Started 2023-12-07

2023/day7/exp

@@ -0,0 +1,11 @@
+32T3K 765
+T55J5 684
+AA8AA 21
+KK677 28
+AAAAA 123
+KTJJT 220
+23332 1
+A23A4 64
+QQQJA 483
+23456 64
+HNJIP 317

2023/day7/part1.py

@@ -3,11 +3,128 @@
 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)