Source: https://www.youtube.com/watch?v=poWB2UCuozA

Activity Selection - greedy

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 1 : Consider the following 3 activities sorted by
by finish time.
     start[]  =  {10, 12, 20};
     finish[] =  {20, 25, 30};
A person can perform at most two activities. The 
maximum set of activities that can be executed 
is {0, 2} [ These are indexes in start[] and 
finish[] ]

Example 2 : Consider the following 6 activities 
sorted by by finish time.
     start[]  =  {1, 3, 0, 5, 8, 5};
     finish[] =  {2, 4, 6, 7, 9, 9};
A person can perform at most four activities. The 
maximum set of activities that can be executed 
is {0, 1, 3, 4} [ These are indexes in start[] and 
finish[] ]

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 print it.

"""The following implementation assumes that the activities
are already sorted according to their finish time"""

"""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 conatins finish time of all activities

def printMaxActivities(s , f ):
    n = len(f)
    print "The following activities are selected"

    # The first activity is always selected
    i = 0
    print i,

    # Consider rest of the activities
    for j in xrange(n):

        # 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]:
            print j,
            i = j

# Driver program to test above function
s = [1 , 3 , 0 , 5 , 8 , 5]
f = [2 , 4 , 6 , 7 , 9 , 9]
printMaxActivities(s , f)

# This code is contributed by Nikhil Kumar Singh

Time Complexity :

It takes O(n log n) time if input activities may not be sorted. It takes O(n) time when it is given that input activities are always sorted.