You are given n activities with their start and finish times. Select the maximum number of activities that can be performed by a single person, assuming that a person can only work on a single activity at a time.
Example:
Consider the following 6 activities.
start[] = {1, 3, 0, 5, 8, 5};
finish[] = {2, 4, 6, 7, 9, 9};
The maximum set of activities that can be executed
by a single person is {0, 1, 3, 4}
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.
- Sort the activities according to their finishing time
- Select the first activity from the sorted array and print it.
- Do following for remaining activities in the sorted array.
…….a) If the start time of this activity is greater than the finish time of previously selected activity then select this activity and print it.
An example code to clarify it further :
// Prints a maximum set of activities that can be done by a single
// person, one at a time.
// n --> Total number of activities
// s[] --> An array that contains start time of all activities
// f[] --> An array that contains finish time of all activities ( sorted )
void printMaxActivities(int s[], int f[], int n)
{
int i, j;
printf ("Following activities are selected \n");
// The first activity always gets selected
i = 0;
printf("%d ", i);
// Consider rest of the activities
for (j = 1; j < n; j++)
{
// If this activity has start time greater than or equal to the finish
// time of previously selected activity, then select it
if (s[j] >= f[i])
{
printf ("%d ", j);
i = j;
}
}
}
How does Greedy Choice work for Activities sorted according to finish time?
Let the give set of activities be S = {1, 2, 3, ..n}
and activities be sorted by finish time. The greedy choice is to always pick activity 1. How come the activity 1 always provides one of the optimal solutions. We can prove it by showing that if there is another solution B with first activity other than 1, then there is also a solution A of same size with activity 1 as first activity. Let the first activity selected by B be k, then there always exist A = {B – {k}} U {1}
.
(Note that the activities in B are independent and k has smallest finishing time among all. Since k is not 1, finish(k) >= finish(1)
).