# Activity Selection Problem

Greedy AlgorithmAlgorithmsData Structure

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