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 <bits/stdc++.h> using namespace std; void printItinerary(map<string, string> dataSet) { // To store reverse of given map map<string, string> reversemap; map<string, string>::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<string, string> 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 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); } } } |
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 order using System; using System.Collections.Generic; public class printItinerary { // Driver function public static void Main( string [] args) { Dictionary< string , string > dataSet = new Dictionary< string , string >(); 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< string , string > dataSet) { // To store reverse of given map Dictionary< string , string > reverseMap = new Dictionary< string , string >(); // 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 |
Bombay->Delhi, Delhi->Goa, Goa->Chennai, Chennai->Banglore,
Time Complexity: O(n).
Auxiliary Space: O(n), The extra space is used in map.
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