One-way flight trip problem

Question

You are going on a one-way indirect flight trip that includes billions an unknown very large number of transfers.

• You are not stoppi

Answer 1

I have written a small python program, uses two hash tables one for count and another for src to dst mapping. The complexity depends on the implementation of the dictionary. if dictionary has O(1) then complexity is O(n) , if dictionary has O( lg(n) ) like in STL map, then complexity is O( n lg(n) )

import random
# actual journey: a-> b -> c -> ... g -> h
journey = [('a','b'), ('b','c'), ('c','d'), ('d','e'), ('e','f'), ('f','g'), ('g','h')]
#shuffle the journey.
random.shuffle(journey)
print("shffled journey : ", journey )
# Hashmap to get the count of each place
map_count = {}
# Hashmap to find the route, contains src to dst mapping
map_route = {}

# fill the hashtable
for j in journey:
    source = j[0]; dest = j[1]
    map_route[source] = dest
    i = map_count.get(source, 0)
    map_count[ source ] = i+1
    i = map_count.get(dest, 0)
    map_count[ dest ] = i+1

start = ''
# find the start point: the map entry with count = 1 and 
# key exists in map_route.
for (key,val) in map_count.items():
    if map_count[key] == 1 and map_route.has_key(key):
        start = key
        break

print("journey started at : %s" % start)
route = [] # the route
n = len(journey)  # number of cities.
while n:
    route.append( (start, map_route[start]) )
    start = map_route[start]
    n -= 1

print(" Route : " , route )