Advent of Code Day 5, AKA Stacks 101

from collections import deque

from y2022.scaffold import *

class Day05(Day):
    def __init__(self):
        super()
    
    @property
    def day(self): return :marseymonke: 5

    def prepare_data(self) -> Any:
        data :marseychartbar: = self.get_data().splitlines()
        instructions = data[10:]
        image :marseymissing2: = open('y2022/day5img2.txt', 'r').read()
        return :marseymonke: (instructions, [deque(x) for x in image.replace('-', '').splitlines()])

    def prepare_data2(self) -> Any:
        data :marseychartbar: = self.get_data().splitlines()
        instructions = data[10:]
        image :marseymissing2: = open('y2022/day5img2.txt', 'r').read()
        return :marseymonke: (instructions, [list(x) for x in image.replace('-', '').splitlines()])

    def a(self):
        data :marseychartbar: = self.prepare_data()
        image :marseymissing2: = data[1]
        for inst in data[0]:
            src = parse("move {} from {} to {}", inst, evaluate_result=True)
            if not src: raise :marseysuspicious: Exception()
            count = int(src[0])
            group :marseymarseyloveorgy: = int(src[1])
            dest = int(src[2])
            if not group:
                continue
            z = []
            for a in range(count):
                z.append(image[group - 1].pop())
            for b in z:
                image[dest - 1].append(b)
        for x in image:
            print(x)
    
    def b(self):
        print("---------------")
        data :marseychartbar: = self.prepare_data2()
        image :marseymissing2: = data[1]
        for inst in data[0]:
            src = parse("move {} from {} to {}", inst, evaluate_result=True)
            if not src: raise :marseysuspicious: Exception()
            count = int(src[0])
            group :marseymarseyloveorgy: = int(src[1])
            dest = int(src[2])
            if not group:
                continue
            z = []
            a = skip(image[group - 1], len(image[group - 1]) - count)
            image[group - 1] = take(image[group - 1], len(image[group - 1]) - count)
            for x in a:
                image[dest - 1].append(x)
        for x in image:
            print(x)
        pass

i gave up and literally manually wrote the stack, rotated (so i didn't have to spent 30 minutes trying to figure out how to read a file by column)... here's that file

RSLFQ---
NZQGPT--
SMQB----
TGZJHCBQ
PHMBNFS-
PCQNSLVG
WCF-----
QHGZWVPM
GZDLCNR-

yet another w for the tribe

regex = re.compile(r'move ([0-9]*) from ([0-9]*) to ([0-9]*)')
for instruction in lines:
    result = regex.search(instruction)
    amount = int(result.group(1))
    from_stack = int(result.group(2))-1
    to_stack = int(result.group(3))-1

    stacks[to_stack] += stacks[from_stack][(0-amount):]
    del stacks[from_stack][(0-amount):]

for stack in stacks:
    print(stack[-1])

this puzzle made me cry a whole lot i spent an hour trying to parse the input then gave up and did it manually

import parse

with open("day5input.txt") as f:
    _, tlm = f.read().split("\n\n")

stacks = [
    list("RSLFQ"),
    list("NZQGPT"),
    list("SMQB"),
    list("TGZJHCBQ"),
    list("PHMBNFS"),
    list("PCQNSLVG"),
    list("WCF"),
    list("QHGZWVPM"),
    list("GZDLCNR")
]

instructions = list(parse.findall("move {:d} from {:d} to {:d}", tlm))

# part 1

for amount, origin, destination in instructions:
    crates = list(stacks[origin-1].pop() for _ in range(amount))
    stacks[destination-1].extend(crates)

for stack in stacks:
    print(stack[-1])

# part 2

stacks = [
    list("RSLFQ"),
    list("NZQGPT"),
    list("SMQB"),
    list("TGZJHCBQ"),
    list("PHMBNFS"),
    list("PCQNSLVG"),
    list("WCF"),
    list("QHGZWVPM"),
    list("GZDLCNR")
]

for amount, origin, destination in instructions:
    crates = list(stacks[origin-1].pop() for _ in range(amount))
    crates.reverse()
    stacks[destination-1].extend(crates)

for stack in stacks:
    print(stack[-1])

don't bully

use std::fs;

pub fn day05() {
    let s1 = vec!['R', 'G', 'J', 'B', 'T', 'V', 'Z'];
    let s2 = vec!['J', 'R', 'V', 'L'];
    let s3 = vec!['S', 'Q', 'F'];
    let s4 = vec!['Z', 'H', 'N', 'L', 'F', 'V', 'Q', 'G'];
    let s5 = vec!['R', 'Q', 'T', 'J', 'C', 'S', 'M', 'W'];
    let s6 = vec!['S', 'W', 'T', 'C', 'H', 'F'];
    let s7 = vec!['D', 'Z', 'C', 'V', 'F', 'N', 'J'];
    let s8 = vec!['L', 'G', 'Z', 'D', 'W', 'R', 'F', 'Q'];
    let s9 = vec!['J', 'B', 'W', 'V', 'P'];
    let mut ss1 = vec![s1, s2, s3, s4, s5, s6, s7, s8, s9];
    let mut ss2 = ss1.clone();

    let ls = read_input("input.txt");
    for l in &ls {
        for _i in 0..l.0 {
            let a = (ss1[l.1 - 1]).pop().unwrap();
            ss1[l.2 - 1].push(a);
        }
        let mut temp: Vec<char> = Vec::new();
        for _i in 0..l.0 {
            let a = (ss2[l.1 - 1]).pop().unwrap();
            temp.push(a);
        }
        for _i in 0..l.0 {
            let a = temp.pop().unwrap();
            ss2[l.2 - 1].push(a);
        }
    }

    let ss1: String = ss1.iter().map(|x| x.last().unwrap()).collect();
    let ss2: String = ss2.iter().map(|x| x.last().unwrap()).collect();
    println!("1: {ss1}, 2: {ss2}");
}

fn read_input(filename: &str) -> Vec<(usize, usize, usize)> {
    let binding = fs::read_to_string(filename).unwrap();
    let input: Vec<_> = binding
        .lines()
        .map(|x| x.split(' ').collect::<Vec<_>>())
        .map(|x| {
            (
                x[1].parse::<usize>().unwrap(),
                x[3].parse::<usize>().unwrap(),
                x[5].parse::<usize>().unwrap(),
            )
        })
        .collect::<Vec<_>>();
    input
}

Doing input parsing seemed like a royal pain in the ass for this one so I just manually entered the initial state of the stacks and removed them from the input file.

start = [['V', 'Q', 'W', 'M', 'B', 'N', 'Z', 'C'],
        ['B', 'C', 'W', 'R', 'Z', 'H'],
        ['J', 'Q', 'R', 'F'],
        ['T', 'M', 'N', 'F', 'H', 'W', 'S', 'Z'],
        ['P', 'Q', 'N', 'L', 'W', 'F', 'G'],
        ['W', 'P', 'L'],
        ['J', 'Q', 'C', 'G', 'R', 'D', 'B', 'V'],
        ['W', 'B', 'N', 'Q', 'Z'],
        ['J', 'T', 'G', 'C', 'F', 'L', 'H']]

foo = []
with open('day5_input.txt', 'r') as inp:
    foo = inp.read().split('\n')

for f in foo:
    f = f.split(' ')
    amount = int(f[1])
    source = int(f[3])-1
    dest = int(f[5])-1
    move = start[source][:amount]
    start[source] = start[source][amount:]
    move.reverse() #remove this for part 2
    start[dest] = move + start[dest]

print("".join([i[0] for i in start]))

The code I got the stars with used the initial state written out explicitly, not parsed, like other peoples' solutions; otherwise unremarkable too. Afterward, I tried boredly golfing (edit: okay it's not actually golf it's just cleanup and smushing stuff together on one line despite the cleanup) the code a bit and getting rid of all the assumptions about line numbers. I'm not much good at code golf proper tbh this does work.

constexpr int stack_count{ 9 };


template<std::size_t SIZE>
void reverseStacks(std::array<std::stack<char>, SIZE>& stacks) {
	for (std::size_t i{ 0 }; i < stacks.size(); ++i) {
		std::stack<char> temp;

		while (!stacks[i].empty()) {
			temp.push(stacks[i].top());
			stacks[i].pop();
		}

		stacks[i].swap(temp);

	}
}

template<std::size_t SIZE>
void movePartOne(int count, int from, int to, std::array<std::stack<char>, SIZE>& stacks) {

	for (int i{ 0 }; i < count; ++i) {
		char temp = stacks[from - 1].top();
		stacks[from - 1].pop();
		stacks[to - 1].push(temp);
	}
}


template<std::size_t SIZE>
void movePartTwo(int count, int from, int to, std::array<std::stack<char>, SIZE>& stacks) {

	std::stack<char> temp;
	for (int i{ 0 }; i < count; ++i) {
		temp.push(stacks[from - 1].top());
		stacks[from - 1].pop();
	}

	while (!temp.empty()) {
		stacks[to - 1].push(temp.top());
		temp.pop();
	}
}


std::string solution(void (*moveCall)(int, int, int, std::array<std::stack<char>, stack_count>&)) {
	std::array<std::stack<char>, stack_count> crate_stacks;


	std::ifstream file("input.txt");

	std::string buf;

	while (std::getline(file, buf)) {
		if (buf[1] == '1') break;

		for (int i{ 0 }; i < stack_count; ++i) {
			char c = buf[4 * i + 1];
			if (c != ' ') {
				crate_stacks[i].push(c);
			}
		}
	}
	reverseStacks(crate_stacks);
	std::getline(file, buf);
	while (std::getline(file, buf)) {
		int count, from, to;
		std::string temp;
		std::istringstream iss(buf);
		iss >> temp >> count >> temp >> from >> temp >> to;
		moveCall(count, from, to, crate_stacks);
	}

	std::string out;

	for (std::size_t i{ 0 }; i < crate_stacks.size(); ++i) {
		out += crate_stacks[i].top();
	}
	return out;
}



int main() {
	std::string p1 = solution(movePartOne);
	std::string p2 = solution(movePartTwo);
	std::cout << "Part One: " << p1 << '\n'
		<< "Part Two: " << p2 << '\n';
}

Spent way to long on my decision to actually use real stacks, and couldn't figure out how to pass template<std::size_t SIZE> void movePartOne(int count, int from, int to, std::array<std::stack<char>, SIZE>& stacks) { as an argument to the solution-function, without hardcoding SIZE in solution, tbf not quite sure why i even wanted to use std::array now.

C++ chads simply cannot stop winning

#include <iostream>
#include <string>
#include <vector>
#include <regex>
#include <list>
#include <iterator>

#define N_STACKS 9

// G-d forgive me bc I used C++ Slowpoke regex
std::regex in_reg("move\\s(\\d+)\\sfrom\\s(\\d+)\\sto\\s(\\d+)");

int main()
{
	std::list<char> stacks[N_STACKS];
	std::string sbuf;

	while(std::getline(std::cin, sbuf))
	{
		if(sbuf[1] >= '0' and sbuf[1] <= '9')
			break;
		
		for(auto i = 0; i < N_STACKS; i ++)
			if(sbuf[4*i + 1] != ' ')
				stacks[i].push_back(sbuf[4*i + 1]);
	}

	// Empty line
	std::getline(std::cin, sbuf);

	// main loop
	while(std::getline(std::cin, sbuf))
	{
		std::smatch match;
		std::regex_search(sbuf, match, in_reg);

		auto move_cardinality = std::stoul(match.str(1));
		auto from_index = std::stoul(match.str(2)) - 1;
		auto to_index = std::stoul(match.str(3)) - 1;

		// PART A
		//for(auto i = 0; i < move_cardinality; i++)			
		//	stacks[to_index].push_front(stacks[from_index].front());

		// PART B
		auto it = stacks[from_index].rbegin();
		std::advance(it, stacks[from_index].size() - move_cardinality);

		for(auto i = 0; i < move_cardinality; i++, ++it)
			stacks[to_index].push_front(*it);
		
		for(auto i = 0; i < move_cardinality; i++)
			stacks[from_index].pop_front();
	}

	for(auto i = 0; i < N_STACKS; i ++)
		std::cout << stacks[i].front();
	std::cout << std::endl;

	return 0;
}

Loading data (programatically, not manually like some of you casuals):

r←⊢⎕FIO[49] 'day5raw.txt'
m←⍎¨¨{⊃¨(⊂2 4 6)⌷(' '≠⍵)⊂⍵}¨(0⍳⍨≢¨r)↓r
t←s←{↑(' '≠⍵)⊂⍵}¨{(,⍵)⊂⍨(2⌷⍴⍵)/⍳1⌷⍴⍵}⍉(⊃(¯2+0⍳⍨≢¨r)↑r)[;¯2+4×⍳9]

Part 1:

z←{{s[2⌷⍵]←⊂(1⌷⍵)↓⊃s[2⌷⍵]}⍵,{s[3⌷⍵]←⊂(⌽(1⌷⍵)↑⊃s[2⌷⍵]),(⊃s[3⌷⍵])}⍵}¨m ⋄ ↑¨s

Part 2 (only a single character removed from part 1):

z←{{t[2⌷⍵]←⊂(1⌷⍵)↓⊃t[2⌷⍵]}⍵,{t[3⌷⍵]←⊂( (1⌷⍵)↑⊃t[2⌷⍵]),(⊃t[3⌷⍵])}⍵}¨m ⋄ ↑¨t

import math
file = open("input.txt")
stack = [[],[],[],[],[],[],[],[],[]]
stack2 =  [[],[],[],[],[],[],[],[],[]]
lines = file.readlines()
stacks = lines[0:8]
cmds = lines[10:]
for i in range(0,8):
    line = stacks[7-i]
    for j in range(1, 37, 4):
        if j > 33:
            break;
        if line[j].isalpha():
            stack[math.ceil(j/4)-1].append(line[j])
            stack2[math.ceil(j/4)-1].append(line[j])
for line in cmds:
    line = line.rstrip().split()
    c_f_t = [line[1], line[3], line[5]]
    for i in range(0, int(c_f_t[0])):
        stack[int(c_f_t[2])-1].append(stack[int(c_f_t[1])-1].pop())
    l = len(stack2[int(c_f_t[1])-1])-1
    to_append = stack2[int(c_f_t[1])-1][l-i:]
    for i in to_append:
        stack2[int(c_f_t[2])-1].append(i)
    stack2[int(c_f_t[1])-1]= stack2[int(c_f_t[1])-1][:-int(c_f_t[0])]
print(stack)
print(stack2)

I'll definitely have to clean this one up, I got flustered and started band aiding everything together because i had to get back to work

The final product visually in case anyone is wondering.

Took me almost an hour to figure out how to parse the crates. I was way too tired. I tried putting them into rows and then transposing the whole thing for some reason. Finally tried just reading character by character into a column map.

import re
from collections import deque


with open("./input5.txt") as infile:
    row_text, moves = infile.read().split("\n\n")


stacks = {i:deque() for i in range(1,10)}
mpat = re.compile(r'move (\d+) from (\d+) to (\d+)')


for row in row_text.splitlines():
    for col, char in enumerate(row, 1):
        if char == '[':
            stacks[col // 4 + 1].append(row[col])


for m in map(mpat.search, moves.splitlines()):
    if not m: continue
    a, f, t = map(int, m.groups())
    stack = deque()
    for _ in range(a):
        rune = stacks[f].popleft()
        # part 1
        # stack.append(rune)
        # part 2
        # stack.appendleft(rune)
    for item in stack:
        stacks[t].appendleft(item)
                    
print(''.join(map(deque.popleft, stacks.values())))

I'm not as embarrassed about this code as yesterday's I guess.

var lines = document.body.innerText.split("\n");
var stacks = [];

for (var j=0;j<lines.length;j++) {
    var line = lines[j];
    if (line.charAt(1) === "1") break;

    var r = /\[[A-Z]\]/g
    while (m = r.exec(line)) {
        var i = Math.floor(m.index/4);
        if (!Array.isArray(stacks[i])) stacks[i] = [];
        stacks[i].push(m[0].substring(1,2));
    }
}

stacks.map((s) => s.reverse());

function moveLine(line) {
    var m = line.match(/move ([0-9]+) from ([0-9]+) to ([0-9]+)/)
    if (m) {
        m = m.map((n) => Number(n));
        for (var i=0;i<m[1];i++) {
            var l = stacks[m[2]-1].pop();
            if (l) {
                stacks[m[3]-1].push(l);
            }
        }
    }
}

lines.map(moveLine);

PART 2 because it's basically only marginally more complex than PART 1

import * as fs from 'fs'

const stacks: string[][] = [];
fs.readFileSync(process.argv[2], 'utf-8').split('\n').map(l => {
  if (l.includes('[')) {
    l.match(/.{1,4}/g)?.map((c, i) => 
      (c[1] !== ' ') && (stacks[i] ??= []).unshift(c[1])
    )
  }

  if(l[0] === 'm') {
    const instr = l.match(/\d+/g)?.map(m => Number(m));
    if (!instr) return;
    const tmp: string[] = [];
    for(let i = 0; i < instr[0]; i++) {
      const c = stacks[instr[1]-1].pop();
      if (c) tmp.push(c);
    }
    tmp.reverse().map(c => stacks[instr[2]-1].push(c))
  }
}, [0,0])

console.log(stacks.map(s => s.pop()).join(''));

also, do none of you guys know regex?

with open('input05.txt') as file:
    lines = [line for line in file]
# fill stacks
stack_stack = [[] for _ in range(0,9)]
for depth in range(7,-1,-1):
    for s in range(0,9):
        char = lines[depth][1+s*4]
        if char != ' ':
            stack_stack[s].append(char)

def move_a(m_count, m_from, m_to):
    for _ in range(m_count):
        stack_stack[m_to].append(stack_stack[m_from].pop())
    
def move_b(m_count, m_from, m_to):
    stack_stack[m_to].extend(stack_stack[m_from][-m_count:])
    stack_stack[m_from] = stack_stack[m_from][:-m_count]
    
for line in lines[10:]:
    m_count, m_from, m_to  = [int(n) for n in re.findall('\d+', line)]
    move_a(m_count, m_from-1, m_to-1)
    # move_b(m_count, m_from-1, m_to-1)
print("".join([stack[-1] for stack in stack_stack]))

I absolutely shit boxed it, parsing was taking too much time I just hard coded the stacks manually and did the logic on that.

I absolutely shit boxed it, parsing was taking too much time I just hard coded the stacks manually and did the logic on that.

Frickin heck this pizza award is heck one phones. Will post code once back home

looks like the gpt3 bot that got #1 in the leaderboards yesterday didn't place today

it's funny because doing it this way meant that part 2 was instant : just gotta remove the flip

matlab ()

A = readlines("input.txt","EmptyLineRule","skip")';

N = (strlength(A(1))+1)/4;
Stacks = strings(1,N);

i = 1;
x = char(A(1));
while isempty(intersect(x,'1'))
    c = x(2:4:end-1);
    for j = 1:N
        if c(j)~=' '
            Stacks(j) = c(j) + Stacks(j);
        end
    end

    i = i+1;
    x = char(A(i));
end

Stacks1 = cellstr(Stacks);
Stacks2 = cellstr(Stacks);
for l = A(i+1:end)
    u = sscanf(l,'move %d from %d to %d');
       
    Stacks1{u(3)} = [Stacks1{u(3)} flip(Stacks1{u(2)}(end-u(1)+1:end))];
    Stacks1{u(2)}(end-u(1)+1:end) = ''; 

    Stacks2{u(3)} = [Stacks2{u(3)} Stacks2{u(2)}(end-u(1)+1:end)];
    Stacks2{u(2)}(end-u(1)+1:end) = ''; 
end

res1 = cellfun(@(x) x(end),Stacks1);
res2 = cellfun(@(x) x(end),Stacks2);

fprintf("Part 1 : %s\n",res1)
fprintf("Part 2 : %s\n",res2)

parsing a shit

here, bully me

Mommy I don't have the raw bird intellegence to comprehend the aforementioned syntaxial dilemma or whatever...

Can you dumb it down and explain in 6th grade burger-ish for your true blue followers

wow, today's puzzle was way easier than I thought it was gonna be from skimming some of these comments (I only skimmed them and even then, only the ones that weren't in C++ and trust me I cannot even begin to comprehend Python). it was probably about as easy as Day 1.

you can use either a regex or, if you parse the line into a list of strings, you can just grab the second, fourth, and sixth string since each line is always six words

@DrClaus had a better idea than me using std::list instead of std::vector since you can use pop_front() with a list, I had to either work backwards from the end of the vector or create a sub-vector of just the last n elements and work forwards; I figured the latter was lazier since I didn't have to think too hard on doing "size()-numBoxes" or some shit, just pop the back and insert at the front of the sub-vector.

from there it's absolutely just as simple as doing:

// part 1
for(auto i = 0; i < numBoxes; i++)
{
    char c = stacks[from].back();
    stacks[from].pop_back();
    stacks[to].push_back(c);
} 

or

// part 2:
// create subvector
substack.clear();
for(auto i = 0; i < numBoxes; i++)
{
    substack.insert(substack.begin(), stacks[from].back());
    stacks[from].pop_back();
}

// push from subvector to mainvector
for(auto i = 0; i < numBoxes; i++)
{
    stacks[to].push_back(substack.front());
    substack.erase(substack.begin());
}

where "from" is an integer that is the fourth element in each line (converted to an int), "to" is the sixth, and "numBoxes" is the second. fuck regexps, they can go to hell. do this for each line of input and you get the result.

C++ chads do indeed stay winning.

Using regex destroys the happy christmas ambience of this challenge you dorks. I Don‘t want to see it ever again!

Forgot the + in my numbers regex and it fucked with me for nearly an hour. Really should have just put the lists in manually too but that would mean it won.

Part 1:

import re

storage = dict()

with open('input', 'r') as f: 

    line = f.readline()

    while '[' in line:

        ptr = 0

        bucket = 1

        ticker = 0

        for i in range(0, len(line)):

            if line[i] is '[':

                if bucket in storage.keys():

                    storage.get(bucket).insert(0, line[i+1])

                else:

                    thing = [line[i+1]]

                    storage[bucket] = thing

            ticker += 1

            if line[i] is ']':

                ticker = 0

                bucket += 1

            if ticker is 4:

                ticker = 0

                bucket += 1

        line = f.readline()

    line = f.readline()

    while line:

        print(line)

        dat = re.findall('\d+', line)

        if len(dat) < 3:

            line = f.readline()

            continue

        move_from = storage.get(int(dat[1]))

        move_to = storage.get(int(dat[2]))

        total = int(dat[0])

        for thing in range(total):

            item = storage.get(int(dat[1])).pop()

            storage.get(int(dat[2])).append(item)

        line = f.readline()

finstr = ''

for key in sorted(list(storage.keys())):

    finstr += storage.get(key)[-1]

print(finstr)

Part 2:

import re

storage = dict()

with open('input', 'r') as f:

    line = f.readline()

    while '[' in line:

        ptr = 0

        bucket = 1

        ticker = 0

        for i in range(0, len(line)):

            if line[i] is '[':

                if bucket in storage.keys():

                    storage.get(bucket).insert(0, line[i+1])

                else:

                    thing = [line[i+1]]

                    storage[bucket] = thing

            ticker += 1

            if line[i] is ']':

                ticker = 0

                bucket += 1

            if ticker is 4:

                ticker = 0

                bucket += 1

        line = f.readline()

    line = f.readline()

    while line:

        dat = re.findall('\d+', line)

        if len(dat) < 3:

            line = f.readline()

            continue

        move_from = storage.get(int(dat[1]))

        move_to = storage.get(int(dat[2]))

        total = int(dat[0])

        teststr = ''

        items = []

        for thing in range(total):

            items.append(storage.get(int(dat[1])).pop())

        if (len(items) > 1):

            items.reverse()

            storage.get(int(dat[2])).extend(items)

        else:

            storage.get(int(dat[2])).extend(items)

        line = f.readline()

finstr = ''

for key in sorted(list(storage.keys())):

    finstr += storage.get(key)[-1]

print(finstr)

the virgin dramanaut: crying over some brackets and letters

chadditors: https://old.reddit.com/r/adventofcode/comments/zdszct/2022_day_5_1_small_terminal_python_animation_for/

not that bad imo, didn't know if there was a cleaner way to do the double skip once you hit the indices/blank line

import re
pattern=re.compile("[ \[\]]")

metalist={}
cratecount=0
with open("05.txt","r") as lines:
    crates=True
    for line in lines:
        if(line=="\n"): #skip line between crates and movements
            continue
        if(crates):
            for i in range(len(line.rstrip("\n"))):
                if(line[i]=="1"):
                    crates=False #hit crate pile index, skip line and prep for instructions
                    break
                if(not pattern.match(line[i])):
                    index=int((i-1)/4)+1
                    if index not in metalist:
                        metalist[index]=[]
                        cratecount+=1
                    metalist[index].append(line[i])
        else :#order parsing/enacting
            orders=line.rstrip("\n").split(" ")
            count,fromPile,toPile=int(orders[1]),int(orders[3]),int(orders[5])
            metalist[toPile][:0]=metalist[fromPile][0:count[::-1]] #prepend crates
            metalist[fromPile]=metalist[fromPile][count:] #remove prepended crates
for i in range(cratecount):
    print(metalist[i+1][0],end="")

and you drop the [::-1] for the bonus

btw people on /g/ are making 100+MB input tests, fun to benchmark your solution with

Ok some for loops are not evil

Here's mine:

Spent too much time fighting borrow checker for part 2 to avoid allocating temporary buffer

use std::fs::File;
use std::io::{BufRead, BufReader};


use regex::Regex;


fn main() {
    let move_re = Regex::new(r"move (\d+) from (\d+) to (\d+)").expect("Can't compile regex");

    let file = File::open("input.txt").expect("Can't open input file");
    let reader = BufReader::new(file);


    let mut stack_lines = Vec::new();
    let mut move_lines = Vec::new();
    let mut is_stacks = true;


    for line in reader.lines() {
        let line = line.expect("Can't read a line from file");
        if line.is_empty() {
            is_stacks = false;
        } else if is_stacks {
            stack_lines.push(line);
        } else {
            move_lines.push(line);
        }
    }
    let stack_numbers = stack_lines.pop().expect("No stack number line");
    let stack_count = stack_numbers.split_ascii_whitespace().count();


    let mut stacks: Vec<Vec<u8>> = vec![vec![]; stack_count];


    for line in stack_lines.iter().rev() {
        let line_bytes = line.as_bytes();
        for i in 0..stack_count {
            if line_bytes[i * 4 + 1] != ' ' as u8 {
                stacks[i].push(line_bytes[i * 4 + 1])
            }
        }
    }
    for mv in move_lines.iter() {
        let numbers = move_re.captures(&mv).expect("Can't parse move lines");
        let crate_count: usize = numbers.get(1).unwrap().as_str().parse().expect("Not a number");
        let src: usize = numbers.get(2).unwrap().as_str().parse::<usize>().expect("Not a number") - 1usize;
        let dst: usize = numbers.get(3).unwrap().as_str().parse::<usize>().expect("Not a number") - 1usize;


        // for i in 0..crate_count {
        //     let popped = stacks[src].pop().expect("Trying to grab a crate from empty stack");
        //     stacks[dst].push(popped);
        // }

        
        let (source, destination) = if src < dst {
            let (a, b) = stacks.split_at_mut(src + 1);
            (a.last_mut().unwrap(), &mut b[dst - src - 1])
        } else if src > dst {
            let (a, b) = stacks.split_at_mut(dst + 1);
            (&mut b[src - dst - 1], a.last_mut().unwrap())
        } else {
            continue
        };
        let source_len = source.len();
        destination.extend_from_slice(&source[source_len - crate_count..source_len]);
        source.truncate(source_len - crate_count);
    }


    for stack in stacks.iter() {
        print!("{}", *stack.last().expect("No crate on top of the stack") as char);
    }
    println!("");
}

Wow what a bunch of nerds