\n The meeting room allocation problem is an algorithmic problem that efficiently allocates meeting rooms based on a given list of times.
\n There are N meetings, and each meeting has a start time and an end time.
\n A meeting room can only hold one meeting at a time, and if there are meetings occurring at the same time, those meetings cannot be allocated.
\n Therefore, write an algorithm that maximizes the number of meetings that can be allocated.\n <\/p>\n
\n - The first line contains N (1 \u2264 N \u2264 100,000).\n - The next N lines provide the start and end times of each meeting.\n - The start time is less than the end time, and all times are integers between 0 and 10^6.\n <\/code><\/pre>\nOutput Format<\/h2>\n\n - Output the maximum number of meetings that can be allocated.\n <\/code><\/pre>\nExample<\/h2>\n
Input Example:<\/strong><\/p>\n\n 5\n 1 3\n 2 4\n 3 5\n 5 6\n 4 7\n <\/code><\/pre>\nOutput Example:<\/strong><\/p>\n\n 4\n <\/code><\/pre>\nProblem Solving Process<\/h2>\n
\n This problem can be approached using a greedy algorithm.
\n A greedy algorithm is a methodology that makes the optimal choice at each step to find the overall optimal solution.
\n In the case of the meeting room allocation problem, sorting meetings by their end times and allocating as many meetings as possible is the most effective method.\n <\/p>\n
1. Sorting Meetings<\/h3>\n
\n Sorting needs to be done based on the end times of the meetings.
\n By prioritizing meetings that end earlier, you can use the start times of later meetings to allocate more meetings.\n <\/p>\n
\n For example, when the meeting data is as follows:\n <\/p>\n
\n 1 3\n 2 4\n 3 5\n 5 6\n 4 7\n <\/code><\/pre>\n\n If we sort this data based on end times, it will look like this:\n <\/p>\n
\n 1 3\n 2 4\n 3 5\n 4 7\n 5 6\n <\/code><\/pre>\n2. Calculating Possible Meetings<\/h3>\n
\n Based on the sorted list of meetings, we iterate through each meeting to calculate how many meetings can be allocated.
\n We check the next meeting that can start after the fastest ending meeting.\n <\/p>\n
\n To do this, we use a variable lastEnd<\/code> to record the end time of the last allocated meeting.
\n Only those meetings whose start time is greater than or equal to lastEnd<\/code> can be allocated.\n <\/p>\n3. Code Implementation<\/h3>\n
\n The following code implements the above logic.
\n Let\u2019s write the Kotlin code as follows.\n <\/p>\n
\n fun maxMeetings(meetings: Array>): Int {\n \/\/ Sort meetings by end time\n val sortedMeetings = meetings.sortedBy { it.second }\n var count = 0\n var lastEndTime = 0\n\n for (meeting in sortedMeetings) {\n \/\/ Allocate meeting if the start time is greater than or equal to the end time of the last meeting\n if (meeting.first >= lastEndTime) {\n count++\n lastEndTime = meeting.second\n }\n }\n return count\n }\n\n fun main() {\n val n = readLine()!!.toInt()\n val meetings = Array(n) {\n val input = readLine()!!.split(\" \")\n Pair(input[0].toInt(), input[1].toInt())\n }\n println(maxMeetings(meetings))\n }\n <\/pair<\/code><\/pre>\nConclusion<\/h2>\n
\n In this post, we solved the meeting room allocation problem using Kotlin.
\n By using a greedy algorithm, we sorted the meeting data based on their end times and checked the start times of each meeting to allocate as many as possible.
\n This problem serves as a good example of how a greedy algorithm can be used to find an optimal solution efficiently.\n <\/p>\n
\n Applying such logic and approaches to various problems could lead to excellent performance in coding tests.
\n Now, we encourage you to take on similar problems!\n <\/p>\n
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