\n There are N meetings that need to be conducted, each with a start time and an end time.
\n How can we assign meeting rooms to maximize the number of meetings?
\n In other words, an algorithm needs to be implemented to assign meeting rooms such that the meetings do not overlap.\n <\/p>\n
\n Input: A list is given containing the start and end times of each meeting. \n Output: Print the maximum number of meetings that can be scheduled.\n <\/p>\n<\/section>\n \n This problem is a typical greedy algorithm problem. \n This way, we can make optimal meeting assignments without overlaps. \n First, we use a structure to store the start and end times of the meetings. \n After sorting, we use a loop to check each meeting. \n After checking all meetings, the final counted value will be the maximum number of meetings that can be scheduled.\n <\/p>\n<\/section>\n \n We can create several test cases to test the above code.\n <\/p>\n
\n For example, there can be a list like [(0, 30), (5, 10), (15, 20)]<\/code>.\n <\/p>\nApproach to the Problem<\/h2>\n
\n First, sort the meetings based on their end times,
\n then use the approach of proceeding with the meeting that ends the quickest and selecting the next meeting accordingly.\n <\/p>\n
\n A meeting can only be conducted if its start time is greater than or equal to the end time of the previous meeting.\n <\/p>\nStep-by-Step Solution Process<\/h2>\n
\n
C++ Code Implementation<\/h2>\n
\n#include
Explanation for Each Step<\/h2>\n
\n An important point in greedy algorithms is that we need to sort each meeting by their end times.\n <\/p>\n
\n If the start time of the current meeting is greater than or equal to the end time of the previous meeting,
\n we can conduct this meeting, so we increase the count and update the last end time.\n <\/p>\nTest Cases<\/h2>\n
\nvector