Abstraction

src/Year_2015/P10.py

@@ -21,18 +21,28 @@
 # Starting with the digits in your puzzle input, apply this process 40 times.
 # What is the length of the result?
 
-from itertools import groupby
 
 with open("files/P10.txt", "r") as f:
-    number = f.read().strip().split()[0]
+    number = f.read().strip()
 
 
-def part_1() -> None:
+def look_and_say(iterations: int) -> None:
-    num = number
+    value = number
-    for _ in range(40):
+    for _ in range(iterations):
-        num = "".join(str(len(list(g))) + k for k, g in groupby(num))
+        result = []
+        current = value[0]
+        count = 1
+        for ch in value[1:]:
+            if ch == current:
+                count += 1
+            else:
+                result.append(f"{count}{current}")
+                current = ch
+                count = 1
+        result.append(f"{count}{current}")
+        value = "".join(result)
 
-    print(f"After 40 iterations, the length is {len(num)}")
+    print(f"After {iterations} iterations, the length is {len(value)}")
 
 
 # --- Part Two ---
@@ -44,14 +54,6 @@ def part_1() -> None:
 # 50 times. What is the length of the new result?
 
 
-def part_2() -> None:
-    num = number
-    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()
+    look_and_say(40)
-    part_2()
+    look_and_say(50)