From 1613813d7a3b977e08e75b03b2e6fb392a1b498f Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?David=20Doblas=20Jim=C3=A9nez?=
 <david.doblas-jimenez@xfel.eu>
Date: Tue, 15 Mar 2022 14:33:14 +0100
Subject: [PATCH] Solution to problem 7 in Python

---
 src/Year_2018/P7.py | 83 +++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 83 insertions(+)

diff --git a/src/Year_2018/P7.py b/src/Year_2018/P7.py
index 5590b3d..587d62d 100644
--- a/src/Year_2018/P7.py
+++ b/src/Year_2018/P7.py
@@ -60,6 +60,8 @@
 
 # In what order should the steps in your instructions be completed?
 
+from string import ascii_uppercase
+
 import networkx as nx
 
 with open("files/P7.txt") as f:
@@ -76,5 +78,86 @@ def part_1() -> None:
     print(f"The correct order will be {order}")
 
 
+# --- Part Two ---
+
+# As you're about to begin construction, four of the Elves offer to help. "The
+# sun will set soon; it'll go faster if we work together." Now, you need to
+# account for multiple people working on steps simultaneously. If multiple
+# steps are available, workers should still begin them in alphabetical order.
+
+# Each step takes 60 seconds plus an amount corresponding to its letter: A=1,
+# B=2, C=3, and so on. So, step A takes 60+1=61 seconds, while step Z takes
+# 60+26=86 seconds. No time is required between steps.
+
+# To simplify things for the example, however, suppose you only have help from
+# one Elf (a total of two workers) and that each step takes 60 fewer seconds
+# (so that step A takes 1 second and step Z takes 26 seconds). Then, using the
+# same instructions as above, this is how each second would be spent:
+
+# Second   Worker 1   Worker 2   Done
+#    0        C          .
+#    1        C          .
+#    2        C          .
+#    3        A          F       C
+#    4        B          F       CA
+#    5        B          F       CA
+#    6        D          F       CAB
+#    7        D          F       CAB
+#    8        D          F       CAB
+#    9        D          .       CABF
+#   10        E          .       CABFD
+#   11        E          .       CABFD
+#   12        E          .       CABFD
+#   13        E          .       CABFD
+#   14        E          .       CABFD
+#   15        .          .       CABFDE
+
+# Each row represents one second of time. The Second column identifies how many
+# seconds have passed as of the beginning of that second. Each worker column
+# shows the step that worker is currently doing (or . if they are idle). The
+# Done column shows completed steps.
+
+# Note that the order of the steps has changed; this is because steps now take
+# time to finish and multiple workers can begin multiple steps simultaneously.
+
+# In this example, it would take 15 seconds for two workers to complete these
+# steps.
+
+# With 5 workers and the 60+ second step durations described above, how long
+# will it take to complete all of the steps?
+
+
+def part_2() -> None:
+    G = nx.DiGraph()
+    for step in steps:
+        parent, child = step[5:6], step[36:37]
+        G.add_edge(parent, child)
+
+    n_workers = 5
+
+    # Add amount of work for each node
+    for node in G.nodes:
+        G.nodes[node]["work"] = 61 + ord(node) - ord("A")
+
+    time = 0
+    while G.nodes:
+        # Find nodes available for work
+        available_nodes = [node for node in G.nodes if G.in_degree(node) == 0]
+        # Sort available nodes: Work on nodes with least remaining work
+        available_nodes.sort(key=lambda node: G.nodes[node]["work"])
+
+        # Reduce remaining work for n_workers of the available nodes
+        for worker, node in zip(range(n_workers), available_nodes):
+            G.nodes[node]["work"] -= 1
+            # Remove finished nodes
+            if G.nodes[node]["work"] == 0:
+                G.remove_node(node)
+
+        time += 1
+
+    print(f"Finishing all the work took {time} seconds")
+
+
 if __name__ == "__main__":
     part_1()
+    part_2()