Nim

Bruteforce, my beloved.

Wasted too much time on part 2 trying to make combinations iterator (it was very slow). In the end solved both parts with 3^n and toTernary.

Runtime: 1.5s

func digits(n: int): int =
  result = 1; var n = n
  while (n = n div 10; n) > 0: inc result

func concat(a: var int, b: int) =
  a = a * (10 ^ b.digits) + b

func toTernary(n: int, len: int): seq[int] =
  result = newSeq[int](len)
  if n == 0: return
  var n = n
  for i in 0..<len:
    result[i] = n mod 3
    n = n div 3

proc solve(input: string): AOCSolution[int, int] =
  for line in input.splitLines():
    let parts = line.split({':',' '})
    let res = parts[0].parseInt
    var values: seq[int]
    for i in 2..parts.high:
      values.add parts[i].parseInt

    let opsCount = values.len - 1
    var solvable = (p1: false, p2: false)
    for s in 0 ..< 3^opsCount:
      var sum = values[0]
      let ternary = s.toTernary(opsCount)
      for i, c in ternary:
        case c
        of 0: sum *= values[i+1]
        of 1: sum += values[i+1]
        of 2: sum.concat values[i+1]
        else: raiseAssert"!!"
      if sum == res:
        if ternary.count(2) == 0:
          solvable.p1 = true
        solvable.p2 = true
        if solvable == (true, true): break
    if solvable.p1: result.part1 += res
    if solvable.p2: result.part2 += res

Codeberg repo