Today's challenge was flashback from Data Structure and Algorithms class during BSC. The first part was not that harder and again I took help from some people's code, specially here. I have updated my repository here.

**Please share your solution too.**

# Part 1

```
from collections import deque
class LinkedListNode:
def __init__(self, item):
self.item = item
self.left = None
self.right = None
my_input = [int(i) for i in "459672813"]
def part1():
my_cups = deque(my_input)
for _ in range(100):
orig = my_cups[0]
dest = my_cups[0]-1
if dest < 1:
dest = 9
my_cups.rotate(-1)
cup1 = my_cups.popleft()
cup2 = my_cups.popleft()
cup3 = my_cups.popleft()
while dest in (cup1, cup2, cup3):
dest = dest - 1 if dest > 1 else dest + 8
while my_cups[0] != dest:
my_cups.rotate(-1)
my_cups.rotate(-1)
my_cups.append(cup1)
my_cups.append(cup2)
my_cups.append(cup3)
while my_cups[0] != orig:
my_cups.rotate(-1)
my_cups.rotate(-1)
while my_cups[0] != 1:
my_cups.rotate(-1)
my_cups.popleft()
return ''.join([str(i) for i in my_cups])
print(f"Part 1 solution: {part1()}")
```

# Part 2

```
def part2():
my_nodes = {}
last_node = None
for i in my_input:
curr_node = LinkedListNode(i)
my_nodes[i] = curr_node
if last_node is not None:
last_node.right = curr_node
curr_node.left = last_node
last_node = curr_node
for i in range(len(my_input)+1, 1_000_001):
curr_node = LinkedListNode(i)
my_nodes[i] = curr_node
if last_node is not None:
last_node.right = curr_node
curr_node.left = last_node
last_node = curr_node
# Complete the circle
ptr = my_nodes[my_input[0]]
last_node.right = ptr
ptr.left = last_node
ptr = my_nodes[my_input[0]]
for i in range(10000000):
p_val = ptr.item
cup1 = ptr.right
cup2 = cup1.right
cup3 = cup2.right
ptr.right = cup3.right
ptr.right.left = ptr
d_val = p_val - 1 or 1000000
while d_val in (cup1.item, cup2.item, cup3.item):
d_val = d_val - 1 or 1000000
d_node = my_nodes[d_val]
cup3.right = d_node.right
cup3.right.left = cup3
d_node.right = cup1
cup1.left = d_node
ptr = ptr.right
while ptr.item != 1:
ptr = ptr.right
return ptr.right.item * ptr.right.right.item
print(f"Part 2 solution: {part2()}")
```

## Top comments (0)