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-rw-r--r-- | reschedule_meetings_max_free2.c | 50 | ||||
-rw-r--r-- | reschedule_meetings_max_free2.py | 46 |
2 files changed, 96 insertions, 0 deletions
diff --git a/reschedule_meetings_max_free2.c b/reschedule_meetings_max_free2.c new file mode 100644 index 0000000..2ad0165 --- /dev/null +++ b/reschedule_meetings_max_free2.c @@ -0,0 +1,50 @@ +// 3440. Reschedule Meetings for Maximum Free Time II +#include "leetcode.h" + +/* + * you are given an integer 'eventTime' denoting the duration of an event. you + * are also given two integer arrays 'startTime' and 'endTime', each of length + * 'n'. these represent the start and end times of 'n' non-overlapping meetings + * that occur during the event between time 't = 0' and time 't = eventTime' + * where the i't meeting occurs during the time '[startTime[i], endTime[i]]'. + * you can reschedule at most one meeting by moving its start time while + * maintaining the same duration such that the meetings remain non-overlapping, + * to maximise the longest continuous period of free time during the event. + * return the maximum amount of free time possible after rearranging the + * meetings. + */ + +int maxFreeTime(int eventTime, int *startTime, int startTimeSize, int *endTime, + int endTimeSize) { + int n = startTimeSize, ans = 0, left = 0, curr; + int *gap = (int *)malloc((n + 1) * sizeof(int)); + int *largest = (int *)malloc((n + 1) * sizeof(int)); + gap[0] = startTime[0]; + for (int i = 1; i < n; i++) + gap[i] = startTime[i] - endTime[i - 1]; + gap[n] = eventTime - endTime[n - 1]; + largest[n] = 0; + for (int i = 1; i >= 0; i--) + largest[i] = fmax(largest[i + 1], gap[i + 1]); + for (int i = 1; i <= n; i++) { + curr = endTime[i - 1] - startTime[i - 1]; + if (curr <= fmax(left, largest[i])) + ans = fmax(ans, gap[i - 1] + gap[i] + curr); + ans = fmax(ans, gap[i - 1] + gap[i]); + left = fmax(left, gap[i - 1]); + } + free(gap), free(largest); + return ans; +} + +int main() { + int st1[] = {1, 3}, et1[] = {2, 5}; + int st2[] = {0, 7, 9}, et2[] = {1, 8, 10}; + int st3[] = {0, 3, 7, 9}, et3[] = {1, 4, 8, 10}; + printf("%d\n", maxFreeTime(5, st1, ARRAY_SIZE(st1), et1, + ARRAY_SIZE(et1))); // expect: 2 + printf("%d\n", maxFreeTime(10, st2, ARRAY_SIZE(st2), et2, + ARRAY_SIZE(et2))); // expect: 7 + printf("%d\n", maxFreeTime(10, st3, ARRAY_SIZE(st3), et3, + ARRAY_SIZE(et3))); // expect: 6 +} diff --git a/reschedule_meetings_max_free2.py b/reschedule_meetings_max_free2.py new file mode 100644 index 0000000..d508bc7 --- /dev/null +++ b/reschedule_meetings_max_free2.py @@ -0,0 +1,46 @@ +# 3440. Reschedule Meetings for Maximum Free Time II + +""" +you are given an integer 'eventTime' denoting the duration of an event. you +are also given two integer arrays 'startTime' and 'endTime', each of length +'n'. these represent the start and end times of 'n' non-overlapping meetings +that occur during the event between time 't = 0' and time 't = eventTime' +where the i't meeting occurs during the time '[startTime[i], endTime[i]]'. +you can reschedule at most one meeting by moving its start time while +maintaining the same duration such that the meetings remain non-overlapping, +to maximise the longest continuous period of free time during the event. +return the maximum amount of free time possible after rearranging the +meetings. +""" + + +class Solution(object): + def maxFreeTime(self, eventTime, startTime, endTime): + """ + :type eventTime: int + :type startTime: List[int] + :type endTime: List[int] + :rtype: int + """ + ans, left = 0, 0 + gap = [startTime[0]] + largest = [0] * len(gap) + for i in range(1, len(startTime)): + gap.append(startTime[i] - endTime[i - 1]) + gap.append(eventTime - endTime[-1]) + for i in range(len(gap) - 2, -1, -1): + largest[i] = max(largest[i + 1], gap[i + 1]) + for i in range(1, len(gap)): + curr = endTime[i - 1] - startTime[i - 1] + if curr <= max(left, largest[i]): + ans = max(ans, gap[i - 1] + gap[i] + curr) + ans = max(ans, gap[i - 1] + gap[i]) + left = max(left, gap[i - 1]) + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.maxFreeTime(eventTime=5, startTime=[1, 3], endTime=[2, 5])) + print(obj.maxFreeTime(eventTime=10, startTime=[0, 7, 9], endTime=[1, 8, 10])) + print(obj.maxFreeTime(eventTime=10, startTime=[0, 3, 7, 9], endTime=[1, 4, 8, 10])) |