leetcode_gym/_problems/reconstructIternary.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Reconstruct Iternary		dfs, graph	https://leetcode.com/problems/reconstruct-itinerary	Not Solved

Problem

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

Note:

If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
All airports are represented by three capital letters (IATA code).
You may assume all tickets form at least one valid itinerary.
One must use all the tickets once and only once.
Example 1:

Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]
Example 2:

Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].
             But it is larger in lexical order.

Solution

Solution that I came up with myself. It works for the first test case but doesn't handle the lexical duplicate tickets.

{% highlight python %}

tickets = "MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO" tickets = "JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"

def findTicketIndForStartCity(startCity):

# TODO: store all ticket indexes that have the same start city
sameStartCityIndex = {}

for ticketInd, cities in enumerate(tickets):
    if cities[0]==startCity:
        return ticketInd

output = ['JFK'] for i in range(0, len(tickets)):

currTicketIndex = findTicketIndForStartCity( output[-1] )
nextCity = tickets[currTicketIndex][1]
output.append(nextCity)

print(output)

{% endhighlight %}

---

Here is the solution in Python using DFS and a stack. The solution is explained in this video.

Visualization here

I don't fully get why it's DFS but you can revisit it. For now, just take a few things as a lesson like how they use a hashmap to build an adjacency list (as seen in other similar graph problems).

{% highlight python %}

def findItinerary(self, tickets: List[List[str]]) -> List[str]: graph = {}

tickets.sort(key=lambda x: x[1])

# build graph as adjacency list
for u, v in tickets:
    if u in graph:
        graph[u].append(v)
    else:
        graph[u] = [v]

itinerary, stack = [], [("JFK")]

while stack:
    curr = stack[-1]
    
    if curr in graph and len(graph[curr]) > 0:
        stack.append(graph[curr].pop(0))
    else:
        itinerary.append(stack.pop())
return itinerary[::-1]

{% endhighlight %}