{"id":33824,"date":"2024-11-01T09:20:51","date_gmt":"2024-11-01T09:20:51","guid":{"rendered":"http:\/\/atmokpo.com\/w\/?p=33824"},"modified":"2024-11-01T11:46:31","modified_gmt":"2024-11-01T11:46:31","slug":"python-coding-test-course-assigning-meeting-rooms","status":"publish","type":"post","link":"https:\/\/atmokpo.com\/w\/33824\/","title":{"rendered":"python coding test course, assigning meeting rooms"},"content":{"rendered":"

<\/p>\n

Problem Description<\/h2>\n

The problem of assigning meeting rooms involves optimally assigning meeting rooms based on the start and end times of multiple meetings. If the given meetings overlap, additional meeting rooms need to be assigned, so the goal is to maximize the number of meetings that can be held without overlaps.<\/p>\n

Problem Definition<\/h3>\n

Given the start and end times of meetings in list format, calculate how many meetings can be conducted simultaneously through room assignment.<\/p>\n

Input Format<\/h3>\n
\n[[start_time1, end_time1], [start_time2, end_time2], ...]\n    <\/pre>\n
Output Format<\/h3>\n
\nMaximum number of possible meetings\n    <\/pre>\n
Example<\/h3>\n
\nInput: [[1, 3], [2, 4], [3, 5], [6, 8]]\nOutput: 3\n    <\/pre>\n
Solution Process<\/h2>\n
This problem can be solved using a greedy algorithm. First, sort the meetings based on their end times, then select the meeting that ends first, and continue to select subsequent meetings that do not overlap with the last selected meeting.<\/p>\n
Step 1: Data Organization<\/h2>\n
First, sort the given list of meetings based on their end times. This allows for conducting meetings while using the least amount of resources.<\/p>\n
Step 2: Determine Meeting Attendance<\/h2>\n
Select the first meeting from the sorted list, and if the next meeting starts at or after the end time of this meeting, select that meeting. Repeat this process to select as many meetings as possible.<\/p>\n
Step 3: Implementation<\/h2>\n
Now, let’s implement the above process in Python code.<\/p>\n
python\ndef max_conferences(conferences):\n    # Sort the conference list based on end times\n    conferences.sort(key=lambda x: x[1])\n    \n    # Select the first conference\n    count = 1\n    last_end_time = conferences[0][1]\n    \n    # Iterate over the remaining conferences\n    for i in range(1, len(conferences)):\n        if conferences[i][0] >= last_end_time:  # If start time is greater than or equal to the last end time\n            count += 1\n            last_end_time = conferences[i][1]  # Update the last end time\n    \n    return count\n\n# Example input\nmeetings = [[1, 3], [2, 4], [3, 5], [6, 8]]\nresult = max_conferences(meetings)\nprint(\"Maximum number of meetings:\", result)\n    <\/code><\/pre>\n
Step 4: Code Explanation<\/h2>\n
Code Explanation<\/h3>\n