AdventOfCode/2023/day7/part1.py

#!/bin/python3

import re
import sys
from pprint import pprint 
from collections import Counter

input_f  = sys.argv[1]

d =[]
rank = 0

strength = 'AKQJT98765432'
print(strength)

with open(input_f) as file:
    for line in file:
        d.append(line.split())
pprint(d)
print()

def same_kind(x):
    s = Counter(x)
    for i in s.values():
        if i == 4:
            return True
    return False

def five_of_kind(x):
    if len(Counter(x)) == 1:
            return True
    return False

def full_house(x):
    #print('full_house')
    s = Counter(x)
    if len(s) != 2:
        return False
    else:
        tmp = list(s.values())
        if (tmp[0] == 2 or tmp[0] == 3) and (tmp[1] == 2 or tmp[1] == 3):
            return True
        else:
            return False

def three_of_kind(x):
    s = Counter(x)
    for i in s.values():
        if i == 3:
            return True
    return False

def pairs(x):
    s = Counter(x)
    pairs = 0
    for i in s.values():
        if i == 2:
            pairs += 1
    return pairs


for idx,i in enumerate(d):
    d[idx].append(rank)
    
    # Five of a kind, where all five cards have the same label: AAAAA
    if five_of_kind(d[idx][0]) == True:
        d[idx][2] = 7
        continue

    #Four of a kind, where four cards have the same label and one card has a different label: AA8AA
    if same_kind(d[idx][0]) == True:
        d[idx][2] = 6
        continue

    #Full house, where three cards have the same label, and the remaining two cards share a different label: 23332
    if full_house(d[idx][0]) == True:
        d[idx][2] = 5
        continue

    #Three of a kind, where three cards have the same label, and the remaining two cards are each different from any other card in the hand: TTT98
    if three_of_kind(d[idx][0]) == True:
        d[idx][2] = 4
        continue

    #Two pair, where two cards share one label, two other cards share a second label, and the remaining card has a third label: 23432
    if pairs(d[idx][0]) == 2:
        d[idx][2] = 3
        continue

    #One pair, where two cards share one label, and the other three cards have a different label from the pair and each other: A23A4
    if pairs(d[idx][0]) == 1:
        d[idx][2] = 2
        continue

    #High card, where all cards' labels are distinct: 23456
    if pairs(d[idx][0]) == 0:
        d[idx][2] = 1
        continue

#d.sort()

d = sorted(d, key=lambda x: x[2])

print()
pprint(d)
print()
for idx,i in enumerate(d):
    if idx == len(d)-1:
        break
    else:
        if d[idx][2] == d[idx+1][2]:
            #print('Comparing ' + str(d[idx][0]) + ' to ' + str(d[idx+1][0]))
            tmp = list(d[idx][0])
            tmp2 = list(d[idx+1][0])
            for x in range(0,5):
                if tmp[x] != tmp2[x]:
                    #print(tmp[x],tmp2[x])
                    if strength.find(tmp[x]) < strength.find(tmp2[x]):
                        d[idx],d[idx+1] = d[idx+1],d[idx]
                        break

result = 0

for idx,i in enumerate(d):
    x = i[1]
    result += (int(x) * (idx+1))

pprint(d)

print(result)