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+# 3439. Reschedule Meetings for Maximum Free Time I
+
+"""
+you are given an integer 'eventTime' denoting the duration of an event, where
+the event occurs from time 't = 0' to time 't = eventTime'. you are also
+given two integer arrays 'startTime' and 'endTime' each of length 'n'. these
+represent the start and end time of 'n' non-overlapping meetings where the
+i'th meeting occurs during the time '[startTime[i], endTime[i]]'. you can
+reschedule at most 'k' meetings by moving their start time while maintaining
+the same duration, to maximise the longest continuous period of free time
+during the event. the relative order of all the meetings should stay the same
+and they should remain non-overlapping. return the maximum amount of free
+time possible after rearranging the meetings. note that the meetings can not
+be rescheduled to a time outside of the event.
+"""
+
+
+class Solution(object):
+ def maxFreeTime(self, eventTime, k, startTime, endTime):
+ """
+ :type eventTime: int
+ :type k: int
+ :type startTime: List[int]
+ :type endTime: List[int]
+ :rtype: int
+ """
+ ans, n = 0, len(startTime)
+ gaps = [0] * (n + 1)
+ pre = [0] * (n + 2)
+ gaps[0] = startTime[0]
+ gaps[n] = eventTime - endTime[-1]
+ for i in range(1, n):
+ gaps[i] = startTime[i] - endTime[i - 1]
+ for i in range(1, n + 2):
+ pre[i] = pre[i - 1] + gaps[i - 1]
+ for i in range(k + 1, n + 2):
+ ans = max(ans, pre[i] - pre[i - k - 1])
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.maxFreeTime(eventTime=5, k=1, startTime=[1, 3], endTime=[2, 5]))
+ print(obj.maxFreeTime(eventTime=10, k=1, startTime=[0, 2, 9], endTime=[1, 4, 10]))
+ print(
+ obj.maxFreeTime(
+ eventTime=5, k=2, startTime=[0, 1, 2, 3, 4], endTime=[1, 2, 3, 4, 5]
+ )
+ )