Solution to problem 10 in Python

src/Year_2015/P10.py

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+# --- Day 10: Elves Look, Elves Say ---
+
+# Today, the Elves are playing a game called look-and-say. They take turns
+# making sequences by reading aloud the previous sequence and using that
+# reading as the next sequence. For example, 211 is read as "one two, two
+# ones", which becomes 1221 (1 2, 2 1s).
+
+# Look-and-say sequences are generated iteratively, using the previous value as
+# input for the next step. For each step, take the previous value, and replace
+# each run of digits (like 111) with the number of digits (3) followed by the
+# digit itself (1).
+
+# For example:
+
+#     1 becomes 11 (1 copy of digit 1).
+#     11 becomes 21 (2 copies of digit 1).
+#     21 becomes 1211 (one 2 followed by one 1).
+#     1211 becomes 111221 (one 1, one 2, and two 1s).
+#     111221 becomes 312211 (three 1s, two 2s, and one 1).
+
+# Starting with the digits in your puzzle input, apply this process 40 times.
+# What is the length of the result?
+
+from itertools import groupby
+
+
+def part_1() -> None:
+    num = "3113322113"
+    for _ in range(40):
+        num = "".join(str(len(list(g))) + k for k, g in groupby(num))
+
+    print(f"After 40 iterations, the length is {len(num)}")
+
+
+# --- Part Two ---
+
+# Neat, right? You might also enjoy hearing John Conway talking about this
+# sequence (that's Conway of Conway's Game of Life fame).
+
+# Now, starting again with the digits in your puzzle input, apply this process
+# 50 times. What is the length of the new result?
+
+
+def part_2() -> None:
+    num = "3113322113"
+    for _ in range(50):
+        num = "".join(str(len(list(g))) + k for k, g in groupby(num))
+
+    print(f"After 50 iterations, the length is {len(num)}")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()