# Java Programming – Find Itinerary from a given list of tickets

We can also use hashing to avoid building a graph. The idea is to first find the starting point. A starting point would never be on ‘to’ side of a ticket.

Given a list of tickets, find itinerary in order using the given list.

Example:

```Input:
"Chennai" -> "Banglore"
"Bombay" -> "Delhi"
"Goa"    -> "Chennai"
"Delhi"  -> "Goa"

Output:
Bombay->Delhi, Delhi->Goa, Goa->Chennai, Chennai->Banglore,```

It may be assumed that the input list of tickets is not cyclic and there is one ticket from every city except final destination.

One Solution is to build a graph and do Topological Sorting of the graph. Time complexity of this solution is O(n).

We can also use hashing to avoid building a graph. The idea is to first find the starting point. A starting point would never be on ‘to’ side of a ticket. Once we find the starting point, we can simply traverse the given map to print itinerary in order. Following are steps.

```1) Create a HashMap of given pair of tickets.  Let the created
HashMap be 'dataset'. Every entry of 'dataset' is of the form
"from->to" like "Chennai" -> "Banglore"

2) Find the starting point of itinerary.
a) Create a reverse HashMap.  Let the reverse be 'reverseMap'
Entries of 'reverseMap' are of the form "to->form".
Following is 'reverseMap' for above example.
"Banglore"-> "Chennai"
"Delhi"   -> "Bombay"
"Chennai" -> "Goa"
"Goa"     ->  "Delhi"

b) Traverse 'dataset'.  For every key of dataset, check if it
is there in 'reverseMap'.  If a key is not present, then we
found the starting point. In the above example, "Bombay" is
starting point.

3) Start from above found starting point and traverse the 'dataset'
to print itinerary.
```

All of the above steps require O(n) time so overall time complexity is O(n).

Java Programming
``````// Java program to print itinerary in order
import java.util.HashMap;
import java.util.Map;

public class printItinerary
{
// Driver function
public static void main(String[] args)
{
Map<String, String> dataSet = new HashMap<String, String>();
dataSet.put("Chennai", "Banglore");
dataSet.put("Bombay", "Delhi");
dataSet.put("Goa", "Chennai");
dataSet.put("Delhi", "Goa");

printResult(dataSet);
}

// This function populates 'result' for given input 'dataset'
private static void printResult(Map<String, String> dataSet)
{
// To store reverse of given map
Map<String, String> reverseMap = new HashMap<String, String>();

// To fill reverse map, iterate through the given map
for (Map.Entry<String,String> entry: dataSet.entrySet())
reverseMap.put(entry.getValue(), entry.getKey());

// Find the starting point of itinerary
String start = null;
for (Map.Entry<String,String> entry: dataSet.entrySet())
{
if (!reverseMap.containsKey(entry.getKey()))
{
start = entry.getKey();
break;
}
}

// If we could not find a starting point, then something wrong
// with input
if (start == null)
{
System.out.println("Invalid Input");
return;
}

// Once we have starting point, we simple need to go next, next
// of next using given hash map
String to = dataSet.get(start);
while (to != null)
{
System.out.print(start +  "->" + to + ", ");
start = to;
to = dataSet.get(to);
}
}
}``````

Output:

`Bombay->Delhi, Delhi->Goa, Goa->Chennai, Chennai->Banglore,`
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