n-meetings.md

Links

GFG

Expected Output

start[] = [8, 0, 1, 5, 3, 5]
end[]   =  [9, 6, 2, 9, 4, 7]
op = 4

Greedy

1. Sort the meetings [ [start[i], end[i], i], ....] based on (end[i], i), if two end match choose the one with lower i value (basically the meeting with shorter duration using (end[i], i) as key for sorting)

Efficiency:

• Time: O(NlogN), Where N is .
• Space: O(N), Where N is

code:

class Solution:
    def maximumMeetings(self,n,start,end):
        meetings = [ [start[i], end[i]] for i in range(n) ]
        meetings.sort(key=lambda x : x[1])
        
        res = 0
        prev_end = -1
        for start, end in meetings:
            if start > prev_end:
                res += 1
                prev_end = end
        
        return res

Note: if meetings[i][0] > prev_end:: current meeting start_time must be strictly greater than the previous_meeting_end_time else the meeting cannot start.