1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
|
// 3439. Reschedule Meetings for Maximum Free Time I
#include "leetcode.h"
/*
* 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.
*/
int maxFreeTime(int eventTime, int k, int *startTime, int startTimeSize,
int *endTime, int endTimeSize) {
int ans = 0, n = startTimeSize;
int *gap = (int *)malloc((n + 1) * sizeof(int));
int *pre = (int *)malloc((n + 2) * 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];
pre[0] = 0;
for (int i = 1; i < n + 2; i++)
pre[i] = pre[i - 1] + gap[i - 1];
for (int i = k + 1; i < n + 2; i++)
ans = fmax(ans, pre[i] - pre[i - k - 1]);
free(pre), free(gap);
return ans;
}
int main() {
int st1[] = {1, 3}, et1[] = {2, 5};
int st2[] = {0, 2, 9}, et2[] = {1, 4, 10};
int st3[] = {0, 1, 2, 3, 4}, et3[] = {1, 2, 3, 4, 5};
printf("%d\n", maxFreeTime(5, 1, st1, ARRAY_SIZE(st1), et1,
ARRAY_SIZE(et1))); // expect: 2
printf("%d\n", maxFreeTime(10, 1, st2, ARRAY_SIZE(st2), et2,
ARRAY_SIZE(et2))); // expect: 6
printf("%d\n", maxFreeTime(5, 2, st3, ARRAY_SIZE(st3), et3,
ARRAY_SIZE(et3))); // expect: 0
}
|