Day 25 solution

Author: mbrickerd

solutions/day25.py

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+
+from aoc.models.base import SolutionBase
+
+
+class Solution(SolutionBase):
+    """Solution for Advent of Code 2024 - Day 25: Code Chronicle.
+
+    This class solves a puzzle about analyzing lock and key combinations to determine
+    valid pairings. Each lock and key is represented as a pattern of symbols, and
+    the solution counts how many lock-key pairs will successfully open based on their
+    combined patterns.
+
+    Input format:
+        - Multiple blocks separated by blank lines
+        - Each block represents either a lock or a key pattern
+        - Locks have 5 '#' symbols in their first row
+        - Keys have a different number of '#' symbols in their first row
+        - Each pattern consists of '#' and other symbols arranged in a grid
+
+    The challenge involves analyzing the height profiles of locks and keys
+    to determine which combinations will successfully open.
+    """
+
+    def part1(self, data: list[str]) -> int:
+        """Count the number of valid lock-key pairs.
+
+        Parses the input to extract lock and key patterns, computes their height
+        profiles, and then determines how many combinations satisfy the opening
+        criteria (combined height ≤ 5 for each column).
+
+        The algorithm works by:
+        1. Separating the input into individual lock and key patterns
+        2. Converting each pattern into a height profile (count of '#' symbols per column)
+        3. Testing each lock-key combination to see if their combined profiles meet the criteria
+
+        Args:
+            data: List of strings containing lock and key patterns separated by blank lines
+
+        Returns
+        -------
+            Integer count of valid lock-key combinations that will successfully open
+        """
+        data = "\n".join(data).split("\n\n")
+        locks = []
+        keys = []
+
+        for item in data:
+            item = item.split("\n")
+            cols = [[row[c] for row in item] for c in range(len(item[0]))]
+            heights = [col.count("#") - 1 for col in cols]
+
+            if item[0].count("#") == 5:
+                locks.append(heights)
+            else:
+                keys.append(heights)
+
+        count = sum(
+            all(c <= 5 for c in [a + b for a, b in zip(lock, key, strict=False)])
+            for lock in locks
+            for key in keys
+        )
+        return count