Count Days Without Meetings.py

'''
You are given a positive integer days representing the total number of days an employee is available for work (starting from day 1). You are also given a 2D array meetings of size n where, meetings[i] = [start_i, end_i] represents the starting and ending days of meeting i (inclusive).

Return the count of days when the employee is available for work but no meetings are scheduled.

Note: The meetings may overlap.
'''

#i know its pretty easy, its only 1438 diff but i m too hectic

#bad solution

class Solution:
    def countDays(self, days: int, meetings: List[List[int]]) -> int:
        
        res = [0 for f in range(days)]
        newm = []
        meetings.sort(key = lambda x: x[0])

        for i in range(len(meetings)-1):

            cur = meetings[i]
            for j in range(i, len(meetings)):
                nxt = meetings[j]

                if cur[1]>=nxt[0]:
                    newm.append(cur[0], nxt[1])
                else:
                    newm.append(cur)
            

        for st,end in newm:
            for i in range(st,end+1):
                res[i-1] = 0
        


        res = sum([1 for x in res if x == 0])
        return res

----------------------------------------------------------------------

class Solution:
    def countDays(self, days: int, meetings: List[List[int]]) -> int:
        meetings.sort()
        
        merged_meetings = []
        for meeting in meetings:
            if not merged_meetings or meeting[0] > merged_meetings[-1][1]:
                merged_meetings.append(meeting)
            else:
                merged_meetings[-1][1] = max(merged_meetings[-1][1], meeting[1])
        
        meeting_days_count = 0
        for start, end in merged_meetings:
            meeting_days_count += end - start + 1
        
        return days - meeting_days_count