
                    // Find the bounds of the number
                    val numberOffset = sb.indexNotOf(SPACE)
    
                    if (numberOffset != -1) {
                        // A number was found, parse it and add it to the list.
                        val endIndex = sb.indexOfAny(STOP_CHARACTERS, numberOffset + 1)
                        val number = java.lang.Long.parseLong(sb, numberOffset, endIndex, 10)
                        numbers.add(number)
    
                        // Check whether there is an operation in the last column.
                        // IF so, that's the next relevant operation
                        val lastColumn = sb[sb.lastIndex]
                        if (lastColumn != SPACE) {
                            operation = lastColumn
                        }
                    } else {
                        // No number was found, that's the separator for two calculations.
                        // Finalize the collection and clear the numbers.
                        // `toLongArray` creates a neat copy of the Longs in the list.
                        add(Problem(operation, numbers.toLongArray()))
                        numbers.clear()
                    }
                }
                // Add the last remaining problem to the list
                add(Problem(operation, numbers.toLongArray()))
            }
    
            // Reduce all problems to their solutions and sum them up.
            val grandTotal = problems.sumOf { problem ->
                when (problem.operation) {
                    ADD -> problem.numbers.sum()
                    MULTIPLY -> problem.numbers.reduce { acc, value -> acc * value }
                    else -> 0
                }
            }
    
            return grandTotal.toString()
        }
    
        private companion object {
            private const val ADD = '+'
            private const val MULTIPLY = '*'
            private const val SPACE = ' '
            private val STOP_CHARACTERS = charArrayOf(SPACE, ADD, MULTIPLY)
    
            @JvmRecord
            @Suppress("ArrayInDataClass")
            private data class Problem(val operation: Char, val numbers: LongArray)
        }
    
  • skissue@programming.dev
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    9 days ago

    Uiua

    I’m new to Uiua, so probably not the best way to express the solution!

    ParseRows ← (
      ⊜∘⊸≠@\n
      ⍜⇌°⊂ # Get the row of operations.
    )
    ParseOperations ← ⊜⊢⊸≠@\s
    SumCalculations ← (
      ≡(
        ⍣(◇/×°@*
        | ◇/+°@+
        )
      )
      /+
    )
    
    Part₁ ← (
      ParseRows
      ⊓(ParseOperations
      | ⍉ ≡(⊜⋕⊸≠@\s) # Parse columns.
      )
      SumCalculations
    )
    
    Part₂ ← (
      ParseRows
      ⊓(ParseOperations
      | ⊜(□≡(⋕▽⊸≠@\s)) ≡/↥⊸≠@\s⍉ # Parse numbers.
      )
      SumCalculations
    )
    
    &fras "6.txt"
    ⊃Part₁ Part₂
    
  • Amy@piefed.blahaj.zone
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    10 days ago

    Haskell

    There’s probably a really clever way of abstracting just the difference between the two layouts.

    import Data.Char (isSpace)  
    import Data.List (transpose)  
    import Data.List.Split (splitWhen)  
    
    op '+' = sum  
    op '*' = product  
    
    part1 =  
      sum  
        . map ((op . head . last) <*> (map read . init))  
        . (transpose . map words . lines)  
    
    part2 =  
      sum  
        . map ((op . last . last) <*> map (read . init))  
        . (splitWhen (all isSpace) . reverse . transpose . lines)  
    
    main = do  
      input <- readFile "input06"  
      print $ part1 input  
      print $ part2 input  
    
  • Deebster@programming.dev
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    edit-2
    9 days ago

    nushell

    I was afk when the puzzle went up so I had another go at doing it on my phone in Turmux with my shell’s scripting language. It’s quite nice how your shell is also a REPL so you can build up the answer in pieces, although I wrote a file for the second part.

    Phone screenshot of my solution being developed

    open input.txt | str replace --all --regex ' +' ' ' |
            lines | each { $in | str trim } | to text |
            from csv --noheaders --separator ' ' |
            reverse | transpose --ignore-titles |
            each {
                    |list| transpose | skip 1 | if $list.column0 == '+' { math sum } else { math product }
            } |
            math sum
    

    Part 2

    let input = open input.txt | lines | each { $in | split chars }
    let last_row = ($input | length) - 1
    let last_col = ($input | first | length) - 1
    
    mut op = ' '
    mut numbers = []
    mut grand_tot = 0
    for x in $last_col..0 {
      if $op == '=' {
        $op = ' '
        continue
      }
      let n = 0..($last_row - 1) | each { |y| $input | get $y | get $x } | str join | into int
      $numbers = ($numbers | append $n)
    
      $op = $input | get $last_row | get $x
      if $op != ' ' {
        $grand_tot += $numbers | if $op == '+' { math sum } else { math product }
        $numbers = []
        $op = '='
      }
    }
    $grand_tot
    
  • Quant@programming.dev
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    7 days ago

    Uiua

    This was fun :D

    I had a fun experience just throwing the strings with both numbers and spaces at the parse function. In the online pad, everything worked out fine but running the same code on my input locally gave me a “invalid float literal” error.
    I thought I’d missed some edge case in the real input again, like is often the case.
    Turns out that the Uiua version I used locally had a bug that’s fixed in the latest build. For once it wasn’t directly my fault ^^

    Run with example input

    Code
    $ 123 328  51 64 
    $  45 64  387 23 
    $   6 98  215 314
    $ *   +   *   +  
    
    # &fras "input-6.txt" ◌
    
    Calc ← (
      ↘₂⊛⊂"+*"/◇⊂
      ≡◇⨬(/+|/×)
      /+
    )
    
    P₁ ← (
      ⊜(⊜□⊸≠@ )⊸≠@\n
      ⊃⊣↘₋₁
      ⊙(⍉≡₀◇⋕)
      Calc
    )
    
    P₂ ← (
      ⊜∘⊸≠@\n
      ⟜⧻
      ⊓⍉(˜↯@ )
      ⊜(⊙(□≡⋕)⍜⍉(⊃(⊢⊣)↘₋₁))¬⤚≡⌟≍
      Calc
    )
    
    1_2 [⊃P₁P₂]
    ≡(&p &pf $"Part _: ")
    
    
  • eco_game@discuss.tchncs.de
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    9 days ago

    Kotlin

    I’m not fully happy with my parsing today, but oh well. I also thought about just plain building the grid and then rotating it, but “normal input parsing” works too.

    Solution
    class Day06 : Puzzle {
    
        val numsPartOne = mutableListOf<MutableList<Long>>()
        val numsPartTwo = mutableListOf<List<Long>>()
        val ops = mutableListOf<(Long, Long) -> Long>()
    
        override fun readFile() {
            val input = readInputFromFile("src/main/resources/a2025/day06.txt")
            val lines = input.lines().filter { it.isNotBlank() }
    
            // parse part1 input
            for (line in lines.dropLast(1)) {
                for ((c, num) in line.trim().split(" +".toRegex()).withIndex()) {
                    if (numsPartOne.getOrNull(c) == null) numsPartOne.add(mutableListOf())
                    numsPartOne[c].add(num.toLong())
                }
            }
    
            // parse part2 input
            var numList = mutableListOf<Long>()
            for (c in 0..<lines.maxOf { it.length }) {
                var numStr = ""
                for (r in 0..<lines.size - 1) {
                    numStr += lines[r].getOrElse(c) { ' ' }
                }
                if (numStr.isBlank()) {
                    numsPartTwo.add(numList)
                    numList = mutableListOf()
                } else {
                    numList.add(numStr.trim().toLong())
                }
            }
            numsPartTwo.add(numList)
    
            // parse operators
            ops.addAll(
                lines.last().split(" +".toRegex())
                    .map { it.trim()[0] }
                    .map {
                        when (it) {
                            '*' -> { a: Long, b: Long -> a * b }
                            '+' -> { a: Long, b: Long -> a + b }
                            else -> throw IllegalArgumentException("Unknown operator: $it")
                        }
                    }
            )
        }
    
        override fun solvePartOne(): String {
            return numsPartOne.mapIndexed { c, list -> list.reduce { a, b -> ops[c](a, b) } }.sum().toString()
        }
    
        override fun solvePartTwo(): String {
            return numsPartTwo.mapIndexed { c, list -> list.reduce { a, b -> ops[c](a, b) } }.sum().toString()
        }
    }
    

    full code on Codeberg

    • chunkystyles@sopuli.xyz
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      9 days ago

      I also thought about trying to rotate, but not for very long. Mine would be a bit simpler if I’d done what you did and build the number string and then check if it’s blank.

      fun main() {
          val input = getInput(6)
          val output = parseInput2(input)
          var total = 0L
          for ((numbers, operator) in output) {
              when (operator) {
                  '+' -> { total += numbers.sum() }
                  '*' -> { total += numbers.reduce { acc, number -> acc * number }}
              }
          }
          println(getElapsedTime())
          println(total)
      }
      
      fun parseInput2(input: String): List<Pair<List<Long>, Char>> {
          val rows = input.lines()
              .filter { it.isNotBlank() }
              .map { it.toCharArray() }
          val output: MutableList<Pair<List<Long>, Char>> = mutableListOf()
          val numberRowCount = rows.size - 1
          var isNewProblem = true
          var currentNumbers: MutableList<Long> = mutableListOf()
          var operator = ' '
          for (column in rows[0].indices) {
              if (!isNewProblem && isColumnEmpty(rows, column)) {
                  isNewProblem = true
                  output.add(currentNumbers to operator)
                  continue
              }
              if (isNewProblem) {
                  isNewProblem = false
                  currentNumbers = mutableListOf()
                  operator = rows.last()[column]
              }
              var number = ""
              for (row in 0..<numberRowCount) {
                  if (rows[row][column] != ' ') {
                      number += rows[row][column]
                  }
              }
              currentNumbers.add(number.toLong())
          }
          if (!isNewProblem) {
              output.add(currentNumbers to operator)
          }
          return output
      }
      
      fun isColumnEmpty(rows: List<CharArray>, column: Int): Boolean {
          for (i in rows.indices) {
              if (rows[i][column] != ' ') {
                  return false
              }
          }
          return true
      }
      
  • ystael@beehaw.org
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    9 days ago

    Ironically for Lisp, a good chunk of the work here is type conversion, because strings, vectors, multidimensional arrays, characters, and numbers don’t have implicit conversions between them; you have to specify what you want explicitly. I also found it easier to manually transpose the character array for part 2 rather than traverse in column-major order, because that makes the relationship between input and output data structure more transparent.

    (ql:quickload :str)
    (ql:quickload :array-operations)
    
    (defun parse-line-1 (line)
      (let ((broken-line (str:split " " (str:collapse-whitespaces (str:trim line)))))
        (mapcar #'(lambda (s)
                    (cond ((equal s "+") #'+)
                          ((equal s "*") #'*)
                          (t (parse-integer s))))
                broken-line)))
    
    (defun read-inputs-1 (filename)
      (let* ((input-lines (uiop:read-file-lines filename)))
        (mapcar #'parse-line-1 input-lines)))
    
    (defun main-1 (filename)
      (let* ((problems (read-inputs-1 filename))
             (arguments (apply #'mapcar #'list (butlast problems))))
        (reduce #'+ (mapcar #'apply (car (last problems)) arguments))))
    
    (defun parse-operands-2 (lines)
      (let* ((initial-rows (length lines))
             (initial-cols (length (car lines)))
             (flat-chars (make-array (list (* initial-rows initial-cols))
                                     :initial-contents (apply #'concatenate 'string lines)))
             (box-chars (make-array (list initial-rows initial-cols) :displaced-to flat-chars))
             (transposed-chars (aops:each-index (i j) (aref box-chars j i))))
        (loop for cv across (aops:split transposed-chars 1)
              for s = (str:trim (coerce cv 'string))
              collect (if (zerop (length s)) nil (parse-integer s)))))
    
    (defun list-split (xs sep &optional (predicate #'equal))
      (let ((current nil)
            (result nil))
        (loop for x in xs
              do (if (funcall predicate x sep)
                     (progn
                       (setf result (cons (reverse current) result))
                       (setf current nil))
                     (setf current (cons x current)))
              finally (setf result (cons (reverse current) result)))
        (reverse result)))
    
    (defun main-2 (filename)
      (let* ((lines (uiop:read-file-lines filename))
             (operators (parse-line-1 (car (last lines))))
             (operands (parse-operands-2 (butlast lines))))
        (loop for rator in operators
              for rands in (list-split operands nil)
              sum (apply rator rands))))
    
  • LeixB@lemmy.world
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    9 days ago

    Haskell

    import Control.Arrow
    import Data.Char
    import Data.List
    import Text.ParserCombinators.ReadP
    
    op "*" = product
    op "+" = sum
    
    part1 s = sum $ zipWith ($) (op <$> a) (transpose $ fmap read <$> as)
      where
        (a : as) = reverse . fmap words . lines $ s
    
    parseGroups = fst . last . readP_to_S (sepBy (endBy int eol) eol) . filter (/= ' ')
      where
        eol = char '\n'
        int = read <$> munch1 isDigit :: ReadP Int
    
    part2 s = sum $ zipWith ($) (op <$> words a) (parseGroups . unlines $ reverse <$> transpose as)
      where
        (a : as) = reverse $ lines s
    
    main = getContents >>= print . (part1 &&& part2)
    
  • reboot6675@sopuli.xyz
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    9 days ago

    Go

    Part 2: Read the whole input in a rune matrix. Scan it column by column, store the numbers as you go, ignoring all spaces, and store the operand when you find it. When you hit an empty column or the end, do the operation and add it to the total.

    spoiler
    func part2() {
    	// file, _ := os.Open("sample.txt")
    	file, _ := os.Open("input.txt")
    	defer file.Close()
    	scanner := bufio.NewScanner(file)
    
    	chars := [][]rune{}
    	for scanner.Scan() {
    		chars = append(chars, []rune(scanner.Text()))
    	}
    
    	m := len(chars)
    	n := len(chars[0])
    	var op rune
    	nums := []int{}
    	total := 0
    
    	for j := range n {
    		current := []rune{}
    		for i := range m {
    			if chars[i][j] == '+' || chars[i][j] == '*' {
    				op = chars[i][j]
    			} else if chars[i][j] != ' ' {
    				current = append(current, chars[i][j])
    			}
    		}
    		if len(current) > 0 {
    			x, _ := strconv.Atoi(string(current))
    			nums = append(nums, x)
    		}
    		if len(current) == 0 || j == n-1 {
    			result := 0
    			if op == '*' {
    				result = 1
    			}
    			for _, x := range nums {
    				if op == '+' {
    					result = result + x
    				} else {
    					result = result * x
    				}
    			}
    			total += result
    			nums = []int{}
    		}
    	}
    
    	fmt.Println(total)
    }
    
  • Pyro@programming.dev
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    10 days ago

    Python

    For part1, regex is king. Thought about rotating the grid for part2, but going column by column is simple enough.

    view code
    import re
    from operator import add, mul
    
    def part1(data: str):
        # split into row, but do not split into cells yet
        rows = data.splitlines()
        m = len(rows)   # number of rows
    
        # initialize the result and operator arrays
        # the operators array will store the operator function for each column block
        # the result array will store the initial value for each column block
        operators = []
        res = []
        # using regex to skip variable number of spaces
        for symbol in re.findall(r'\S', rows[-1]):
            if symbol == '+':
                operators.append(add)
                res.append(0)
            elif symbol == '*':
                operators.append(mul)
                res.append(1)
    
        n = len(res)    # number of columns
        # iterate through each row, except the last one
        for i in range(m-1):
            # use regex to find all numbers in the row
            for j, num in enumerate(map(int, re.findall(r'\d+', rows[i]))):
                # apply the operator to update the result for the appropriate column
                res[j] = operators[j](res[j], num)
    
        return sum(res)
    
    def part2(data: str):
        # completely split into grid
        grid = [list(line) for line in data.splitlines()]
        m, n = len(grid), len(grid[0])
        
        res = 0
        
        curr = None     # current value of the block
        op = None       # operator for the block
        for j in range(n):
            if curr is None:
                # we just started a new block
                # update the current value and operator based on the symbol
                symbol = grid[-1][j]
                curr = 0 if symbol == '+' else 1
                op = add if symbol == '+' else mul
            
            # read the number from the column
            num = 0
            for i in range(m-1):
                if grid[i][j] != ' ':
                    num = num * 10 + int(grid[i][j])
            # if there is no number, we are at the end of a block
            if num == 0:
                # add the block value to the result
                #   and reset the current value and operator
                res += curr
                curr = None
                op = None
                continue
            
            # otherwise, update the current value using the operator
            curr = op(curr, num)
        # finally, don't forget to add the last block value that's being tracked
        res += curr
    
        return res
    
    sample = """123 328  51 64 
     45 64  387 23 
      6 98  215 314
    *   +   *   +  """
    assert part1(sample) == 4277556
    assert part2(sample) == 3263827
    
  • Camille@lemmy.ml
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    edit-2
    10 days ago

    Go

    Damn, I actually reeaally enjoyed this one! I didn’t expect the twist of part 2, but somehow it wasn’t that hard to manage.

    Here is my modern solution:

    day06.go
    package main
    
    import (
    	"aoc/utils"
    	"fmt"
    	"regexp"
    	"slices"
    	"strconv"
    	"strings"
    )
    
    type operation int
    
    func (o operation) compute(values []int) int {
    	switch o {
    	case add:
    		sum := 0
    		for _, val := range values {
    			sum += val
    		}
    		return sum
    	case mul:
    		product := 1
    		for _, val := range values {
    			product *= val
    		}
    		return product
    	}
    
    	return 0
    }
    
    const (
    	add operation = iota
    	mul
    )
    
    var allOperationSymbols = []string{"+", "*"}
    
    func operationFromSymbol(sym string) operation {
    	switch sym {
    	case "+":
    		return add
    	case "*":
    		return mul
    	default:
    		panic(fmt.Sprintf("wtf is a %s?", sym))
    	}
    }
    
    type problems struct {
    	values     [][]int
    	operations []operation
    }
    
    func (p *problems) feed(column string) {
    	last := string(column[len(column)-1])
    	done := false
    
    	if slices.Contains(allOperationSymbols, last) {
    		p.operations = append(p.operations, operationFromSymbol(last))
    		column = column[:len(column)-1]
    		done = true
    	}
    
    	val, _ := strconv.Atoi(strings.TrimSpace(column))
    	idx := len(p.values) - 1
    	p.values[idx] = append(p.values[idx], val)
    
    	if done {
    		p.values = append(p.values, []int{})
    	}
    }
    
    func (p *problems) addLine(line string) (done bool) {
    	parts := strings.Split(line, " ")
    	parts = slices.DeleteFunc(parts, func(elem string) bool {
    		return elem == ""
    	})
    
    	if slices.Contains(allOperationSymbols, parts[0]) {
    		p.operations = make([]operation, len(parts))
    		for idx, sym := range parts {
    			p.operations[idx] = operationFromSymbol(sym)
    		}
    		done = true
    	} else {
    		if len(p.values) == 0 {
    			lenparts := len(parts)
    			p.values = make([][]int, lenparts)
    			for idx := range lenparts {
    				p.values[idx] = []int{}
    			}
    		}
    
    		for idx, part := range parts {
    			num, _ := strconv.Atoi(part)
    			p.values[idx] = append(p.values[idx], num)
    		}
    		done = false
    	}
    
    	return done
    }
    
    func (p problems) solve() []int {
    	solutions := make([]int, len(p.values))
    
    	for idx, values := range p.values {
    		op := p.operations[idx]
    		solutions[idx] = op.compute(values)
    	}
    
    	return solutions
    }
    
    func stepOne(input chan string) (int, error) {
    	modernProblems := problems{}
    	for line := range input {
    		done := modernProblems.addLine(line)
    		if done {
    			break
    		}
    	}
    
    	modernSolutions := modernProblems.solve()
    	sum := 0
    	for _, solution := range modernSolutions {
    		sum += solution
    	}
    
    	return sum, nil
    }
    
    func transposeInputChan(input chan string) []string {
    	lines := [][]rune{}
    	for line := range input {
    		lines = append(lines, []rune(line))
    	}
    
    	linecount := len(lines)
    	columncount := len(lines[0])
    	transposed := make([][]rune, columncount)
    
    	for idx := range transposed {
    		transposed[idx] = make([]rune, linecount)
    	}
    
    	for row, line := range lines {
    		for col, char := range line {
    			transposed[col][row] = char
    		}
    	}
    
    	columns := make([]string, len(transposed))
    	for idx, col := range transposed {
    		columns[idx] = string(col)
    	}
    
    	return columns
    }
    
    func stepTwo(input chan string) (int, error) {
    	transposedInput := transposeInputChan(input)
    	slices.Reverse(transposedInput)
    
    	// problem-set with one empty problem.
    	modernProblems := problems{
    		values: [][]int{[]int{}},
    	}
    
    	for _, column := range transposedInput {
    		if matched, _ := regexp.MatchString("^\\s*$", column); matched {
    			continue
    		}
    
    		modernProblems.feed(column)
    	}
    
    	// Remove last useless empty problem.
    	modernProblems.values = modernProblems.values[:len(modernProblems.values)-1]
    
    	modernSolutions := modernProblems.solve()
    	sum := 0
    	for _, solution := range modernSolutions {
    		sum += solution
    	}
    
    	return sum, nil
    }
    
    func main() {
    	inputFile := utils.FilePath("day06.txt")
    	utils.RunStep(utils.ONE, inputFile, stepOne)
    	utils.RunStep(utils.TWO, inputFile, stepTwo)
    }
    
  • Gobbel2000@programming.dev
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    10 days ago

    Rust

    Mainly difficult parsing today.

    View on github

    fn part1(input: String) {
        let mut nums: Vec<Vec<u64>> = Vec::new();
        let mut mul: Vec<bool> = Vec::new();
        for l in input.lines() {
            if l.chars().next().unwrap().is_ascii_digit() {
                let row = l
                    .split_ascii_whitespace()
                    .map(|s| s.parse::<u64>().unwrap())
                    .collect();
                nums.push(row);
            } else {
                mul = l.split_ascii_whitespace().map(|s| s == "*").collect();
            }
        }
        let mut sum = 0;
        for (idx, op_mul) in mul.iter().enumerate() {
            let col = nums.iter().map(|row| row[idx]);
            sum += if *op_mul {
                col.reduce(|acc, n| acc * n)
            } else {
                col.reduce(|acc, n| acc + n)
            }
            .unwrap();
        }
        println!("{sum}");
    }
    
    fn part2(input: String) {
        let grid: Vec<&[u8]> = input.lines().map(|l| l.as_bytes()).collect();
        let n_rows = grid.len() - 1; // Not counting operator row
        let mut op_mul = grid[n_rows][0] == b'*';
        let mut cur = if op_mul { 1 } else { 0 };
        let mut sum = 0;
        for x in 0..grid[0].len() {
            let digits: Vec<u8> = (0..n_rows).map(|y| grid[y][x]).collect();
            if digits.iter().all(|d| *d == b' ') {
                sum += cur;
                op_mul = grid[n_rows][x + 1] == b'*';
                cur = if op_mul { 1 } else { 0 };
                continue;
            }
            let n = String::from_utf8(digits)
                .unwrap()
                .trim()
                .parse::<u64>()
                .unwrap();
            if op_mul {
                cur *= n;
            } else {
                cur += n;
            }
        }
        sum += cur;
        println!("{sum}");
    }
    
    util::aoc_main!();
    
  • Strlcpy@1@lemmy.sdf.org
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    ·
    9 days ago

    C

    Well so much for reading a grid of ints in part 1! For part 2, initially I reworked the parsing to read into a big buffer, but then thought it would be fun to try and use memory-mapped I/O as not to use any more memory than strictly necessary for the final version:

    Code
    #include <stdio.h>
    #include <stdlib.h>
    #include <inttypes.h>
    #include <ctype.h>
    #include <assert.h>
    #include <sys/mman.h>
    #include <unistd.h>
    #include <err.h>
    
    #define GH	5
    
    int
    main()
    {
    	char *data, *g[GH], *p;
    	uint64_t p1=0,p2=0, acc;
    	int len, h=0, i, x,y, val;
    	char op;
    
    	if ((len = (int)lseek(0, 0, SEEK_END)) == -1)
    		err(1, "<stdin>");
    	if (!(data = mmap(NULL, len, PROT_READ, MAP_SHARED, 0, 0)))
    		err(1, "<stdin>");
    	for (i=0; i<len; i++)
    		if (!i || data[i-1]=='\n') {
    			assert(h < GH);
    			g[h++] = data+i;
    		}
    
    	for (x=0; g[h-1]+x < data+len; x++) {
    		if ((op = g[h-1][x]) != '+' && op != '*')
    			continue;
    
    		for (acc = op=='*', y=0; y<h-1; y++) {
    			val = atoi(&g[y][x]);
    			acc = op=='+' ? acc+val : acc*val;
    		}
    
    		p1 += acc;
    
    		for (acc = op=='*', i=0; ; i++) {
    			for (val=0, y=0; y<h-1; y++) {
    				p = &g[y][x+i];
    				if (p < g[y+1] && isdigit(*p))
    					val = val*10 + *p-'0';
    			}
    			if (!val)
    				break;
    			acc = op=='+' ? acc+val : acc*val;
    		}
    
    		p2 += acc;
    	}
    
    	printf("06: %"PRIu64" %"PRIu64"\n", p1, p2);
    }