Imagine! You, from the future, decide to travel to a specific time using a time machine. However, this time machine can only operate through a complex algorithm. The problem is to calculate the minimum number of jumps needed to reach the given time.<\/p>\n
Two integers S and E are given. S is the starting time and E is the arrival time. The time machine operates according to the following rules:<\/p>\n
In the first line, two integers S (1 \u2264 S \u2264 105<\/sup>) and E (1 \u2264 E \u2264 105<\/sup>) are given.<\/p>\n Output the minimum number of jumps required for the time machine to reach E.<\/p>\n This problem can be solved using the BFS (Breadth-First Search) algorithm. BFS starts from a specific node and explores all nodes connected to it before exploring the nodes of the next layer. In this problem, each time zone can be represented as a node in a graph, while each jump can be represented as an edge.<\/p>\n The time complexity of this algorithm is O(n). Since all time zones will be visited once, performing BFS for up to 100,000 nodes is efficient.<\/p>\n In this lecture, we explored how to apply the BFS algorithm to solve the time machine problem and calculate the shortest distance. Understanding and applying the advantages of implementing it in Kotlin as well as the basic principles of BFS will help you in coding tests. In the next session, we will prepare a lecture to explore other problems and more diverse algorithms!<\/p>\n <\/body><\/p>\n","protected":false},"excerpt":{"rendered":"Output<\/h3>\n
Example<\/h2>\n
\n Input:<\/strong>\n 4 7\n\n Output:<\/strong>\n 2\n <\/pre>\n
Problem Solving Process<\/h2>\n
Step-by-Step Solution<\/h3>\n
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Code Implementation<\/h3>\n
\nfun minJumps(S: Int, E: Int): Int {\n val queue: QueueTime Complexity Analysis<\/h2>\n
Conclusion<\/h2>\n