13.md

Day 13

	Part 1	Part 2	Total
Time	47:21	8:50	56:11

Another long one for me which is frustrating. Andrew Kritzler, my nemesis in the NJIT ACM leaderboard, completed Part 1 in 12:03 and Part 2 in 5:11. He's too good!

Part 1

Unfortunately, I failed to handle the cases where the recursive input didn't have any conclusive output. For example, if the lists ran out at the same time, I was returning True, completely ignoring whatever came next in the checks. I switched from using booleans to -1, 0, and 1 representing False, Inconclusive, and True respectively. If one of the recursive calls is inconclusive, just keep on going!

Took me really long to try that. In my mind, I knew that I needed to do something like that from really early on, I think I had the bulk of it done in 5 minutes (thanks eval!) but I was too scared to try it. This has cost me time yesterday as well, and it's a trend I'd like to break.

from helpers.datagetter import data_in, submit

ans = 0
din = data_in(split=True, numbers=False)


def check(left, right):
    for i in range(max(len(left), len(right))):
        # When lists run out
        if i >= len(left) and i < len(right):
            return 1
        elif i < len(left) and i >= len(right):
            return -1
        elif i >= len(left) and i >= len(right):
            return 0

        # When comparing ints
        if type(left[i]) is int and type(right[i]) is int:
            if left[i] < right[i]:
                return 1
            elif left[i] > right[i]:
                return -1

        # When comparing lists
        if type(left[i]) is list and type(right[i]) is list:
            ch = check(left[i], right[i])
            if ch != 0:
                return ch

        # When comparing one int and one list
        if type(left[i]) is int and type(right[i]) is list:
            ch = check([left[i]], right[i])
            if ch != 0:
                return ch
        if type(left[i]) is list and type(right[i]) is int:
            ch = check(left[i], [right[i]])
            if ch != 0:
                return ch

    # If all tests are inconclusive, return 0
    return 0


for line in range(0, len(din), 3):
    left = eval(din[line])
    right = eval(din[line+1])
    
    if check(left, right) == 1:
        ans += (line // 3) + 1

print(ans)
submit(ans)

Part 2

Simple insertion sort.

# ADD THIS AFTER FUNCTION DEFINITION

packets = []

for line in range(0, len(din), 3):
    packets.append(eval(din[line]))
    packets.append(eval(din[line+1]))

packet_list = [[[2]], [[6]]]

# Insertion sort
for packet in packets:
    for i in range(len(packet_list)):
        if check(packet, packet_list[i]) == 1:
            packet_list.insert(i, packet)
            break
    packet_list.append(packet)

ans = (packet_list.index([[2]])+1) * (packet_list.index([[6]])+1)
print(ans)
submit(ans)