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Journey Scheduling
Journey Scheduling
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import java.io.; import java.math.; import java.text.; import java.util.; import java.util.regex.*;
public class Solution { private static class Node { final int index; final Set neighbors = new HashSet<>();
// System.out.println(String.format("Diameter: %d, distances: %s", diameter, Arrays.toString(distances)));
// try { result = journeyScheduling(n, roads, journeys); /* } catch (StackOverflowError e) { result = new long[1]; }*/
}
Here, I think the main aim is to indentify the fact that the maximum distance for visiting successing cities after visiting the first city, will always be equal to maximum possible distance between any two nodes, which has to be only calculated for once. Hence, we only have to calculate the distance to the starting city (V), and then add the previously calculated maximum distance for the remaining count of cities.
Here is my solution in jav, javascript, python, C, C++, Csharp HackerRank Journey Scheduling Problem Solution
Here is the solution of Journey Scheduling Click Here
https://zeroplusfour.com/journey-scheduling-hackerrank-solution/
Here's how I did in all languages Java 8 , C++ , C , Python 3, Python 2.