Refactor 2021 for new UI.

src/holt59/aoc/2021/day8.py

@@ -1,8 +1,7 @@
 import itertools
-import os
-import sys
+from typing import Any, Iterator
 
-VERBOSE = os.getenv("AOC_VERBOSE") == "True"
+from ..base import BaseSolver
 
 digits = {
     "abcefg": 0,
@@ -17,71 +16,74 @@ digits = {
     "abcdfg": 9,
 }
 
-lines = sys.stdin.read().splitlines()
 
-# part 1
-lengths = {len(k) for k, v in digits.items() if v in (1, 4, 7, 8)}
-answer_1 = sum(
-    len(p) in lengths for line in lines for p in line.split("|")[1].strip().split()
-)
-print(f"answer 1 is {answer_1}")
+class Solver(BaseSolver):
+    def solve(self, input: str) -> Iterator[Any]:
+        lines = input.splitlines()
 
-# part 2
-values: list[int] = []
+        # part 1
+        lengths = {len(k) for k, v in digits.items() if v in (1, 4, 7, 8)}
+        yield sum(
+            len(p) in lengths
+            for line in lines
+            for p in line.split("|")[1].strip().split()
+        )
 
-for line in lines:
-    parts = line.split("|")
-    broken_digits = sorted(parts[0].strip().split(), key=len)
+        # part 2
+        values: list[int] = []
 
-    per_length = {
-        k: list(v)
-        for k, v in itertools.groupby(sorted(broken_digits, key=len), key=len)
-    }
+        for line in lines:
+            parts = line.split("|")
+            broken_digits = sorted(parts[0].strip().split(), key=len)
 
-    # a can be found immediately
-    a = next(u for u in per_length[3][0] if u not in per_length[2][0])
+            per_length = {
+                k: list(v)
+                for k, v in itertools.groupby(sorted(broken_digits, key=len), key=len)
+            }
 
-    # c and f have only two possible values corresponding to the single entry of
-    # length 2
-    cf = list(per_length[2][0])
+            # a can be found immediately
+            a = next(u for u in per_length[3][0] if u not in per_length[2][0])
 
-    # the only digit of length 4 contains bcdf, so we can deduce bd by removing cf
-    bd = [u for u in per_length[4][0] if u not in cf]
+            # c and f have only two possible values corresponding to the single entry of
+            # length 2
+            cf = list(per_length[2][0])
 
-    # the 3 digits of length 5 have a, d and g in common
-    adg = [u for u in per_length[5][0] if all(u in pe for pe in per_length[5][1:])]
+            # the only digit of length 4 contains bcdf, so we can deduce bd by removing cf
+            bd = [u for u in per_length[4][0] if u not in cf]
 
-    # we can remove a
-    dg = [u for u in adg if u != a]
+            # the 3 digits of length 5 have a, d and g in common
+            adg = [
+                u for u in per_length[5][0] if all(u in pe for pe in per_length[5][1:])
+            ]
 
-    # we can deduce d and g
-    d = next(u for u in dg if u in bd)
-    g = next(u for u in dg if u != d)
+            # we can remove a
+            dg = [u for u in adg if u != a]
 
-    # then b
-    b = next(u for u in bd if u != d)
+            # we can deduce d and g
+            d = next(u for u in dg if u in bd)
+            g = next(u for u in dg if u != d)
 
-    # f is in the three 6-length digits, while c is only in 2
-    f = next(u for u in cf if all(u in p for p in per_length[6]))
+            # then b
+            b = next(u for u in bd if u != d)
 
-    # c is not f
-    c = next(u for u in cf if u != f)
+            # f is in the three 6-length digits, while c is only in 2
+            f = next(u for u in cf if all(u in p for p in per_length[6]))
 
-    # e is the last one
-    e = next(u for u in "abcdefg" if u not in {a, b, c, d, f, g})
+            # c is not f
+            c = next(u for u in cf if u != f)
 
-    mapping = dict(zip((a, b, c, d, e, f, g), "abcdefg"))
+            # e is the last one
+            e = next(u for u in "abcdefg" if u not in {a, b, c, d, f, g})
 
-    value = 0
-    for number in parts[1].strip().split():
-        digit = "".join(sorted(mapping[c] for c in number))
-        value = 10 * value + digits[digit]
+            mapping = dict(zip((a, b, c, d, e, f, g), "abcdefg"))
 
-    if VERBOSE:
-        print(value)
+            value = 0
+            for number in parts[1].strip().split():
+                digit = "".join(sorted(mapping[c] for c in number))
+                value = 10 * value + digits[digit]
 
-    values.append(value)
+            self.logger.info(f"value for '{line}' is {value}")
 
+            values.append(value)
 
-answer_2 = sum(values)
-print(f"answer 2 is {answer_2}")
+        yield sum(values)