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// 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
}
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