Solution to problem 5 in Python

src/P5.py

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+# --- Day 5: Binary Boarding ---
+
+# You board your plane only to discover a new problem: you dropped your
+# boarding pass! You aren't sure which seat is yours, and all of the flight
+# attendants are busy with the flood of people that suddenly made it through
+# passport control.
+
+# You write a quick program to use your phone's camera to scan all of the
+# nearby boarding passes (your puzzle input); perhaps you can find your seat
+# through process of elimination.
+
+# Instead of zones or groups, this airline uses binary space partitioning to
+# seat people. A seat might be specified like FBFBBFFRLR, where F means
+# "front", B means "back", L means "left", and R means "right".
+
+# The first 7 characters will either be F or B; these specify exactly one of
+# the 128 rows on the plane (numbered 0 through 127). Each letter tells you
+# which half of a region the given seat is in. Start with the whole list of
+# rows; the first letter indicates whether the seat is in the front (0 through
+# 63) or the back (64 through 127). The next letter indicates which half of
+# that region the seat is in, and so on until you're left with exactly one row.
+
+# For example, consider just the first seven characters of FBFBBFFRLR:
+
+#     Start by considering the whole range, rows 0 through 127.
+#     F means to take the lower half, keeping rows 0 through 63.
+#     B means to take the upper half, keeping rows 32 through 63.
+#     F means to take the lower half, keeping rows 32 through 47.
+#     B means to take the upper half, keeping rows 40 through 47.
+#     B keeps rows 44 through 47.
+#     F keeps rows 44 through 45.
+#     The final F keeps the lower of the two, row 44.
+
+# The last three characters will be either L or R; these specify exactly one of
+# the 8 columns of seats on the plane (numbered 0 through 7). The same process
+# as above proceeds again, this time with only three steps. L means to keep the
+# lower half, while R means to keep the upper half.
+
+# For example, consider just the last 3 characters of FBFBBFFRLR:
+
+#     Start by considering the whole range, columns 0 through 7.
+#     R means to take the upper half, keeping columns 4 through 7.
+#     L means to take the lower half, keeping columns 4 through 5.
+#     The final R keeps the upper of the two, column 5.
+
+# So, decoding FBFBBFFRLR reveals that it is the seat at row 44, column 5.
+
+# Every seat also has a unique seat ID: multiply the row by 8, then add the
+# column. In this example, the seat has ID 44 * 8 + 5 = 357.
+
+# Here are some other boarding passes:
+
+#     BFFFBBFRRR: row 70, column 7, seat ID 567.
+#     FFFBBBFRRR: row 14, column 7, seat ID 119.
+#     BBFFBBFRLL: row 102, column 4, seat ID 820.
+
+# As a sanity check, look through your list of boarding passes. What is the
+# highest seat ID on a boarding pass?
+
+with open("files/P5.txt", "r") as f:
+    boarding_passes = [code for code in f.read().strip().split("\n")]
+
+
+def binary_search(
+    arr: str, upper_limit: int, var_lower: str, var_upper: str
+) -> int:
+    range_l = 0
+    range_h = upper_limit
+    range_m = 0
+    for char in arr:
+        range_m = (range_h + range_l) // 2
+        if char == var_lower:
+            range_h = range_m
+        elif char == var_upper:
+            range_l = range_m + 1
+    return int(range_l)
+
+
+def part_1() -> None:
+    seat_ID: int = 0
+    for code in boarding_passes:
+        row, col = code[:7], code[-3:]
+
+        _row = binary_search(row, 127, "F", "B")
+        _col = binary_search(col, 7, "L", "R")
+        _seat_ID: int = _row * 8 + _col
+        seat_ID = max(seat_ID, _seat_ID)
+    print(f"The highest seat ID on a boarding pass is {seat_ID}")
+
+
+if __name__ == "__main__":
+    part_1()