#!/usr/bin/env python3

import sys, math
import numpy as np


def dijkstra(graph, v, a):
    Q = [n for n in graph.keys()]
    lq = len(Q)
    while len(Q) > 0:
        u = min([u for u in Q], key=lambda x: a[x])
        Q.remove(u)
        for vo in graph[u]:
            if vo in Q:
                distance_update(graph, u,vo,a,v)
    return v, a


def distance_update(g, u, vo, a, v):
    dist = 1 if vo[2] == u[2] else 1000
    alt = a[u] + dist
    # joining the sets does not work because we never remove the older nodes!
    if alt <= a[vo]:
        if alt < a[vo]:
            v[vo] = set([])
            a[vo] = alt
        v[vo].add(u)
        v[vo] = v[vo].union(set(v[u]))


def count_set(edges, lab):
    edg = set([])
    for e in edges:
        edg.add((e[0], e[1]))
        lab[e[1]][e[0]] = 'O'
    return edg


if __name__ == '__main__':
    lab = [list(line.strip('\n')) for line in open(sys.argv[1])]
    rot90 = {(0,-1): [(-1,0),(1,0)], (1,0): [(0,-1),(0,1)], (0,1): [(1,0),(-1, 0)], (-1,0): [(0,1),(0,-1)]}

    start = (1, len(lab)-2, (1,0))
    end = (len(lab[0])-2, 1, (0,-1))
    graph = {}
    distance = {}
    pred = {}
    for y, l in enumerate(lab):
        for x, f in enumerate(lab[y]):
            if f == '#': continue
            for d in [(0,1), (0,-1), (1,0), (-1,0)]:
                distance[(x,y,d)] = math.inf
                pred[(x,y,d)] = set([])
                graph[(x,y,d)] = []
                for d1 in rot90[d]:
                    if not lab[y+d1[1]][x+d1[0]] == '#': graph[(x,y,d)].append((x,y,d1))
                if lab[y+d[1]][x+d[0]] != '#':
                    graph[(x,y,d)].append((x+d[0], y+d[1], d))
    distance[start] = 0
    v1, a1 = dijkstra(graph, pred, distance)
    d1 = count_set(v1[end], lab)

    # challenge 1
    res1 = a1[end]
    print(f"challenge 1:\n{res1}\n")

    # challenge 2, needed some reddit help again...
    d1.add((end[0],end[1]))
    d1.add((start[0], start[1]))
    res2 = len(d1)
    print(f"challenge 2:\n{res2}")

    for line in lab:
      print(''.join(line))