Solution to problem 1 part 2 in Python

src/Year_2021/P1.py

@@ -74,5 +74,61 @@ def part_1(lst: list[int]) -> int:
     return increments
 
 
+# --- Part Two ---
+
+# Considering every single measurement isn't as useful as you expected: there's
+# just too much noise in the data.
+
+# Instead, consider sums of a three-measurement sliding window. Again
+# considering the above example:
+
+# 199  A
+# 200  A B
+# 208  A B C
+# 210    B C D
+# 200  E   C D
+# 207  E F   D
+# 240  E F G
+# 269    F G H
+# 260      G H
+# 263        H
+
+# Start by comparing the first and second three-measurement windows. The
+# measurements in the first window are marked A (199, 200, 208); their sum is
+# 199 + 200 + 208 = 607. The second window is marked B (200, 208, 210); its sum
+# is 618. The sum of measurements in the second window is larger than the sum
+# of the first, so this first comparison increased.
+
+# Your goal now is to count the number of times the sum of measurements in this
+# sliding window increases from the previous sum. So, compare A with B, then
+# compare B with C, then C with D, and so on. Stop when there aren't enough
+# measurements left to create a new three-measurement sum.
+# In the above example, the sum of each three-measurement window is as follows:
+
+# A: 607 (N/A - no previous sum)
+# B: 618 (increased)
+# C: 618 (no change)
+# D: 617 (decreased)
+# E: 647 (increased)
+# F: 716 (increased)
+# G: 769 (increased)
+# H: 792 (increased)
+
+# In this example, there are 5 sums that are larger than the previous sum.
+
+# Consider sums of a three-measurement sliding window. How many sums are larger
+# than the previous sum?
+
+
+def part_2(lst: list[int]) -> int:
+    increments = sum(
+        1
+        for idx, val in enumerate(lst)
+        if sum(lst[idx + 1 : idx + 4]) > sum(lst[idx : idx + 3])
+    )
+    return increments
+
+
 if __name__ == "__main__":
     print(part_1(report))
+    print(part_2(report))