Activity Selection Problem

There are n different activities are given with their starting time and ending time. Select the maximum number of activities to solve by a single person. We will use the greedy approach to find the next activity whose finish time is minimum among rest activities, and the start time is more than or equal with the finish time of the last selected activity.

• The complexity of this problem is O(n log n) when the list is not sorted.
• When the sorted list is provided the complexity will be O(n).

Input and Output

Input:
A list of different activities with starting and ending times.
{(5,9), (1,2), (3,4), (0,6), (5,7), (8,9)}
Output:
Selected Activities are:
Activity: 0 , Start: 1 End: 2
Activity: 1 , Start: 3 End: 4
Activity: 3 , Start: 5 End: 7
Activity: 5 , Start: 8 End: 9

Algorithm

maxActivity(act, size)

Input: A list of activity, and the number of elements in the list.

Output − The order of activities how they have been chosen.

Begin
   initially sort the given activity List
   set i := 1
   display the ith activity //in this case it is the first activity

   for j := 1 to n-1 do
      if start time of act[j] >= end of act[i] then
         display the jth activity
         i := j
   done
End

Example

#include<iostream>
#include<algorithm>
using namespace std;

struct Activitiy {
   int start, end;
};

bool comp(Activitiy act1, Activitiy act2) {
   return (act1.end < act2.end);
}

void maxActivity(Activitiy act[], int n) {
   sort(act, act+n, comp); //sort activities using compare function

   cout << "Selected Activities are: " << endl;
   int i = 0;// first activity as 0 is selected
   cout << "Activity: " << i << " , Start: " <<act[i].start << " End:
      " << act[i].end <<endl;

   for (int j = 1; j < n; j++) { //for all other activities
      if (act[j].start >= act[i].end) { //when start time is >= end
         time, print the activity
         cout << "Activity: " << j << " , Start: " <<act[j].start << " End: " << act[j].end <<endl;
         i = j;
      }
   }
}

int main() {
   Activitiy actArr[] = {{5,9},{1,2},{3,4},{0,6},{5,7},{8,9}};
   int n = 6;
   maxActivity(actArr,n);
   return 0;
}

Output

Selected Activities are:
Activity: 0 , Start: 1 End: 2
Activity: 1 , Start: 3 End: 4
Activity: 3 , Start: 5 End: 7
Activity: 5 , Start: 8 End: 9