Solution to problem 2 in Python

src/Year_2018/P2.py

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+# --- Day 2: Inventory Management System ---
+
+# You stop falling through time, catch your breath, and check the screen on the
+# device. "Destination reached. Current Year: 1518. Current Location: North
+# Pole Utility Closet 83N10." You made it! Now, to find those anomalies.
+
+# Outside the utility closet, you hear footsteps and a voice. "...I'm not sure
+# either. But now that so many people have chimneys, maybe he could sneak in
+# that way?" Another voice responds, "Actually, we've been working on a new
+# kind of suit that would let him fit through tight spaces like that. But, I
+# heard that a few days ago, they lost the prototype fabric, the design plans,
+# everything! Nobody on the team can even seem to remember important details of
+# the project!"
+
+# "Wouldn't they have had enough fabric to fill several boxes in the warehouse?
+# They'd be stored together, so the box IDs should be similar. Too bad it would
+# take forever to search the warehouse for two similar box IDs..." They walk
+# too far away to hear any more.
+
+# Late at night, you sneak to the warehouse - who knows what kinds of paradoxes
+# you could cause if you were discovered - and use your fancy wrist device to
+# quickly scan every box and produce a list of the likely candidates (your
+# puzzle input).
+
+# To make sure you didn't miss any, you scan the likely candidate boxes again,
+# counting the number that have an ID containing exactly two of any letter and
+# then separately counting those with exactly three of any letter. You can
+# multiply those two counts together to get a rudimentary checksum and compare
+# it to what your device predicts.
+
+# For example, if you see the following box IDs:
+
+#     abcdef contains no letters that appear exactly two or three times.
+#     bababc contains two a and three b, so it counts for both.
+#     abbcde contains two b, but no letter appears exactly three times.
+#     abcccd contains three c, but no letter appears exactly two times.
+#     aabcdd contains two a and two d, but it only counts once.
+#     abcdee contains two e.
+#     ababab contains three a and three b, but it only counts once.
+
+# Of these box IDs, four of them contain a letter which appears exactly twice,
+# and three of them contain a letter which appears exactly three times.
+# Multiplying these together produces a checksum of 4 * 3 = 12.
+
+# What is the checksum for your list of box IDs?
+
+from itertools import combinations
+from typing import Dict
+
+with open("files/P2.txt") as f:
+    IDs = [line for line in f.read().strip().split()]
+
+
+def part_1() -> None:
+    twice, three_times = 0, 0
+    for word in IDs:
+        twice_added, three_times_added = False, False
+        while not twice_added and not three_times_added:
+            for letter in word:
+                counts = word.count(letter)
+                if counts == 2 and not twice_added:
+                    twice += 1
+                    twice_added = True
+                elif counts == 3 and not three_times_added:
+                    three_times += 1
+                    three_times_added = True
+
+    print(f"The checksum for my list is {twice*three_times}")
+
+
+# --- Part Two ---
+
+# Confident that your list of box IDs is complete, you're ready to find the
+# boxes full of prototype fabric.
+
+# The boxes will have IDs which differ by exactly one character at the same
+# position in both strings. For example, given the following box IDs:
+
+# abcde
+# fghij
+# klmno
+# pqrst
+# fguij
+# axcye
+# wvxyz
+
+# The IDs abcde and axcye are close, but they differ by two characters (the
+# second and fourth). However, the IDs fghij and fguij differ by exactly one
+# character, the third (h and u). Those must be the correct boxes.
+
+# What letters are common between the two correct box IDs? (In the example
+# above, this is found by removing the differing character from either ID,
+# producing fgij.)
+
+
+def part_2() -> None:
+    # each entry has word as key and a dict of all letters as value
+    dict_IDs: Dict[str, Dict[int, str]] = {}
+    for word in IDs:
+        dict_IDs[word] = {}
+        for pos, letter in enumerate(word):
+            dict_IDs[word][pos] = letter
+
+    # create combinations of two values for later comparison
+    pairs = list(combinations(dict_IDs.values(), 2))
+
+    for d1, d2 in pairs:
+        set1 = set(d1.items())
+        set2 = set(d2.items())
+        # only one letter on each set is different
+        if len(set1 ^ set2) == 2:
+            # remove extra letter from first set
+            d1.pop(list(set1 ^ set2)[1][0])
+            print(f"Common letters are {''.join(d1.values())}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()