Activity selection problem

#1

this problem is from the algorithm stl section.
how to approach it?

#2

The greedy choice is to always pick the next activity whose finish time is least among the remaining activities and the start time is more than or equal to the finish time of previously selected activity. We can sort the activities according to their finishing time so that we always consider the next activity as minimum finishing time activity.

  1. Sort the activities according to their finishing time
  2. Select the first activity from the sorted array and print it.
  3. Do following for remaining activities in the sorted array.
    …….a) If the start time of this activity is greater than or equal to the finish time of previously selected activity then select this activity and count it

Reference Code

if you have more doubts regarding this feel free to ask
i hope this helps
if yes hit a like and don’t forgot to mark doubt as resolved :grinning:

#3

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