Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Reconstruct Itinerary in C++
Suppose we have a list of airline tickets represented by pairs of departure and arrival airports like [from, to], we have to reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. So, the itinerary must begin with JFK.
So if the input is like [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]], then the output will be ["JFK", "MUC", "LHR", "SFO", "SJC"].
To solve this, we will follow these steps −
Define array ret and a map called graph.
Define a method called visit. This will take airport name as input
while size of the graph[airport] is not 0
x := first element of graph[airport]
delete the first element from graph[airport]
call visit(x)
insert airport into ret
Now from the main method, do the following −
for i in range 0 to size of tickers array
u := tickets[i, 0], v := tickets[i, 1], insert v into graph[u]
visit(“JFK”) as this is the first airport
reverse the list ret and return
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector <string> ret;
map < string, multiset <string> > graph;
vector<string> findItinerary(vector<vector<string>>& tickets) {
for(int i = 0; i < tickets.size(); i++){
string u = tickets[i][0];
string v = tickets[i][1];
graph[u].insert(v);
}
visit("JFK");
reverse(ret.begin(), ret.end());
return ret;
}
void visit(string airport){
while(graph[airport].size()){
string x = *(graph[airport].begin());
graph[airport].erase(graph[airport].begin());
visit(x);
}
ret.push_back(airport);
}
};
main(){
Solution ob;
vector<vector<string>> v = {{"MUC","LHR"},{"JFK","MUC"},{"SFO","SJC"},{"LHR","SFO"}};
print_vector(ob.findItinerary(v));
}Input
[["MUC","LHR"],["JFK","MUC"],["SFO","SJC"],["LHR","SFO"]]
Output
[JFK, MUC, LHR, SFO, SJC, ]
144 Views
