Open In App

# Find Itinerary from a given list of tickets

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 the final destination.

One Solution is to build a graph and do Topological Sorting of the graph. The 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. The 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).

Below is Java implementation of above idea.

## C++

 #include using namespace std; void printItinerary(map dataSet){    // To store reverse of given map    map reversemap;    map::iterator it;     // To fill reverse map, iterate through the given map    for (it = dataSet.begin(); it!=dataSet.end(); it++)        reversemap[it->second] = it->first;     // Find the starting point of itinerary    string start;     for (it = dataSet.begin(); it!=dataSet.end(); it++)    {        if (reversemap.find(it->first) == reversemap.end())        {            start = it->first;            break;        }    }     // If we could not find a starting point, then something wrong with input     if (start.empty())     {        cout << "Invalid Input" << endl;        return;     }     // Once we have starting point, we simple need to go next,    //next of next using given hash map    it = dataSet.find(start);    while (it != dataSet.end())    {        cout << it->first << "->" << it->second << endl;        it = dataSet.find(it->second);    } } int main(){    map dataSet;    dataSet["Chennai"] = "Banglore";    dataSet["Bombay"] = "Delhi";    dataSet["Goa"] = "Chennai";    dataSet["Delhi"] = "Goa";     printItinerary(dataSet);     return 0;}// C++ implementation is contributed by Aditya Goel

## Java

 // Java program to print itinerary in orderimport java.util.HashMap;import java.util.Map; public class printItinerary{    // Driver function    public static void main(String[] args)    {        Map dataSet = new HashMap();        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 dataSet)    {        // To store reverse of given map        Map reverseMap = new HashMap();         // To fill reverse map, iterate through the given map        for (Map.Entry entry: dataSet.entrySet())            reverseMap.put(entry.getValue(), entry.getKey());         // Find the starting point of itinerary        String start = null;        for (Map.Entry 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);        }    }}

## Python3

 class Solution():    #Solution class carries method for printing itinerary    def __init__(self):        pass    #method for printing itinerary    def printItinerary(self,d):        # First step : create a reversed mapping. Here also for storing key value pairs dictionary is used.        reverse_d = dict()        for i in d:            reverse_d[d[i]] = i        # Second step : find the starting point. Starting point will be that value which is not present in 'd' as key.        for i in reverse_d:            if reverse_d[i] not in reverse_d:                starting_pt = reverse_d[i]                break;        #Third step : simply proceed one by one to print whole route. Assuming that there exist Starting point.        while(starting_pt in d):            print(starting_pt,"->",d[starting_pt],end=", ")            starting_pt = d[starting_pt]        #method prints here only. Does not return anything.  if __name__=="__main__":    # Mapping using inbuilt data structure 'dictionary'    d = dict()    d["Chennai"] = "Banglore"    d["Bombay"] = "Delhi"    d["Goa"] = "Chennai"    d["Delhi"] = "Goa"     # call for method that would print itinerary.    obj = Solution()    obj.printItinerary(d)

## C#

 // C# program to print itinerary in orderusing System;using System.Collections.Generic; public class printItinerary{  // Driver function  public static void Main(string[] args)  {    Dictionary dataSet = new Dictionary();    dataSet["Chennai"] = "Banglore";    dataSet["Bombay"] = "Delhi";    dataSet["Goa"] = "Chennai";    dataSet["Delhi"] = "Goa";     printResult(dataSet);  }   // This function populates 'result' for given input 'dataset'  private static void printResult( Dictionary dataSet)  {     // To store reverse of given map    Dictionary reverseMap = new  Dictionary();     // To fill reverse map, iterate through the given map    foreach (var entry in dataSet)      reverseMap[entry.Value] = entry.Key;     // Find the starting point of itinerary    string start = null;    foreach (var entry in dataSet)    {      if (!reverseMap.ContainsKey(entry.Key))      {        start = entry.Key;        break;      }    }     // If we could not find a starting point, then something wrong    // with input    if (start == null)    {      Console.WriteLine("Invalid Input");      return;    }     // Once we have starting point, we simple need to go next, next    // of next using given hash map    string to = dataSet[start];    while (true)    {      Console.Write(start +  "->" + to + ", ");      start = to;      if (!dataSet.ContainsKey(to))        break;      to = dataSet[to];    }  }}  // This code is contributed by phasing17

## Javascript

 // JavaScript approach to sollve the problem function printItinerary(dataSet){    // To store reverse of given map    let reversemap = new Map();     // To fill reverse map, iterate through the given map    for (const[key,value] of dataSet)        reversemap.set(value,key);     // Find the starting point of itinerary    let start = "";     for (const key of dataSet.keys())    {        if (!reversemap.has(key))        {            start = key;            break;        }    }     // If we could not find a starting point, then something wrong with input     if (start.length == 0)     {        console.log("Invalid Input");        return;     }     // Once we have starting point, we simple need to go next,    //next of next using given hash map    let it = start;    while (dataSet.has(it))    {        console.log(it+"->"+dataSet.get(it));        it = dataSet.get(it);    } } // driver code let dataSet = new Map();dataSet.set("Chennai","Banglore");dataSet.set("Bombay","Delhi");dataSet.set("Goa","Chennai");dataSet.set("Delhi","Goa");printItinerary(dataSet); //code is contributed by shinjanpatra

Output

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

Time Complexity: O(n).
Auxiliary Space: O(n), The extra space is used in map.