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+// 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
+}